Visit : www.Civildatas.

com

Thermodynamics
Contents

m
1. Basic Concepts
Thermodynamic System and Control Volume

co
Open and Closed systems
Thermodynamic Equilibrium
Quasi-Static Process
Concept of Continuum

.
Zeroth Law of Thermodynamics

Work a path function

PdV-work or Displacement Work
tas
International Practical Temperature Scale

Free Expansion with Zero Work Transfer

Heat Transfer
Heat Transfer-A Path Function
lda
2. FIRST LAW OF THERMODYNAMICS
First Law of Thermodynamics
Application of First Law to a Process
Internal Energy--A Property of System
vi

Perpetual Motion Machine of the First Kind-PMM1

Enthalpy
Application of First Law of Thermodynamics to Non-flow or Closed System
Ci

Application of First Law to Steady Flow Process S.F.E.E

Variable Flow Processes
Discharging and Charging a Tank

3. SECOND LAW OF THERMODYNAMICS

w.

Qualitative Difference between Heat and Work

Kelvin-Planck Statement of Second Law
Clausius' Statement of the Second Law
Clausius' Theorem
ww

Refrigerator and Heat Pump [with RAC]

Equivalence of Kelvin-Planck and Clausius Statements
Carnot Engine with same efficiency or same work output

4. ENTROPY
Two Reversible Adiabatic Paths cannot Intersect Each Other
The Property of Entropy
Temperature-Entropy Plot

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

The Inequality of Clausius

Entropy Change in an Irreversible Process
Entropy Principle
Applications of Entropy Principle
Entropy Transfer with Heat Flow
Entropy Generation in a Closed System
Entropy Generation in an Open System
Reversible Adiabatic Work in a Steady Flow System

m
Entropy and Direction: The Second Law a Directional law of Nature

5. AVAILABILITY, IRREVERSIBILITY

co
Available Energy
Available Energy Referred to a Cycle
Quality of Energy
Maximum Work in a Reversible Process

.
Reversible Work by an Open System Exchanging Heat only with the Surroundings
Useful Work
Dead State
Availability
tas
Irreversibility and Gouy-Stodola theorem
Second Law efficiency
lda
6. TdS RELATIONS, CLAPERYRON AND REAL GAS
EQUATIONS
Highlight
Some Mathematical Theorems
vi

Maxwell's Equations
TdS Equations
Difference in Heat Capacities and Ratio of Heat Capacities Cp ,Cv and γ
Ci

Energy Equation
Joule-Kelvin Effect
Clausius-Clapeyron Equation
Mixtures of Variable Composition
w.

Conditions of Equilibrium of a Heterogeneous System

Gibbs Phase Rule
Types of Equilibrium
Local Equilibrium Conditions
ww

Conditions of Stability

7. PURE SUBSTANCES
p-v Diagram for a Pure Substance
Triple point
p-T Diagram for a Pure Substance
p-v-T Surface
T-s Diagram for a Pure Substance

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Critical Point
h-s Diagram or Mollier Diagram for a Pure Substance
Quality or Dryness Fraction
Steam Tables
Charts of Thermodynamic Properties
Measurement of Steam Quality
Throttling

m
8. PROPERTIES OF GASSES AND GAS MIXTURE
Avogadro's Law

co
Ideal Gas
Equation of State of a Gas
Van der Waals equation
Beattie-Bridgeman equation

.
Virial Expansions
Compressibility
Critical Properties
Boyle temperature
tas
Adiabatic process
Isothermal Process
lda
Polytropic process
Constant Pressure or Isobaric Process
Constant volume or isochoric Process
Properties of Mixtures of Gases
vi

VAPOUR POWER CYCLES

(With Power Plant)
Ci

GAS POWER CYCLE

(With IC Engine)
w.

REFRIGERATION CYCLE
(With RAC)

PSYCHROMETRICS
ww

(With RAC)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Basic Concepts
1. Which of the following are intensive properties? [IES-2005]
1. Kinetic Energy 2. Specific Enthalpy 3. Pressure 4. Entropy
Select the correct answer using the code given below:
(a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4
1. Ans. (b)

m
2. List I List II [GATE-1998]
(A) Heat to work (1) Nozzle
(B) Heat to lift weight (2) Endothermic chemical reaction
(C) Heat to strain energy (3) Heat engine

co
(D) Heat to electromagnetic energy (4) Hot air balloon/evaporation
(5) Thermal radiation
(6) Bimetallic strips
2. Ans. (A) -3, (B) -4, (C) -6, (D)-5

.
Thermodynamic System and Control Volume
tas
3. Assertion (A): A thermodynamic system may be considered as a quantity of working substance
with which interactions of heat and work are studied. [IES-2000]
Reason (R): Energy in the form of work and heat are mutually convertible.
3. Ans. (b)
lda
4. Which one of the following is the extensive property of a thermodynamic system? [IES-1999]
(a) Volume (b) Pressure (c) Temperature (d) Density
4. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive
property.
vi

5. The following are examples of some intensive and extensive properties:

1. Pressure 2. Temperature
3. Volume 4. Velocity
5. Electric charge 6. Magnetisation
7. Viscosity 8. Potential energy
Ci

[IAS-1995]
Which one of the following sets gives the correct combination of intensive and extensive
properties?
Intensive Extensive
(a) 1, 2, 3, 4 5, 6, 7, 8
(b) 1, 3, 5, 7 2, 4, 6, 8
w.

(c) 1, 2, 4, 7 3, 5, 6, 8
(d) 2, 3, 6, 8 1, 4, 5, 7
5. Ans. (c)
Intensive properties, i.e. independent of mass are pressure, temperature, velocity and viscosity. Extensive
properties, i.e. dependent on mass of system are volume, electric charge, magnetisation, and potential
ww

energy. Thus correct choice is (c).

Open and Closed systems

6. A closed thermodynamic system is one in which [IES-1999]
(a) there is no energy or mass transfer across the boundary
(b) there is no mass transfer, but energy transfer exists

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(c) there is no energy transfer, but mass transfer exists

(d) both energy and mass transfer take place across the boundary, but the mass transfer is
controlled by valves
6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transfer exists.

7. Which of the following are intensive properties?

1. Kinetic energy 2. Thermal conductivity
3. Pressure 4. Entropy
Select the correct answer using the code given below:

m
(a) 1 and 2 (b) 2 and 3 only
(c) 2, 3 and 4 (d) 1, 3 and 4 [IES 2007]
7. Ans. (b)

co
8. Which of the following is/are reversible process (es)? [IES-2005]
1. Isentropic expansion 2. Slow heating of water from a hot source
3. Constant pressure heating of an ideal gas from a constant temperature source

.
4. Evaporation of a liquid at constant temperature

(a) 1 only
8. Ans. (b)
(b) 1 and 2
tas
Select the correct answer using the code given below:
(c) 2 and 3
Isentropic means reversible adiabatic.
(d) 1 and 4

9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a reversible

process is the most efficient process. [IES-2001]
lda
Reason (R): The energy transfer as heat and work during the forward process is always
identically equal to the energy transfer as heat and work during the reversal or the process.
9. Ans. (a)

10. An isolated thermodynamic system executes a process, choose the correct statement(s) form
the following [GATE-1999]
vi

(a) No heat is transferred (b) No work is done

(c) No mass flows across the boundary of the system
(d) No chemical reaction takes place within the system
10. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system.
Ci

dQ = 0, dW = 0, ∴dE = 0 or E = Cons tan t

w.

Zeroth Law of Thermodynamics

11. Consider the following statements: [IES-2003]
1. Zeroth law of thermodynamics is related to temperature
2. Entropy is related to first law of thermodynamics
3. Internal energy of an ideal gas is a function of temperature and pressure
ww

4. Van der Waals' equation is related to an ideal gas

Which of the above statements is/are correct?
(a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4
11. Ans. (d) Entropy - related to second law of thermodynamics.
Internal Energy (u) = f (T) only
Van der Wall's equation related to => real gas.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

12. Two blocks which are at different states are brought into contact with each other and allowed
to reach a final state of thermal equilibrium. The final temperature attained is specified by the
(a) Zeroth law of thermodynamics (b) First law of thermodynamics
(c) Second law of thermodynamics (d) Third law of thermodynamics [IES-1998]
12. Ans. (a)

13. Zeroth Law of thermodynamics states that [IES-1996]

(a) two thermodynamic systems are always in thermal equilibrium with each other.
(b) if two systems are in thermal equilibrium, then the third system will also be in thermal

m
equilibrium
(c) two systems not in thermal equilibrium with a third system are also not in thermal equilibrium
with
(d) When two systems are in thermal equilibrium with a third system, they are in thermal

co
equilibrium
13. Ans. (d) Statement at (d) is correct definition of Zeroth law of thermodynamics

14. Match List-I with List-II and select the correct answer using the codes given below
the lists:

.
[IAS-2004]
List-I tas List-II
A. Reversible cycle 1. Measurement of temperature
B. Mechanical work 2. Clapeyron equation
C. Zeroth Law 3. Clausius Theorem
D. Heat 4. High grade energy
5. 3rd law of thermodynamics
lda
6. Inexact differential
Codes: A B C D A B C D
(a) 3 4 1 6 (b) 2 6 1 3
(c) 3 1 5 6 (d) 1 4 5 2
14. Ans. (a)
vi

15. Match List I with List II and select the correct answer: [IAS-2000]
List I List II
A. The entropy of a pure crystalline 1. First law of thermodynamics
Ci

substance is zero at absolute zero temperature

B. Spontaneous processes occur 2. Second law of thermodynamics
in a certain direction
C. If two bodies are in thermal 3. Third law of thermodynamics
equilibrium with a third body,
w.

then they are also in thermal

equilibrium with each other
D. The law of conservation of 4. Zeroth law of thermodynamics
energy.
A B C D A B C D
ww

(a) 2 3 4 1 (b) 3 2 1 4
(c) 3 2 4 1 (d) 2 3 1 4
15. Ans. (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

International Temperature Scale

17. Which one of the following correctly defines 1 K, as per the internationally accepted
definition of temperature scale? [IES-2004]
th
(a) 1/100 of the difference between normal boiling point and normal freezing point of
water
(b) 1/273.15th of the normal freezing point of water
(c) 100 times the difference between the triple point of water and the normal freezing
point of water

m
(d) 1/273.15th of the triple point of water
17. Ans. (d)

co
18. In a new temperature scale say oρ, the boiling and freezing points of water at one atmosphere
are 100°ρ and 300°ρ respectively. Correlate this scale with the Centigrade scale. The reading of
0°ρ on the Centigrade scale is [IES-2001]
(a) 0°C (b) 50°C (c) 100°C (d) 150°C
18. Ans. (d)

.
tas
20. Assertion (a): If an alcohol and a mercury thermometer read exactly 0oC at the ice point and
100°C at the steam point and the distance between the two points is divided into 100 equal parts
in both thermometers, the two thermometers will give exactly the same reading at 50°C.
Reason (R): Temperature scales are arbitrary. [IES-1995]
20. Ans. (a) Both A and R are correct and R is true explanation for A.
lda
21. A new temperature scale in degrees N is to be defined. The boiling and freezing on this scale
are 4000N and 1000N respectively. What will be the reading on new scale corresponding to 600C?
(a) 1200N (b) 1800N (c) 2200N (d) 2800N. [IAS-1995]
21. Ans. (d)

22. Match List I with II and select the correct answer using the code given below the
vi

List I List II
(Type of Thermometer) (Thermometric Property)
A. Mercury-in-glass 1. Pressure
Ci

B. Thermocouple 2.Electrical resistant

C. Thermistor 3.Volume
D. Constant volume gas 4.Induced electric voltage
Code: [IES 2007]
w.

A B C D A B C D
(a) 1 4 2 3 (b) 3 2 4 1
(c) 1 2 4 3 (d) 3 4 2 1
22. Ans. (d)
ww

23. Pressure reaches a value of absolute zero [IES-2002]

(a) at a temperature of - 273 K (b) under vacuum condition
(c) at the earth's centre (d) when molecular momentum of system becomes zero
23. Ans. (d)

24. The time constant of a thermocouple is the time taken to attain:

(a) the final value to he measured (b) 50% of the value of the initial temperature difference
(c) 63.2% of the value of the initial temperature difference
(d) 98.8% of the value of the initial temperature difference [IES-1997]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

24. Ans. (c) Time constant of a thermocouple is the time taken to attain 63.2% of the value of the
initial temperature difference

Work a path function

25. Assertion (A): Thermodynamic work is path-dependent except for an adiabatic
process. [IES-2005]
Reason(R): It is always possible to take a system from a given initial state to any final

m
state by performing adiabatic work only.
25. Ans. (c)

co
Free Expansion with Zero Work Transfer
26. In free expansion of a gas between two equilibrium states, the work transfer involved
(a) can be calculated by joining the two states on p-v coordinates by any path and estimating the
area below [IAS-2001]

.
(b) can be calculated by joining the two states by a quasi-static path and then finding the area
below tas
(c) is zero
(d) is equal to heat generated by friction during expansion.
26. Ans. (c)

27. Work done in a free expansion process is [IAS-2002]

lda
(a) positive (b) negative (c) zero (d) maximum
27. Ans. (c) Since vacuum does not offer any resistance , there is no work transfer involved in
free expansion.

28. In the temperature-entropy diagram of a vapour

shown in the given figure, the
thermodynamic process shown by the dotted line
vi

AB represents
(a) hyperbolic expansion(b) free expansion (c)
constant volume expansion(d) polytropic expansion
[IAS-1995]
Ci

28. Ans. (b)

29. Match items in List-I (Process) with those in List-II (Characteristic) and select the correct
w.

answer using the codes given below the lists: [IES-2001]

List-I (Process) List-II (Characteristic)
A. Throttling process 1. No work done
B. Isentropic process 2. No change in entropy
C. Free expansion 3. Constant internal energy
ww

D. Isothermal process 4. Constant enthalpy

Codes: A B C D A B C D
(a) 4 2 1 3 (b) 1 2 4 3
(c) 4 3 1 2 (d) 1 3 4 2
29. Ans. (a)

30. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The
balloon ruptures and the gas fills up the entire room. Which one of the following statements is
TRUE at the end of above process? [GATE-2008]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(A) The internal energy of the gas decreases from its initial value, but the enthalpy
remains constant (B) The internal energy of the gas increases from its initial value, but the
enthalpy remains constant
(C) Both internal energy and enthalpy of the gas remain constant
(D) Both internal energy and enthalpy of the gas increase
30. Ans. (C) It is free expansion. Since vacuum does not offer any resistance, there is no work
transfer involved in free expansion.
2
Here ∫ ₫W=0 and Q1-2=0 therefore Q1-2= ΔU +W1-2 so ΔU =0

m
1
31. A free bar of length ‘l’ uniformly heated from 0°C to a temperature t° C. α is the coefficient of
linear expansion and E is the modulus of elasticity. The stress in the bar is [GATE-1995]
(a) α tE (b) α tE/2 (c) zero (d) None of the above

co
31. Ans. (c) Ends are not constrained. It is a free expansion problem. Hence there is no stress in
the member.

32. One kg of ice at 00C is completely melted into water at 00C at 1 bar pressure. The
latent heat of fusion of water is 333 kJ/kg and the densities of water and ice at 00C are

.
999.0 kg/m3 and 916.0 kg/ m3, respectively. What are the approximate values of the work
tas
done and energy transferred as heat for the process, respectively?
(a) -9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ
(c) 333.o kJ and -9.4 J (d) None of the above [IES 2007]
⎛1 1 ⎞
32. Ans. (a) Work done (W) = P Δ V = 100 × (V1-V2) = 100 × ⎜⎜ − ⎟
ρ 2 ⎟⎠
lda
⎝ ρ1
⎛ 1 1 ⎞
= 100 × ⎜ − ⎟ = -9.1J
⎝ 999 916 ⎠
33. Which one of the
following is the correct
vi

sequence of the three

processes A, B and C in the
increasing order of the
Ci

amount of work done by a gas

following ideal-gas
expansions by these
processes?
w.

(a) A - B - C
(b) B - A – C
(c) A - C - B
(d) C - A – B
ww

[IES-2006]
33. Ans. (d)
WA = ∫ pdV = 4 × (2 − 1) = 4kJ
1
WB = ∫ pdV = × 3 × (7 − 4) = 4.5kJ
2
WC = ∫ pdV = 1× (12 − 9) = 3kJ

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

34. An ideal gas undergoes an

isothermal expansion from state R to
state S in a turbine as shown in the
diagram given below:
The area of shaded region is 1000
Nm. What is the amount is turbine
work done during the process?
(a) 14,000 Nm (b) 12,000 Nm

m
(c) 11,000Nm
[IES-2004]

co
34. Ans. (c) Turbine work = area under curve R-S

= ∫ P dv
= 1 bar × ( 0.2 − 0.1) m3 + 1000 Nm

.
= 105 × ( 0.2 − 0.1) Nm + 1000Nm
tas
= 11000Nm
35. Identify the process for which the two integrals ∫ pdv and - ∫ vdp evaluated between
any two given states give the same value [IES-2003]
(a) Isenthalpic (b) Isothermal (c) Isentropic (d) Polytropic
lda
35. Ans. (b)

36. Assertion (A): The area 'under' curve on pv plane, ∫ pdv represents the work of reversible
non-flow process. [IES-1992]

∫ Tds
vi

Reason (R): The area 'under' the curve T-s plane represents heat of any reversible
process.
36. Ans. (b)
Ci

37. If ∫ pdv and −∫ vdp for a thermodynamic system of an ideal gas on valuation gives the same
quantity (Positive/negative) during a process, then the process undergone by the system is
(a) isenthalpic (b) isentropic (c) isobaric (d) isothermal [IAS-1997]
37. Ans. (d)
w.

38. For the expression ∫ pdv to represent the work, which of the following conditions should
apply?
(a) The system is closed one and process takes place in non-flow system
ww

(b) The process is non-quasi static [IAS-2002]

(c) The boundary of the system should not move in order that work may be transferred
(d) If the system is open one, it should be non-reversible
38. Ans. (a)

39 Air is compressed adiabatically in a steady flow process with negligible change in

potential and kinetic energy. The Work done in the process is given by
(a) -∫Pdv (b) +∫Pdv (c) -∫vdp (d) +∫vdp [IAS-2000, GATE-1996]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

39. Ans. (c) For closed system W = + ∫ pdv , for steady flow W = − ∫ vdp

40 If ∫Pdv and -∫vdp for a thermodynamic system of an Ideal gas on valuation give same
quantity (positive/negative) during a process, then the process undergone by the
system is [IES-2003, IAS-1997]
(a) Isomeric (b) isentropic (c) isobaric (d) isothermal

m
40. Ans. (d) Isothermal work is minimum of any process.
41. Match list-I with List-II and select the correct answer using the codes given below the
lists:

co
List-I List-II
A. Bottle filling of gas 1. Absolute Zero Temperature
B. Nernst simon Statement 2. Variable flow
C. Joule Thomson Effect 3. Quasi-Static Path

.
D. ∫PdV tas4. Isentropic Process
5. Dissipative Effect [IAS-2004]
6. Low grade energy
Codes: A B C D 7. Process and temperature during phase
(a) 6 5 4 3 change.
(b) 2 1 4 3
lda
(c) 2 5 7 4
(d) 6 1 7 4
41. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out.
Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2.
vi

pdV-work or Displacement Work

42. Thermodynamic work is the product of [IAS-1998]
Ci

(a) two intensive properties

(b) two extensive properties
(c) an intensive property and change in an extensive property
(d) an extensive property and change in an intensive property
42. Ans. (c) W = ∫ pdv where pressure (p) is an intensive property and volume (v) is an
w.

extensive property

43. In a steady state steady flow process taking place in a device with a single inlet and a single
outlet

∫
ww

outlet, the work done per unit mass flow rate is given by w = - vdp, where v is the specific
inlet
volume and p is the pressure. The expression for w given above [GATE-2008]
(A) is valid only if the process is both reversible and adiabatic
(B) is valid only if the process is both reversible and isothermal
(C) is valid for any reversible process
outlet
(D) is incorrect; it must be w = ∫
inlet
pdv

43. (C)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

44. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very
slow,
and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure
of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.0 I m3.
The maximum amount of work that could be utilized from the above process is [GATE-2008]
(A) 0kJ (B)1kJ (C) 2kJ (D) 3kJ
44. Ans. (C) W=P. Δ V =Pgauge. Δ V= (300-200) × 0.1 kJ=2kJ

m
45. For reversible adiabatic compression in a steady flow process, the work transfer per unit
mass is [GATE-1996]
(a ) ∫ pdv (b) ∫ vdp (c) ∫ Tds (d ) ∫ sdT

co
45. Ans. (b) W = − ∫ vdp

.
Heat Transfer-A Path Function

end states.
tas
46. Assertion (A): The change in heat and work cannot be expressed as difference between the

Reason (R): Heat and work are both exact differentials. [IES-1999]
46. Ans. (c) A is true because change in heat and work are path functions and thus can't be
expressed simply as difference between the end states. R is false because both work and heat
are inexact differentials.
lda
47. Match List I with List II and select the correct answer using the codes given below
the lists:
List I (Parameter) List II (Property)
A. Volume 1.Path function [IAS-1999]
vi

B. Density 2. Intensive property

C. Pressure 3. Extensive property
D. Work 4. Point function
Codes: A B C D A B C D
Ci

(a) 3 2 4 1 (b) 3 2 1 4
(c) 2 3 4 1 (d) 2 3 1 4
47. Ans. (a)
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

2. FIRST LAW OF THERMODYNAMICS

First Law of Thermodynamics
30. Which one of the following sets of thermodynamic laws/relations is directly involved in
determining the final properties during an adiabatic mixing process? [IES-2000]
(a) The first and second laws of thermodynamics
(b) The second law of thermodynamics and steady flow relations

m
(c) Perfect gas relationship and steady flow relations
(d) The first law of thermodynamics and perfect gas relationship
30. Ans. (d)

co
40. For a closed system, the difference between the heat added to the system and the work done
by the system is equal to the change in [IES-1992]
(a) enthalpy (b) entropy (c) temperature (d) internal energy
40. Ans. (d)
From First law of thermodynamics, for a closed system the net wnergy transferred as heat Q and as work W

.
is equal to the change in internal energy, U, i.e. Q - W = dU

the p-v and T-s planes. [GATE-2005]

tas
15. The following four figures have been drawn to represent a fictitious thermodynamic cycle, on
lda

According to the first law of thermodynamics, equal areas are enclosed by

(a) figures 1and 2 b) figures 1and 3 c) figures 1and 4 d) figures 2 and 3
15. Ans. (a)
vi

Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power cycle, so area is
not meant something.

76. An ideal cycle is shown in the figure. Its

Ci

thermal efficiency is given by

⎛ v3 ⎞ ⎛ v3 ⎞
⎜ − 1⎟ ⎜ − 1⎟
1 ⎝ v1 ⎠
(a )1 − ⎝ 1 ⎠
v
(b) 1 −
⎛ p2 ⎞ γ ⎛ p2 ⎞
w.

⎜ − 1⎟ ⎜ − 1⎟
⎝ p1 ⎠ ⎝ p1 ⎠

(c)1 − γ
( v3 − v1 ) p1 (b) 1 −
1 ( v3 − v1 ) p1
( p2 − p1 ) v1 γ ( p2 − p1 ) v1
ww

[IES-

76. Ans. (b)

58. Which one of the following is correct?

The cyclic integral of (δQ − δW ) for a process is

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(a) positive (b) negative

(c) Zero (d) unpredictable [IES 2007]
Ans. (c) It is du = đQ – đW, as u is a thermodynamic property and its cyclic integral must
be zero.

71. A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are
+20 kJ and +50 kJ, respectively. If the system is returned to state, 1, and Q2-1 is -10 kJ,
what is the value of the work W2-1? [IES-2005]

m
(a) + 20 kJ (b) -40 kJ (c) - 80 kJ (d) + 40 kJ
71. Ans. (b) ΣdQ = ΣdW or Q1− 2 + Q2 −1 = W1− 2 + W2 −1
or 20 + ( −10 ) = 50 + W2 −1 or W2 −1 = −40kJ

co
75. A gas is compressed in a cylinder by a movable piston to a volume one-half of its
original volume. During the process, 300 kJ heat left the gas and the internal energy
remained same. What is the work done on the gas? [IES-2005]
(a) 100kNm (b) 150 kNm (c) 200 kNm (d) 300 kNm

.
75. Ans. (d) dQ = du + dw as u = const. tas
Therefore du = 0 or dQ = dw = 300kNm

33. In a steady-flow adiabatic turbine, the changes in the internal energy, enthalpy,
lda
kinetic energy and potential energy of the working fluid, from inlet to exit, are -100
kJ/kg, -140 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives
the amount of work developed by the turbine? [IES-2004]
(a) 100 kJ/kg (b) 110 kJ/kg (c) 140 kJ/kg (d) 150 kJ/kg
33. Ans. (d)
⎛ V2 ⎞
vi

Q − Wx = Δ ⎜ h + + gz ⎟
⎝ 2 ⎠
O − Wx = −140 − 10 + 0
or Wx = 150 kJ / kg
Ci

Change of internal energy = -100 kJ/kg is superfluous data.

67. Gas contained in a closed system consisting of piston cylinder arrangement is

expanded. Work done by the gas during expansion is 50 kJ. Decrease in internal energy
w.

of the gas during expansion is 80 kJ. Heat transfer during the process is equal to [IES-
2003]
(a) -20 kJ (b) +20 kJ (c) -80 kJ (d) +80 kJ
67. Ans. (b) Q = Δ E+ Δ W
ww

Δ E = - 30 kJ (decrease in internal energy)

Δ W = + 50 kJ (work done by the system)
Q = - 30 + 50 = + 20 kJ

16. A system while undergoing a cycle [IES-2001]

A - B - C - D - A has the values of heat and work transfers as given in the table:
Process kJ/min kJ/min

A-B +687 +474

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

B-C -269 0
C-D -199 -180
D-A +75 -0
The power developed in kW is, nearly,
(a) 4.9 (b) 24.5 (c) 49 (d) 98
16. Ans. (a)

57. A tank containing air is stirred by a paddle wheel. The work input to the paddle wheel is 9000
kJ and the heat transferred to the surroundings from the tank is 3000 kJ. The external work done

m
by the system is [IES-1999]
(a) zero (b) 3000 kJ (c) 6000 kJ (d) 9000 kJ
57. Ans. (c)

co
74. The values of heat transfer and work transfer for four processes of a thermodynamic cycle are
given below: [IES-1994]
Process Heat Transfer (kJ) Work Transfer (kJ)

1 300 300

.
2 tas Zero 250
3 -100 -100
4 zero -250
The thermal efficiency and work ratio for the cycle will be respectively.
(a) 33% and 0.66 (b) 66% and 0.36. (c) 36% and 0.66 (d) 33% and 0.36.
Work done 300 − 100
74. Ans. (b) ηth = = = 0.66
lda
heat added 300

Work ratio =
∑ ( + w) − ∑ ( − w) = 550 − 350 = 0.36
∑ ( + w) 550
vi

71. A system executes a cycle during which there are four heat transfers: Q12 = 220 kJ, Q23 = -
25kJ, Q34 = -180 kJ, Q41 = 50 kJ. The work during three of the processes is W12 = 5kJ, W23 = -10
kJ, W34 = 60kJ. The work during the process 4 -1 is
(a) - 230 kJ (b) 0 kJ (c) 230 kJ (d) 130 kJ [IAS-2003]
Ci
w.

78. Two ideal heat engine cycles are

represented in the given figure. Assume VQ =
ww

QR, PQ = QS and UP =PR =RT. If the work

interaction for the rectangular cycle (WVUR) is
48 Nm, then the work interaction for the other
cycle PST is
(a) 12Nm (b) 18 Nm
(c) 24 Nm (d) 36 Nm [IAS-2001]

78. Ans. (c) Area under p-v diagram is represent work.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

1 1
Areas Δ PTS= Area (WVUR) ∴ Work PTS= × 48 =24 Nm
2 2
12. A system undergoes a change of state during which 80 kJ of heat is transferred to it and it
does 60 kJ of work. The system is brought back to its original state through a process during
which 100 kJ of heat is transferred to it. The work done by the system is [IAS-1998]
(a) 40 kJ (b) 60 kJ (c) 120 kJ (d) 180 kJ
12. Ans. (c)
Q1− 2 = ΔE1− 2 + W1− 2 or 80 = ΔE1− 2 + 60 or ΔE1− 2 = 20kJ

m
Q2 −1 = ΔE2 −1 + W2 −1 or 100 = −20 + W2−1 or W2−1 = 120kJ

14. A reversible heat engine operating between hot and cold reservoirs delivers a work output of
54 kJ while it rejects a heat of 66 kJ. The efficiency of this engine is

co
(a) 0.45 (b) 0.66 (c) 0.75 (d) 0.82 [IAS-1998]
work output work out put 54
14. Ans. (a) η = = = = 0.45
Heat input work output + heat rejection 54 + 66
24. If a heat engine gives an output of 3 kW when the input is 10,000 J/s, then the

.
thermal efficiency of the engine will be [IAS-1995]
(a) 20%
24. Ans. (b)
(b) 30% tas
(c) 70% (d) 76.7%
lda
88. In an adiabatic process, 5000J of work is performed on a system. The system returns to its
original state while 1000J of heat is added. The work done during the non-adiabatic process is
(a) + 4000J (b) - 4000J (c) + 6000J (d) - 6000J [IAS-1997]
88. Ans. (b)
Q1− 2 = ΔE1− 2 + W1−2
or 0 = ΔE1− 2 + ( −5000 ) or ( ΔE )1− 2 = 5000 J
vi

Q2 −1 = ( ΔE )2 −1 + W2−1 or W2 −1 = Q2 −1 − ( ΔE )2 −1 = 1000 − 5000 = − 4000 J

Ci

26. In a thermodynamic cycle consisting of four processes, the heat and work are as follows:
Q: + 30, - 10, -20, + 5
W: + 3, 10, - 8, 0
The thermal efficiency of the cycle will be [IAS-1996]
(a) Zero (b) 7.15% (c) 14.33% (d) 28.6%
w.

26. Ans. (c) Net work output = 3 + 10 – 8 = 5 unit and Heat added = 30 + 5 = 35 unit
5
Therefore efficiency, η = × 100% = 14.33%
35
27. Match List I (Devices) with List II (Thermodynamic equations) and select the correct answer
using the codes below the lists: [IAS-1996]
ww

List I List II
A. Turbine 1. W=h2-h1
B. Nozzle 2. h1=h2
C. Valve 3. h1=h2+V2/2
D. Compressor 4. W=h1-h2
Codes: A B C D A B C D
(a) 4 3 2 1 (b) 2 3 1 4
(c) 4 3 1 2 (d) 3 2 4 1
27. Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

37. Given that the path 1-2-3, a system absorbs 100kJ as heat and does 60kJ work while along the path 1-4-
3 it does 20kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is [IAS 1994]
(a) - 140 kJ (b) - 80 kJ (c) - 40kJ (d) + 60 kJ

m
Ans. (d) Q123 = U13 + W123 or, 100 = U13 + 60 or, U13 = 40 kJ

co
And Q143 = U13 + W143 = 40+20 = 60 kJ

40. The given figure shows the variation of force in an elementary system which undergoes a process
during which the plunger position changes from 0 to 3 m. lf the internal energy of the system at the end of

.
the process is 2.5 J higher, then the heat absorbed during the process is [IAS 1994]
(a) 15 J
tas
(b) 20 J (c) 25 J (d) 30 J
vi lda

1
Ans. (b) Total work = 5 x 3 + × 5 × 1 = 17.5 J or δW = du + δW = 2.5 + 17.5 = 20 J
2
Ci

45. The efficiency of a reversible cyclic process undergone by a substance as shown in the given diagram is
(a) 0.40 (b) 0.55 (c) 0.60 (d) 0.80 [IAS 1994]
w.
ww

1
Ans. (c)Efficiency = Area under 500 and 1500 × {(5 − 1) + (4 − 2)}× (1500 − 500)
2 3000
= = = 0.6
Area under 0 and 1500 1 5000
× {(5 − 1) + ( 4 − 2)}× (1500 − 500) + (5 − 1) × 500
2

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Internal Energy--A Property of System

60. For a simple closed system of constant composition, the difference between the net
heat and work interactions is identifiable as the change in [IES-2003]
(a) Enthalpy (b) Entropy (c) Flow energy (d) Internal energy
60. Ans. (d)

m
61. Assertion (A): The internal energy depends on the internal state of a body, as
determined by its temperature, pressure and composition. [IES-2006]

co
Reason (R): Internal energy of a substance does not include any energy that it may
possess as a result of its macroscopic position or movement.
61. Ans. (a)

.
69. Change in internal energy in a reversible process occurring in a closed system is

(a) Pressure (b) Volume

69. Ans. (b) dQ = dU + pdV
tas
equal to the heat transferred if the process occurs at constant: [IES-2005]
(c) Temperature
if V is cons tan t ( dQ )v = ( dU)v
(d) Enthalpy

35. 170 kJ of heat is supplied to a system at constant volume. Then the system rejects 180
lda
kJ of heat at constant pressure and 40 kJ of work is done on it. The system is finally
brought to its original state by adiabatic process. If the initial value of internal energy is
100 kJ, then which one of the following statements is correct? [IES-2004]
(a) The highest value of internal energy occurs at the end of the constant volume process
(b) The highest value of internal energy occurs at the end of constant pressure process.
vi

(c) The highest value of internal energy occurs after adiabatic expansion
(d) Internal energy is equal at all points

35. Ans. (a)

Ci

Q2 = 180kJ = Δu + ΔW = Δu + ( −40)
∴U1 = 100kJ, U2 = 100 + 170 = 270 kJ,
U3 = 270 − 180 + 40 = 130 kJ
w.
ww

40. A system undergoes a process during which the heat transfer to the system per degree
increase in temperature is given by the equation: [IES-2004]
dQ/dT = 20 kJ/oC The work done by the system per degree increase in temperature is
given by the equation
dW/dT = 2 – 0.1 T, where T is in °C. If during the process, the temperature of water
varies from 100°C to 150°C, what will be the change in internal energy?

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(a) 125 kJ (b) -250 kJ (c) 625 kJ (d) -1250 kJ

40. Ans. (c)
dQ = du + dw
2.dt = du + ( 2 − 0.1T ) dT
0.1 150 0.1
or ∫ du = ∫ 0.1TdT = × ⎡T 2 ⎤ =
2 ⎣ ⎦100 2 ⎣
⎡1502 − 1002 ⎤⎦ = 625kJ

m
29. When a system is taken from state A to state B along the path
A-C-B, 180 kJ of heat flows into the system and it does 130 kJ of
work (see figure given) :

co
How much heat will flow into the system along the path A-D-B if
the work done by it along the path is 40 kJ?
(a) 40 kJ (b) 60 kJ
(c) 90 kJ (d) 135 kJ
[IES-1997]

.
tas
29. Ans. (c) Change of internal energy from A to B along path ACB = 180 - 130 = 50 kJ. It will
be same even along path ADB. :. Heat flow along ADB = 40 + 50 = 90 kJ
lda
71. The heat transfer, Q, the work done W and the change in internal energy U are all zero in the
case of [IES-1996]
(a) a rigid vessel containing steam at 150°C left in the atmosphere which is at 25oC
(b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly outwards.
(c) a rigid vessel containing ammonia gas connected through a valve to an evacuated rigid
vessel, the vessel, the valve and the connecting pipes being well insulated and the valve being
opened and after a time, conditions through the two vessels becoming uniform.
vi

(d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated bottle
71. Ans. (c) In example of (c), heat transfer, work done, and change in internal energy are all
zero.
Ci

45. The internal energy of a certain system is a function of temperature alone and is given by the
formula E = 25 + 0.25t kJ. If this system executes a process for which the work done by it per
degree temperature increase is 0.75 kN-m, the heat interaction per degree temperature increase, in
kJ, is [IES-1995]
w.

(a) -1.00 (b) -0.50 (c) 0.50 (d ) 1.00.

45. Ans. (d) dQ = du +dw = 0.25 + 0.75 = 1.00 kJ

36. When a gas is heated at constant pressure, the percentage of the energy supplied, which
goes as the internal energy of the gas is [IES-1992]
ww

(a) more for a diatomic gas than for triatomic gas

(b) same for monatomic, diatomic and triatomic gases but less than 100%
(c) 100% for all gases (d) less for triatomic gas than for a diatomic gas
36. Ans. (a)

77. Which one of the following is the correct expression for change in the internal energy
for a small temperature change Δ T for an ideal gas? [IAS-2007]
(a) ΔU = Cv × ΔT (b) ΔU = C p × ΔT

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(d) ΔU = ( C p − Cv ) × ΔT
Cp
(c) ΔU = × ΔT
Cv
77. Ans. (a)

110. The heat transferred in a thermodynamic cycle of a system consisting of four

processes is successively 0, 8, 6 and - 4 units. The net change in the internal energy of the
system will be [IAS-1999]

m
(a) - 8 (b) zero (c) 10 (d) -10
110. Ans. (b) Internal energy is a property of a system so ∫ du = 0

co
112. During a process with heat and work interactions, the internal energy of a system
increases by 30 kJ. The amounts of heat and work interactions are respectively
(a) - 50 kJ and - 80 kJ (b) -50 kJ and 80 kJ [IAS-1999]
(c) 50 kJ and 80 kJ (d) 50 kJ and - 80 kJ

.
112. Ans. (a) dQ = du + dW if du = +30kJ then dQ = −50kJ and dW = −80kJ
tas
35. A mixture of gases expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ
of heat during the process. The change in internal energy of the mixture is [IAS 1994]
(a) 30 kJ (c) 84 kJ (b) 54 kJ (d) 114 kJ
Ans. (b) δW = du + δW = du + pdV
Or 84x103J = du + 1x106x(0.06-0.03) = du +30 kJ or du = 83 – 30 = 54 kJ
lda
20. A gas contained in a cylinder is compressed, the work required for compression being 5000
kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The
change in internal energy of the gas during the process is [GATE-2004]
(a) - 7000 kJ (b) - 3000 kJ (c) + 3000 kJ (d) + 7000 kJ
20. Ans. (c)
vi

dQ = du + dw
Q = u2 − u1 + W or − 2000 = u2 − u1 − 5000 or u2 − u1 = 3000kJ
Ci

50. In an adiabatic process 6000 J of work is performed on a system. In the non-

adiabatic process by which the system returns to its original state 1000J of heat is added
to the system. What is the work done during non-adiabatic process? [IAS-2004]
(a) + 7000 J (b) - 7000 J (c) + 5000 J (d) - 5000 J
50. Ans. (a) Q1-2 = U2 –U1 +W1-2
w.
ww

Perpetual Motion Machine of the First Kind-PMM1

32. Consider the following statements: [IES-2000]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

1. The first law of thermodynamics is a law of conservation of energy.

2. Perpetual motion machine of the first kind converts energy into equivalent work.
3. A closed system does not exchange work or energy with its surroundings.
4. The second law of thermodynamics stipulates the law of conservation of energy and entropy.
Which of the statements are correct?
(a) 1 and 3 (b) 2 and 4 (c) 2, 3 and 4 (d) 1, 2 and 3
32. Ans. (d)

m
Enthalpy
41. The fundamental unit of enthalpy is [IAS 1994]
(a) MLT-2 (b) ML-2T-1 (c) ML2T-2 (d) ML3T-2
Ans. (c)

co
64. Assertion (A): If the enthalpy of a closed system decreases by 25 kJ while the system
receives 30 kJ of energy by heat transfer, the work done by the system is 55 kJ. [IES-2001]
Reason (R): The first law energy balance for a closed system is (notations have their usual
meaning) ΔE = Q − W

.
64. Ans. (a) tas
Application of First Law to Steady Flow Process S.F.E.E
74. Which one of the following is the steady flow energy equation for a boiler?
lda
v12 v2
(a) h1 + = h2 + 2 (b) Q = (h2 − h1 ) [IES-2005]
2 gJ 2 gJ
v12 v2
(c) h1 + + Q = h2 + 2 (d) Ws = (h2 − h1 ) + Q
2 gJ 2 gJ
vi

v12 dQ v2 dw
74. Ans. (b) h1 + + gz1 + = h2 + 2 + gz 2 + =0
2 dm 2 dm
dw
For boiler v1, v2 is negligible and z1 = z2 and =0
dm
Ci

dQ
or = ( h2 − h1 )
dm

90. In a test of a water-jacketed compressor, the shaft work required is 90 kN-m/kg of air
w.

compressed. During compression, increase in enthalpy of air is 30 kJ/kg of air and increase in
enthalpy of circulating cooling water is 40 kJ/ kg of air. The change is velocity is negligible. The
amount of heat lost to the atmosphere from the compressor per kg of air is [IAS-2000]
(a) 20kJ (b) 60kJ (c) 80 kJ (d) 120kJ
ww

90. Ans. (a)

Energy balance gives as
dW dQ
= ( Δh )air + ( Δh ) water +
dm dm
dQ
or = 90 − 30 − 40 = 20kJ / kg of air compressed.
dm

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

92. When air is compressed, the enthalpy is increased from 100 to 200 kJ/kg. Heat lost during
this compression is 50 kJ/kg. Neglecting kinetic and potential energies, the power required for a
mass flow of 2 kg/s of air through the compressor will be [IAS-1997]
(a) 300 kW (b) 200 kW (c) 100 kW (d) 50 kW
92. Ans. (a)
dQ dw
m ( h1 ) + = m ( h2 ) +
dt dt
dw dQ
or = m ( h1 − h2 ) + = 2 × (100 − 200 ) − 50 × 2 = −300kW
dt dt

m
i.e. 300kW work have to given to the system.

co
Variable Flow Processes
80. Match List-I with List-II and select the correct answer using the codes given below
Lists: [IAS-2004]

.
List-I List-II
A. Bottle filling of gas
B. Nernst Simon statement
C. Joule Thomson effect
tas
1. Absolute zero temperature
2. Variable flow
3. Quasistatic path
D. ∫ pdv 4. Isenthalpic process
lda
5. Dissipative effect
6. Low grade energy
7. Process and temperature during phase change
Codes: A B C D A B C D
(a) 6 5 4 3 (b) 2 1 4 3
(c) 2 5 7 4 (d) 6 1 7 4
vi

80. Ans. (b)

93. A gas chamber is divided into two parts by means of a partition wall. On one side, nitrogen
Ci

gas at 2 bar pressure and 20°C is present. On the other side, nitrogen gas at 3.5 bar pressure
and 35°C is present. The chamber is rigid and thermally insulated from the surroundings. Now, if
the partition is removed,
(a) high pressure nitrogen will get throttled [IAS-1997]
(b) mechanical work, will be done at the expense of internal energy
w.

(c) work will be done on low pressure nitrogen

(d) internal energy of nitrogen will be conserved
93. Ans. (a)
ww

Discharging and Charging a Tank

An insulated tank initially contains 0.25 kg of a gas with an internal energy of 200 kJ/kg
.Additional gas with an internal energy of 300 kJ/kg and an enthalpy of 400
kJ/kg enters the tank until the total mass of gas contained is 1 kg. What is the final
internal energy(in kJ/kg) of the gas in the tank?
(a) 250 (b) 275 [IES 2007]
(c) 350 (d) None of the above
Ans. (c) Enthalpy of additional gas will be converted to internal energy.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Uf= miui+(mf-mi)h = 0.25x200+(1-0.25)x400 = 350 kJ As total mass = 1kg, uf=350 kJ/kg

49. A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through
which air (assumed to be ideal gas with constant specific heats) passes at I MPa, 3500 C. A valve
connected with the supply line is opened and the tank is charged with air until the final pressure
inside the tank reaches I MPa. The final temperature inside the tank [GATE-2008]

m
. co
tas
(A) is greater than 3500C (B) is less than 3500C
(C) is equal to 350°C (D) may be greater than, less than, or equal to 350°C,
lda
depending on the volume of the tank

49. Ans (A) The final Temp. (T2)= γT1

vi
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

3. SECOND LAW OF THERMODYNAMICS

59. Which one of the following is correct on basis of the second law of
Thermodynamics?
(a) For any spontaneous process, the entropy of the universe increases
(b) ∆S =qrev/T at constant temperature
(c) Efficiency of the Starling cycle is more than that of a Carnot cycle

m
(d) ∆E=q+w
(The symbols have their usual meaning) [IES 2007]
Ans. (a)

co
37. Assertion (A): Second law of thermodynamics is called the law of degradation of energy.
[IES-1999]
Reason (R): Energy does not degrade each time it flows through a finite temperature difference.
37. Ans. (b) Both A and R are true but R does not give correct reasoning for A.

.
(a) Zeroth Law of Thermodynamics
(c) Second Law of Thermodynamics
tas
1. Heat transfer takes place according to [IES-1996]
(b) First Law of Thermodynamics
(d) Third Law of Thermodynamics.
1. Ans. (c) Heat transfer takes place according to second law of thermodynamics as it tells about
the direction and amount of heat flow that is possible between two reservoirs.
lda
79. Which of the following statements are associated with second law of thermodynamics?
(a) When a system executes a cyclic process, net work transfer is equal to net heat transfer.
(b) It is impossible to construct an engine, that operating in a cycle will produce no other effect
than the extraction of heat from a reservoir and performance of an equivalent amount of work.
(c) It is impossible by any procedure, no matter how idealized, to reduce any system to the
absolute zero of temperature in a finite number of operations.
vi

(d) It is impossible to construct a device that operating in a cycle will produce no effect other than
transfer of heat from a cooler to hotter body. [IAS-2001]
Select the correct answer using the codes given below:
Codes: (a) 1, 2 and 4 (b) 2 and 4 (c) 2, 3 and 4 (d) 2 and 3
Ci

79. Ans. (b)

1.10 A system undergoes a state change from 1 to 2. According the second law of
thermodynamics for the process to be feasible, the entropy change, S2 – S1 of the system
[GATE-1997]
w.

(a) is positive or zero (b) is negative or zero (c) is zero (d) can be positive, negative or zero
1.10 Ans. (d) ( ΔS ) isolated system can never decrease but it is only a process.
ww

Qualitative Difference between Heat and Work

Kelvin-Planck Statement of Second Law

63. Assertion (A): No machine would continuously supply work without expenditure of
some other form of energy. [IAS-2001]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Reason (R): Energy can be neither created nor destroyed, but it can only be transformed
from one form into another.
63. Ans. (a)

42. Consider the following statements:

The definition of
1. temperature is due to Zeroth Law of Thermodynamics. [IES-1993]
2. entropy is due to First Law of Thermodynamics.
3. internal energy is due to Second Law of Thermodynamics.

m
4. reversibility is due to Kelvin-Planck's statement.
Of these statements
(a) 1,2 and 3 are correct (b) 1, 3 and 4 are correct
(c) 1 alone is correct (d) 2 alone is correct

co
42. Ans. (c) Out of 4 definitions given, only first definition is correct and balance three are wrong.

Clausius' Statement of the Second Law

.
36. Assertion (A): Heat cannot spontaneously pass from a colder system to a hotter system
without simultaneously producing other effects in the surroundings. [IES-1999]
tas
Reason (R): External work must be put into heat pump so that heat can be transferred from a
cold to a hot body.
36. Ans. (a) A and R are true. A is the Clausius statement of second law of thermodynamics.
Spontaneously means without change in surroundings. Statement at R provides the correct
reasoning for A, i.e. the work must be done by surroundings on the system for heat to flow from
lower temperature to higher temperature.
lda

Clausius' Theorem
120. A steam power plant is shown in figure,
(a) the cycle violates first and second laws of
vi

thermodynamics.
(b) the cycle does not satisfy the condition of
Clausius inequality.
(c) the cycle only violates the second laws of
Ci

thermodynamics
(d) the cycle satisfies the Clausius inequality.
[IES
w.
ww

120. Ans. (d)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Refrigerator and Heat Pump [with RAC]

Equivalence of Kelvin-Planck and Clausius Statements

81. Assertion (A): Efficiency of a reversible engine operating between temperature limits T1 and
T2 is maximum. [IES-2002]
Reason (R): Efficiency of a reversible engine is greater than that of an irreversible engine.
81. Ans. (b)

m
25. A heat engine is supplied with 250 KJ/s of heat at a constant fixed temperature of
2270C. The heat is rejected at 270C. The cycle is reversible, if the amount of heat
rejected is [IAS-1995]
(a) 273 KJ/s (b) 200 KJ/s (c) 180 KJ/s (d) 150 KJ/s.

co
Q1 Q2
25. Ans. (d) =
T1 T2
29. A reversible engine En as shown in the given
figure draws 300 kcal from 200 K reservoir and

.
does 50 kcal of work during a cycle. The sum of
heat interactions tas
with the other two reservoir is given by
(a)Q1 + Q2 = + 250 kcal
(b) Q1 + Q2 = - 250 kcal
(c) Q1 + Q2 = + 350 kcal
(d)Q1 + Q2 = - 350 kcal [IAS-1996]
lda
29. Ans. (a) Q1 + Q2 = 300 – 50 = 250 Kcal

Carnot Engine with same efficiency or same work output

vi

30. A reversible engine operates between temperatures T1, and T2, The energy rejected by this
engine is received by a second reversible engine at temperature T2 and rejected to a reservoir at
temperature T3. If the efficiencies of the engines are same then the relationship between T1, T2
and T3 is given by [IES-2002]
(T1 + T3 ) (T1 + 2T3 )
Ci

(a) T2 =
2
(b) T2 = (T
1
2
+ T32 ) (c) T2 = T1T3 (d) T2 =
2
30. Ans. (c)
w.

63. A reversible engine operates between temperatures 900 K & T2 (T2 < 900 K), &
another reversible engine between T2 & 400 K (T2 > 400 K) in series. What is the value
of T2 if work outputs of both the engines are equal? [IES-2005]
(a) 600 K (b) 625 K (c) 650 K (d) 675 K
63. Ans. (c) Figure from another question
ww

W1 = W2
T1 + T3 900 + 400
or Q1 − Q2 = Q2 − Q3 or T1 − T2 = T2 − T3 or T2 = = = 650K
2 2

38. Two reversible engine operate between thermal reservoirs at 1200 K, T2K and 300 K
such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal
reservoir at T2K, while the 2nd engine receives heat from thermal reservoir at T2K and
rejects heat to the thermal reservoir at 300 K. The efficiency of both the engines is equal

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

What is the value of temperature T2? [IES-2004]

(a) 400 K (b) 500 K (c) 600 K (d) 700 K

38. Ans. (a)

η1 = η2
T2 300
or 1 − = 1−
1200 T2

m
or T2 = 1200 × 300 = 600K

. co
tas
29. Consider the following statements:
lda
1. Amount of work from cascaded Carnot engines corresponding to fixed
temperature difference falls as one goes to lower absolute level of temperature.
2. On the enthalpy-entropy diagram, constant pressure lines diverge as the entropy
increases. [IAS-2007]
Which of the statements given above is/are correct?
vi

(a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

29. Ans. (b) For reversible cycle

Ci

T1 T2 T3
= =
Q1 Q2 T3
T1 − T2 Q1 − Q2
or =
T2 Q2
w.

T2
or T1 − T2 = (Q1 − Q2 ) ×
Q2
T3
similarly T2 − T3 = ( Q2 − Q3 ) ×
ww

Q3
if T1 − T2 = T2 − T3 then Q1 − Q2 = Q2 − Q3 or W1 = W2

47. One reversible heat engine operates between 1600 K and T2 K, and another reversible heat
engine operates between T2K and 400 K. If both the engines have the same heat input and
output, then the temperature T2 must be equal to [IES-1993]
(a) 1000 (b) 1200 (c) 1400 (d) 800
47. Ans. (d) Two reversible heat engines operate between limits of
1600K and T2 ; T2 and 400K

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Both have the same heat input and output,

T1 − T2 1600 − T2 T2 − 400
i.e. is same for both or = or T2 = 800 K
T1 1600 T2

38. In a cyclic heat engine operating between a source temperature of 600°C and a sink temperature of
20°C, the least rate of heat rejection per kW net output of the engine is [IAS 1994]
(a) 0.460 kW (c) 0.588 kW (b) 0.505 kW (d) 0.650 kW
Q1 Q2 Q1 − Q2 W
Ans. (b) Reversible engine has maximum efficiency where = = =

m
T1 T2 T1 − T2 T1 − T2

W 1
Therefore least heat rejection per kW net output, Q2 = × T2 = × 293 = 0.505kW

co
T1 − T2 873 − 293

.
tas
vi lda
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

4. ENTROPY
Two Reversible Adiabatic Paths cannot Intersect Each Other
76. Which one of the following is the correct statement? [IAS-2007]
Two adiabatic will
(a) intersect at absolute zero temperature (b) never intersect

m
(c) become orthogonal at absolute zero temperature
(d) become parallel at absolute zero temperature
76. Ans. (b)

co
The Property of Entropy
54. Assigning the basic dimensions to mass, length, time and temperature

.
respectively as M, L, T and θ (Temperature), what are the dimensions of
entropy?
(a) M LT-2 θ
2 -2 -1
(c) M L T θ
tas (b) M L2 T-1 θ-1
(d) M L3T-2 θ -1 [IES 2007]
Ans. (c)
lda
54. Heat flows between two reservoirs having temperatures 1000 K and 500 K,
respectively. If the entropy change of the cold reservoir is 10 kJ/K, then what is the
entropy change for the hot reservoir? [IAS-2004]
(a) - 10 kJ/K (b) - 5 kJ/K (c) 5 kJ/K (d) 10 kJ/K
54. Ans. (b)
vi

+Q
S2 = = 10
500
or Q = 5000 kJ
Ci

−Q −5000
S1 = = = −5kJ / k
1000 1000
⎡∴Heat added to the system is +ive ⎤
⎢ Heat rejected from the system is -ive ⎥
w.

⎣ ⎦
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Temperature-Entropy Plot
32. A system comprising of a pure substance
executes reversibly a cycle 1 -2 -3 -4 - 1
consisting of two isentropic and two isochoric
processes as shown in the Fig. 1.
[IES-2002]

m
co
Which one of the following is the correct representation of this cycle on the temperature - entropy
coordinates?

.
tas
vi lda
Ci
w.
ww

32. Ans. (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

51. A cycle of pressure – volume diagram is

shown in the given figure-I, Same cycle on
temperature-entropy diagram will be
represented by
[IES-1

m
. co
tas
51. Ans. (d) Figure at (d) matches with given process on P-V plane.
lda
96. An ideal cycle is shown in the given pressure-volume diagram:
vi

[IAS-1997]
Ci

The same cycle on temperature-entropy diagram will be represented as

w.
ww

96. Ans. (d)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

36. An ideal air standard cycle is shown in the

given temperature-entropy diagram.
[IES

m
co
The same cycle, when represented on the pressure-volume coordinates takes the form

.
36. Ans. (a)
tas
lda
80. Match figures of Column I with those given in Column II and select given below the columns:
Column I (p-v diagram) Column II (T-s diagram)
[IES-1994]
vi
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
. co
tas
lda

Codes: A B C A B C
(a) 1 2 3 (b) 2 3 1
vi

(c) 3 1 2 (d) 3 2 1
80. Ans. (c)
Ci

48. A cyclic process ABCD shown in the V-T diagram performed with a constant mass of an ideal
gas. The process of p-V diagram will be as shown in [IES-1992]
w.
ww

(a) (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(b) (d)

m
48. Ans. (d)

71. Three processes are represented on the p-v and T-s diagrams in the following figures. Match

co
processes in the two diagrams and select the correct answer using the codes given below the
diagrams: [IES-1994]

.
tas
lda

Codes: A B C A B C
(a) 1 2 3 (b) 2 3 1
(c) 3 2 1 (d) 1 3 2
vi

71. Ans. (c)

93. The thermal efficiency of the hypothetical

heat engine cycle shown in the given figure is
Ci

(a) 0.5
(b) 0.45
(c) 0.35
(d) 0.25
w.

[IAS-2000]
ww

1
× ( 5 − 1) × ( 800 − 400 )
Work done area1 − 2 − 3
93. Ans. (d) η = = = 2 = 0.25
Heat added areaunder curve 2 − 3 ( 5 − 1) × 800

65. Which one of the following pairs best expresses a relationship similar to that expressed in the
pair 'pressure-volume' for a thermodynamic system undergoing a process? [IAS-1995]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(a) Enthalpy-entropy (b) Pressure-enthalpy (c) Pressure-temperature (d)Temperature-entropy

65. Ans. (d)

109. An ideal gas contained in a rigid tank is cooled

such that T2 < and P2 <P1 In the given temperature
entropy diagram, this process path is represented by the
line labeled.
(a) A

m
(b) B
(c) C
(d) D
[IAS-1999]

co
109. Ans. (a)

.
54. Assertion (A): If a graph is plotted for absolute temperature as a function of entropy, the area
tas
under the curve would give the amount of heat supplied. [IES-1998]
Reason (R): Entropy represents the maximum fraction of work obtainable from heat per degree
drop in temperature.
54. Ans. (c)

87. In the T-S diagram shown in the figure, which

lda
one of the following is represented by the area
under the curve? [IAS-2004]
(a) Total work done during the process
(b) Total heat absorbed during the process
(c) Total heat rejected during the process
vi

(d) Degree of irreversibility

Ci

87. Ans. (b)

w.

The Inequality of Clausius

80. Clausius inequality is stated as [IAS-2001]
ww

Q Q
(a) ∫ δQ < 0 (b) ∫ δQ = 0 (c) ∫δ T >0 (d) ∫δ T ≤0
80. Ans. (d)

31. For a real thermodynamic cycle, which one of the following is correct? [IES-2005]
dQ dQ
(a) ∫ ds = 0 (b) ∫ T
<0 (c) ∫ T
=0 (d) ∫ ds > 0
31. Ans. (b)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

72. For a thermodynamic cycle to be irreversible, it is necessary that [IES-1998]

δQ δQ δQ δQ
(a) ∫ =0 (b) ∫ <0 (c ) ∫ >0 (d ) ∫ ≥0
T T T T
72. Ans. (b)

dQ
27. When a system undergoes a process such that ∫ T
= 0 and Δs > 0 , the process is [IES-

m
1997]
(a) irreversible adiabatic (b) reversible adiabatic (c) isothermal (d) isobaric
dQ
27. Ans. (d) Since ∫ = 0 , process is reversible. Since Δs > 0 , process is constant pressure or

co
T
isobaric

73. For an irreversible cycle [IES-1994]

dQ dQ dQ dQ

.
(a) ∫ T
≤0 (b) ∫ T
>0
tas (c) ∫ T
<0 (d) ∫ T
≥0
73. Ans. (b)

79. For real thermodynamic cycle [IAS-2003]

dQ dQ dQ dQ
(a) ∫ > 0 but < ∞ (b) ∫ <0 (c) ∫ =0 (d) ∫ =∞
lda
T T T T
79. Ans. (b)

111. If a system undergoes an irreversible adiabatic process, then (symbols have usual
meanings) [IAS-1999]
dQ dQ
(a) ∫ = 0 and ΔS > 0 (b) ∫ = 0 and ΔS = 0
vi

T T
dQ dQ
(c) ∫ > 0 and ΔS = 0 (d) ∫ < 0 and ΔS < 0
T T
Ci

111. Ans. (a)

85. A cyclic heat engine receives 600 kJ of heat from a 1000 K source and rejects 450 kJ to a
dQ
w.

300 K sink. The quantity ∫ T

and efficiency of the engine are respectively [IAS-2001]

(a) 2.1 kJ/K and 70% (b) - 0.9 kJ/K and 25%
(c) + 0.9 kJ/K and 70% (d) - 2.1 kJ/K and 25%
dQ 600 (450)
=∫ − = −0.9kJ / K
ww

85. Ans. (b)

T 1000 300
Q − Q2 Q 450
η= 1 = 1− 2 = 1− = 0.25 = 25% [IAS-2001]
Q1 Q1 600

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Entropy Change in an Irreversible Process

75. Consider the following statements: [IES-1998]
In an irreversible process
1. entropy always increases.
2. the sum of the entropy of all the bodies taking part in a process always increases.
3. once created, entropy cannot be destroyed.
Of these statements
(a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct
75. Ans. (a)

m
28. Consider the following statements: [IES-1997]
When a perfect gas enclosed in a cylinder piston device executes a reversible adiabatic expansion

co
process,
1. its entropy will increase 2. its entropy change will be zero.
3. the entropy change of the surroundings will be zero.
Of these statements
(a) 1 and 3 are correct (b) 2 alone is correct (c) 2 and 3 are correct (d) 1 alone is correct

.
28. Ans. (c) In reversible adiabatic expansion, entropy change is zero and no change in entropy of
surroundings
tas
33. A system of 100 kg mass undergoes a process in which its specific entropy increases from
0.3 kJ/kg-K to 0.4 kJ/kg-K. At the same time, the entropy of the surroundings decreases from
80 kJ/K to 75 kJ/K.
lda
The process is: [IES-1997]
(a) Reversible and isothermal (b) Irreversible (c) Reversible (d) Impossible
33. Ans. (b) Entropy increase in process = 100 (0.4 - 0.3) = 10 kJ/kg [IES-1997]
Entropy change of surroundings = 5 kJ/K
Thus net entropy increases and the process is irreversible.
vi

75. Which one of the following temperature entropy diagrams of steam shows the reversible and
irreversible processes correctly? [IES-1996]
Ci
w.
ww

75. Ans. (c) In reversible process entropy change is zero and in four figures it is represented by
straight vertical line. However, in irreversible process, entropy increases in all processes
(expansion as well as compression).

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Applications of Entropy Principle

73. A Carnot engine operates between 27°C and 327oC. If the engine produces 300 kJ
of Work, What is the entropy change during heat addition? [IES-2005]
(a) 0.5 kJ/K (b) 1.0 kJ/K (c) 1.5 kJ/K (d) 2.0 kJ/K

m
73. Ans. (b)

( T1 − T2 ) ΔS = W

co
300
or ΔS = = 1 kJ / k
600 − 300

.
tas
10. The entropy of a mixture of ideal gases is the sum of the entropies of constituents
evaluated at: [IES-2005]
(a) Temperature and pressure of the mixture
lda
(b) Temperature of the mixture and the partial pressure of the constituents
(c) Temperature and volume of the mixture
(d) Pressure and volume of the mixture
10. Ans. (c)
vi

40. The heat added to a closed system during a reversible process is given by Q = α T + β T ,
2

where α and β are constants. The entropy change of the system as its temperature changes from
T1 to T2 is equal to [IES-2000]
⎡ β ⎤
(b) ⎢α (T2 − T1 ) + (T22 − T12 ) ⎥ / T1
Ci

(a ) α + β (T2 − T1 )
⎣ 2 ⎦
⎡α β ⎤ ⎛T ⎞
(c) ⎢ (T22 − T12 ) + (T23 − T13 ) ⎥ / T12 ( d ) α ln ⎜ 2 ⎟ + 2β (T2 − T1 )
⎣2 2 ⎦ ⎝ T1 ⎠
w.

40. Ans. (c)

72. Which one of the following statements is not correct?

(a) Change in entropy during a reversible adiabatic process is zero [IAS-2003]
ww

(b) Entropy increases with the addition of heat

(c) Throttling is a constant entropy expansion process
(d) Change in entropy when a gas is heated under constant pressure given by
T2
s2 − s1 = mC p log e
T1
72. Ans. (c)

98. Assertion (A): Entropy change for a reversible adiabatic process is zero.
Reason (R): There is no heat transfer in an adiabatic process.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Ans. (b) [IAS 1994]

41. One kg of air is subjected to the following processes: [IES-2004]

1. Air expands isothermally from 6 bar to 3 bar.
2. Air is compressed to half the volume at constant pressure
3. Heat is supplied to air at constant volume till the pressure becomes three fold
In which of the above processes, the change in entropy will be positive?
(a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3

m
. co
tas
37. A reversible heat engine receives 6 kJ of heat from thermal reservoir at temperature 800 K,
and 8 kJ of heat from another thermal reservoir at temperature 600 K. If it rejects heat to a third
thermal reservoir at temperature 100 K, then the thermal efficiency of the engine is approximately
equal to: [IES-2002]
(a) 65% (b) 75% (c) 80% (d) 85%
lda
37. Ans. (d)

119. A reversible engine exceeding 630 cycles per minute drawn heat from two constant
temperature reservoirs at 1200 K and 800 K rejects heat to constant temperature at 400 K. The
engine develops work 100kW and rejects 3200 KJ heat per minute. The ratio of heat drawn from
Q1200
vi

two reservoirs is nearly. [IES-1992]

Q800
(a) 1 (b) 1.5 (c) 3 (d) 10.5
119. Ans. (d)
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
. co
tas
38. In which one of the following situations the entropy change will be negative
(a) Air expands isothermally from 6 bars to 3 bars [IES-2000]
(b) Air is compressed to half the volume at constant pressure
(c) Heat is supplied to air at constant volume till the pressure becomes three folds
lda
(d) Air expands isentropic ally from 6 bars to 3 bars
38. Ans. (a)

2.13. A 1500 W electrical heater is used to heat 20 kg of water (Cp = 4186 J/kg K) in an insulated
bucket, from a temperature of 30°C to 80°C. If the heater temperature is only infinitesimally larger
than the water temperature during the process, the change in entropy for heater is….. J/k and for
vi

water ............. J/K. [GATE-1994]

3.13 Ans. - 11858 J/K, 12787 J/K.
Ci

Entropy Generation in a Closed System

92. 1600 kJ of energy is transferred from a heat reservoir at 800 K to another heat reservoir at
400 K. The amount of entropy generated during the process would be [IAS-2000]
(a) 6 kJ/k (b) 4 kJ/k (c) 2kJ/k (d) zero
w.

dQ dQ 1600 1600
92. Ans. (c) Entropy generated = dsat 400K − dsat 800K = − = − = 2kJ / K
400 800 400 800

21. An electric motor of 5 kW is subjected to a braking test for 1 hour. The heat generated by the
frictional forces in the process is transferred to the surroundings at 20°C. The resulting entropy
ww

change will be [IAS-1998]

(a) 22.1 kJ/K (b) 30.2 kJ/K (c) 61.4 kJ/K (d) 82.1 kJ/K
ΔQ 5 × 3600
21. Ans. (c) ΔS = = kJ / K = 61.4kJ / K
T 293

Data for Q. 85 - 86 are given below. Solve the problems and choose correct answers.
Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of
2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

condition, until the volume becomes 2m3. Heat exchange 42 occurs with the atmosphere at 298 K
during this process.
85. The work interaction for the Nitrogen gas is [GATE-2003]
(a) 200 kJ (b) 138.6 kJ (c) 2 kJ (d) - 200 kJ
⎛ υ2 ⎞ ⎛ υ2 ⎞ ⎛2⎞
85. Ans. (b) w 1− 2 = mRT In ⎜ ⎟ = pυ In ⎜ ⎟ = 200 × 1× In ⎜ ⎟ kJ = 138.6 kJ
⎝ υ1 ⎠ ⎝ υ1 ⎠ ⎝ 1⎠

86. The entropy change for the Universe during the process in kJ/K is [GATE-2003]
(a) 0.4652 (b) 0.0067 (c) 0 (d) - 0.6711

m
86. Ans. (c) It is reversible process so ( ΔS )universe = 0

co
Entropy Generation in an Open System

Reversible Adiabatic Work in a Steady Flow System

.
tas
Entropy and Direction: The Second Law a Directional law of Nature

64. A mass M of a fluid at temperature T1 is mixed with an equal mass of the same fluid at
temperature T2. The resultant change in entropy of the universe is [IES-1992]
(a) zero (b) negligible (c) always negative (d) always positive
64. Ans. (d)
vi lda

81. M1 kg of water at T1 is isobarically and adiabatically mixed with M2 kg of water at T2

(T1 > T2). The entropy change of the universe is [IAS-2004]
Ci

(a) Necessarily positive (b) Necessarily negative.

(c) Always zero (d) Negative or positive but not zero
81. Ans. (a)
w.

30. In which one of the following processes is there an increase in entropy with no degradation of
energy?
(a) Polytropic expansion (b) Isothermal expansion
(c) Isochoric heat addition (d) Isobaric heat addition [IAS-1996]
30. Ans. (b)
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

5. AVAILABILITY, IRREVERSIBILITY

Available Energy
42. What will be the loss of available energy associated with the transfer of 1000 kJ of

m
heat from constant temperature system at 600 K to another at 400 K when the
environment temperature is 300 K? [IAS-1995; IES-2004]
(a) 150 kJ (b) 250 kJ (c) 500 kJ (d) 700 kJ

co
42. Ans. (b) Loss of available energy = To × ( ΔS )univ. = 300 ⎧⎨
1000 1000 ⎫
− ⎬ kJ = 250kJ
⎩ 400 600 ⎭
60. An inventor claims that heat engine has the following specifications:
Power developed = 50 kW; [IAS-2002]
Fuel burned per hour = 3 kg,

.
Heating value of fuel =75,000 kJ per kg
Temperature limits = 627°C and 27oC
Cost of fuel =Rs. 30/kg,
Value of power = Rs. 5/kWh,
tas
(a) possible (b) not possible (c) economical (d) uneconomical
T2 300 2
60. Ans. (b) Maximum possible efficiency (ηmax) = 1 − = 1− =
lda
T1 900 3
Maximum possible Power output with this machine
3 × 75000 2
(Wmax) = Q ×ηmax = × kW 41. 67 KW
3600 3
So above demand is impossible.
vi

94. For a reversible power cycle, the operating temperature limits are 800 K and 300 K. It takes
400kJ of heat. The unavailable work will be [IAS-1997]
(a) 250 kJ (b) 150 kJ (c) 120 kJ (d) 100 kJ
Ci

⎛ T2 ⎞
94. Ans. (b) Available part of the heat (WE) = Q ⎜ 1 − ⎟ = 400 ⎛⎜ 1 −
300 ⎞
=250 kJ
⎝ T1 ⎠ ⎝ 800 ⎟⎠
Unavailable work (Wu) = 400 – 250 = 150 kJ
w.

Available Energy Referred to a Cycle

54. A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during
ww

this process is to be used as a source of energy. The ambient temperature is 303 K and specific
heat of steel is 0.5 kJ/kg K. The available energy of this billet is [GATE-2004]
(a) 490.44 MJ (b) 30.95 MJ (c) 10.35 MJ (d) 0.10 MJ
T2
⎛ To ⎞
T2
⎡ ⎛ T2 ⎞ ⎤
54. Ans. (a) A.E = ∫ mc p ⎜ 1 − ⎟ dT = ∫ mc p ⎢( T2 − T1 ) − To ln ⎜ ⎟ ⎥
T1 ⎝ T⎠ T1 ⎢⎣ ⎝ T1 ⎠ ⎥⎦
⎡ ⎛ 1250 ⎞ ⎤
= 2000 × 0.5 ⎢(1250 − 450 ) − 303ln ⎜ ⎟ ⎥ = 490MJ
⎣ ⎝ 450 ⎠ ⎦

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

83. A heat source H1 can supply 6000kJ/min. at 300°C and another heat source H2 can
supply 60000 kJ/min. at 100oC. Which one of the following statements is correct if the
surroundings are at 27°C? [IES-2006]
(a) Both the heat sources have the same efficiency
(b) The first heat source has lower efficiency
(c) The second heat source has lower efficiency
(d) The first heat source produces higher power
Tource

m
83. Ans. (c) η = 1 − ∴η1 >η2
Tsurroundings

co
Quality of Energy
42. Increase in entropy of a system represents [IAS 1994]
(a) increase in availability of energy (b) increase in temperature
(c) decrease in pressure (d) degradation of energy

.
Ans. (d)

Maximum Work in a Reversible Process

tas
Reversible Work by an Open System Exchanging Heat only with the
lda
Surroundings

Useful Work

Dead State
vi

Availability
1.5 Availability of a system at any given state is [GATE-2000]
(a) a property of the system
Ci

(b) the maximum work obtainable as the system goes to dead state
(c) the total energy of the system
(d) the maximum useful work obtainable as the system goes to dead state
1.5 Ans. (d) maximum useful work, i.e. total work minus pdv work. Not maximum work.
w.

62. Assertion (A): The change in availability of a system is equal to the change in the
Gibbs function of the system at constant temperature and pressure. [IES-2006]
Reason (R): The Gibbs function is useful when evaluating the availability of systems in
which chemical reactions occur.
ww

62. Ans. (a)

72. For a steady flow process from state 1 to 2, enthalpy changes from h1 = 400 kJ/kg to
h2 = 100 kJ/kg and entropy changes from s1 = 1.1 kJ/kg-K to s2 = 0.7 kJ/kg-K.
Surrounding environmental temperature is 800 K. Neglect changes in kinetic and
potential energy. The change in availability of the system is [IES-2003]
(a) 420 kJ/kg (b) 300 kJ/kg (c) 180 kJ/kg (d) 90 kJ/kg
72. Ans. (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

U.E. = To (s1 - s 2 )
= 300 × (1.1- 0.7) = 120 kJ/kg
Change in availability = (h1 - h 2 ) - (U.E.)
= (400 - 100) - 120 = 180 kJ/kg

35. Availability function for a closed system is expressed as: [IES-2002]

(a) φ = u + po v − To S (b) φ = du + po dv − To ds

m
(c) φ = du + po dv + To ds (d) φ = u + po v + To S
35. Ans. (a)

co
22. Consider the following statements: [IES-2001]
1. Availability is the maximum theoretical work obtainable.
dTs ⎡ hg − h f ⎤
2. Clapeyron's equation for dry saturated steam is given by (V g −Vf ) = ⎢
dQ ⎣ Ts ⎦
⎥

.
3. A gas can have any temperature at a given pressure unlike a vapour which has a fixed
tas
temperature at a given pressure.
⎡ ∂s ⎤
4. Joule Thomson coefficient is expressed as μ=⎢ ⎥
⎣ ∂p ⎦ h
Of these statements
lda
(a) 1, 2 and 3 are correct (b) 1, 3 and 4 are correct
(c) 2 and 3 are correct (d) 1, 2 and 4 are correct
22. Ans. (a)

37. 10kg of water is heated from 300 K to 350 K in an insulated tank due to churning action by a
vi

stirrer. The ambient temperature is 300 K. In this context, match List I and List II and select the
correct answer using the codes given below the Lists: [IES-2000]
List I List II
A. Enthalpy change 1. 12.2 kJ/kg
Ci

B. Entropy change/kg 2. 1968 kJ

C. Availability/kg 3. 2090 kJ
D. Loss of availability 4. 656 J/kg-k
Codes: A B C D A B C D
(a) 3 1 4 2 (b) 2 4 1 3
w.

(c) 3 4 1 2 (d) 2 1 4 3
37. Ans. (c)

73. Neglecting changes in kinetic energy and potential energy, for unit mass the availability in a
non-flow process becomes a = ɸ - ɸo, where ɸ is the availability function of the [IES-1998]
ww

(a) open system (b) closed system (c) isolated system (d) steady flow process
73. Ans. (a)

76. Consider the following statements: [IES-1996]

1. Availability is generally conserved 2. Availability can either be negative or positive.
3. Availability is the maximum theoretical work obtainable.
4. Availability can be destroyed in irreversibility.
Of these correct statements are:
(a) 3 and 4 (b) 1 and 2 (c) 1 and 3 (d) 2 and 4

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

76. Ans. (a) Availability is the maximum theoretical work obtainable and it can be destroyed in
irreversibility.

73. If u, T, v, s, hand p refer to internal energy, temperature, volume, entropy, enthalpy and
pressure respectively; and subscript 0 refers to environmental conditions, availability function for
a closed system is given by [IAS-2003]
(a) u+Po v-To s (b) u- Po v+ To s (c) h + Po v – Tos (d) h - Po v + To s
73. Ans. (a)

m
22. Match List - I with List - II and select the correct answer using the codes given below the lists:
List - I List - II
A. Irreversibility 1. Mechanical equivalent
B. Joule Thomson experiment 2. Thermodynamic temperature scale

co
C. Joule's experiment 3. Throttling process
D. Reversible engines 4. Loss of availability

Codes: A B C D A B C D
(a) 1 2 3 4 (b) 1 2 4 3

.
(c) 4 3 2 1 (d) 4 3 1 2
22. Ans. (d) tas
5.10 A heat reservoir at 900 K is brought into contact with the ambient at 300 K for a short time.
During this period 9000 kJ of heat is lost by the heat reservoir. The total loss in availability due to
this process is [GATE-1995]
(a) 18000 kJ (b) 9000 kJ (c) 6000 kJ (d) None of the above
lda
5.10 Ans. (d) The availability of a thermal reservoir is equivalent to the work output of a Carnot
heat engine operating between the reservoir and the environment. Here as there is no change in
the temperatures of source (reservoir) or the sink (atmosphere), the initial and final availabilities
are same, Hence there is no loss in availability.
vi

Irreversibility
77. The irreversibility is defined as the difference of the maximum useful work and actual
Ci

work: I = Wmax,useful- Wactual. How can this be alternatively expressed? [IES-2005]

(a) I = To (ΔS system + ΔS surrounding ) (b) I = To (ΔS system − ΔS surrounding )
(c) I = To ( ΔS system + ΔS surrounding ) (d) I = To ( ΔS system − ΔS surrounding )
I = To × ( ΔS )universe = To × ⎡⎣ ΔSsystem + ΔSsurrounding ⎤⎦
w.

77. Ans. (a)

63. Assertion (A): All constant entropy processes are adiabatic, but all adiabatic
processes are not isentropic. [IES-2006]
ww

Reason (R): An adiabatic process which resists the exchange of energy to the
surroundings may have irreversibility due to friction and heat conduction.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

63. Ans. (d) A is false, For a process

due to irreversibility entropy will
increase and actual process may be 1-
2` but due to heat loss to the
surroundings, may 2` coincide with 2
but the process not adiabatic. So all
isentropic process is not adiabatic.

m
co
4. Which of the following statement is incorrect? [IES-1992]
(a) The greater the pressure difference in throttling the lesser the irreversibility
(b) The primary source of internal irreversibility in power is fluid friction in rotary machines.
(c) The greater the irreversibility, the greater the increase in adiabatic process

.
(d) The entropy of the universe is continually on the increase.
4. Ans. (a) tas
1. The loss due to irreversibility in the
expansion valve of a refrigeration cycle
shown in the given figure is represented by
lda
the area under the line·
(a) GB
(b) AG
(c) AH
(d) BH
vi

[IAS-1999]
1. Ans. (d) Entropy will increase in the process AH is BH.
Therefore Irreversibility (I) = To × ΔS i.e. area under the line BH.
Ci

29. Assertion (A): When a gas is forced steadily through an insulated pipe containing a porous
plug, the enthalpy of gas is the same on both sides of the plug. [IAS-1997]
Reason (R): The gas undergoes an isentropic expansion through the porous plug.
29. Ans. (c) Expansion through the porous plug is adiabatic as no heat added or rejected to the
w.

system. It is not reversible, due to large irreversibility entropy increases so it is not an isentropic
process.

Second Law efficiency

ww

69. Assertion (A): The first-law efficiency evaluates the energy quantity utilization, whereas the
second-law efficiency evaluates the energy quality utilization. [IAS-1998]
Reason (R): The second-law efficiency for a process is defined as the ratio of change of
available energy of the source to the change of available energy of the system.
mimimum energy int ake to perform the given task
69. Ans. (c) ηII =
actual energy int ake to perform the same task

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

6. TdS RELATIONS, CLAPERYRON AND REAL GAS

EQUATIONS
Highlight
2
5. Adiabatic index (γ) =1+ where N is degrees of freedom of molecules

m
N
N=3 for monatomic gas
N=5 for diatomic gas
N=6 for try atomic gas

co
Some Mathematical Theorems
43. Given:
p = pressure, [IES-1993]

.
T = Temperature,
v = specific volume,
tas
Which one of the following can be considered as property of a system?
⎛ dT p.dv ⎞ ⎛ dT v.dp ⎞
(a ) ∫ pdv (b) ∫ vdp (c ) ∫ ⎜ + ⎟ (d ) ∫ ⎜ − ⎟
⎝ T v ⎠ ⎝ T T ⎠
43. Ans. (d) P is a function of v and both are connected by a line path on p and v coordinates.
lda
Thus ∫ pdv and ∫ vdp are not exact differentials and thus not properties.
If X and Y are two properties of a system, then dx and dy are exact differentials. If the differential
⎡ ∂M ⎤ ⎡ ∂N ⎤
is of the form Mdx + Ndy, then the test for exactness is ⎢ ∂y ⎥ = ⎢ ∂x ⎥
⎣ ⎦x ⎣ ⎦ y
vi

Now applying above test for

⎛ dT p.dv ⎞ ⎡ ∂ (1/ T ) ⎤ ⎡ ∂ ( p / v) ⎤ ⎡ ∂ ( RT / v ) ⎤
2
R
∫ ⎜⎝ T v ⎟⎠ ⎢⎣ ∂v ⎥⎦T ⎢⎣ ∂T ⎥⎦ v ⎢⎣ ∂T ⎥⎦ or 0 = v 2
+ , = =
v
Ci

⎛ dT p.dv ⎞
This differential is not exact and hence is not a point function and hence ∫ ⎜⎝ T +
v ⎠
⎟ is not
a point function and hence not a property.
v.dp ⎞ ⎡ ∂ (1/ T ) ⎤
⎛ dT ⎡ ∂ ( −v / T ) ⎤ ⎡ ∂ (− R / P) ⎤
w.

And for ∫ ⎜⎝ T −
⎟⎢ ⎥ =⎢
T ⎠ ⎣ ∂p ⎦T ⎣ ∂T ⎦ P ⎣ ∂T ⎥ =⎢ ⎥⎦ or 0 = 0
P

⎛ dT v.dp ⎞
Thus ∫ ⎜ − ⎟ is exact and may be written as ds, where s is a point function and hence a
⎝ T T ⎠
ww

property

Maxwell's Equations
55. Which thermodynamic property is evaluated with the help of Maxwell equations from
the data of other measurable properties of a system?
(a) Enthalpy (b) Entropy (c) Latent heat (d) Specific heat [IES 2007]
55. Ans. (a) From Maxwell relation Clapeyron equation comes.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

87. Consider the following statements pertaining to the Clapeyron equation:

1. It is useful to estimate properties like enthalpy from other measurable properties.
2. At a change of phase, it can be used to find the latent heat at a given pressure.
⎛ ∂p ⎞ ⎛ ∂s ⎞
3. It is derived from the relationship ⎜ ⎟ = ⎜ ⎟ [IES-2006]
⎝ ∂v ⎠T ⎝ ∂T ⎠V
Which of the statements given above are correct?
(a) 1,2 and 3 (b) Only 1 and 2 (c) Only 1 and 3 (d) Only 2 and 3

m
87. Ans. (b) 3 is false. It is derived from the Maxwell’s 3rd relationship
⎛ ∂p ⎞ ⎛ ∂s ⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂T ⎠v ⎝ ∂v ⎠T

co
89. According to the Maxwell relation, which of the following is/are correct?
⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞
(a) ⎜ ⎟ = −⎜ ⎟ (b) ⎜ ⎟ = ⎜ ⎟ [IAS-2007]
⎝ ∂T ⎠ p ⎝ ∂P ⎠T ⎝ ∂v ⎠T ⎝ ∂T ⎠v
⎛ ∂P ⎞ ⎛ ∂s ⎞

.
(c) ⎜ ⎟ =⎜ ⎟ (d) All of the above
⎝ ∂T ⎠v ⎝ ∂v ⎠T tas
⎛ ∂P ⎞ ⎛ ∂S ⎞
89. Ans. (c) ⎜ ⎟ =⎜ ⎟ To memorize Maxwell’s relation remember T V P S, -ive
⎝ ∂T ⎠V ⎝ ∂V ⎠T
and S S V P see highlights.
lda

TdS Equations
36. T ds equation can be expressed as [IES-2002]
T β dv Tdv
vi

(a) Tds = Cv dt + (b) Tds = Cv dt +

k k
Tk Tβ
(c) Tds = Cv dt + dv (d) Tds = Cv dt + dp
β k
Ci

36. Ans. (a)

56. Which of the following relationships is valid only for reversible processes undergone by a
closed system of simple compressible substance (neglect changes in kinetic and potential
w.

energy? [GATE-2007]
(a) δQ = dU + δW (b) TdS = dU + pdV (c) TdS = dU + δW (d) δQ = dU + pdV
56. Ans. (d)

49. Considering the relationship TdS = dU + pdV between the entropy (S), internal energy (U),
ww

pressure (p), temperature (T) and volume (V), which of the following statements is correct?
[GATE-2003]
(a) It is applicable only for a reversible process (b) For an irreversible process, TdS > dU + pdV
(c) It is valid only for an ideal gas (d) It is equivalent to 1 law, for a reversible process
49. Ans. (d)

73. Which one of the following statements applicable to a perfect gas will also be true for an
irreversible process? (Symbols have the usual meanings). [IES-1996]
(a) dQ = du + pdV (b) dQ = Tds (c) Tds = du + pdV (d) None of the above

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

73. Ans. (c) The relations in (a) and (b) are applicable for a reversible processes and (c) Tds = du
+ pdV is a relation among properties which are independent of the path.

15. Which one of the following expressions for T ds is true for a simple compressible substance?
(Notations have the usual meaning) [IAS-1998]
(a) dh – vdp (b) dh + vdp (c) dh - pdv (d) dh +pdv
15. Ans. (a) dQ = dh - Vdp or Tds = dh - Vdp

m
39. Consider the following thermodynamic relations: [IES-2000]
1.Tds = du + pdv 2.Tds = du − pdv
3.Tds = dh + vdp 4.Tds = dh − vdp

co
Which of these thermodynamic relations are correct?
(a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4
39. Ans. (b)

.
(a) temperature (b) pressure
tas
Difference in Heat Capacities and Ratio of Heat Capacities
1.12 The specific heats of an ideal gas depend on its
(c) volume
[GATE-1996]
(d) molecular weight and structure
1.12 Ans. (a)
lda
70. Match List-l (Terms) with List-II (Relations) and select the correct answer using the
codes given below the Lists: [IES-2003]
List I List II
(Terms) (Relations)
1 ⎛ ∂v ⎞
vi

A. Specific heat at constant volume Cv 1. ⎜ ⎟

v ⎝ ∂T ⎠ p
⎛ ∂p ⎞ ⎛ ∂v ⎞
B. Isothermal compressibility kT 2. T ⎜ ⎟ ⎜ ⎟
Ci

⎝ ∂T ⎠v ⎝ ∂T ⎠ p
⎛ ∂s ⎞
C. Volume expansivity β 3. T ⎜ ⎟
⎝ ∂T ⎠v
w.

D. Difference between specific heats at

1 ⎛ ∂v ⎞
constant pressure and at constant Cp-Cv 4. − ⎜ ⎟
v ⎝ ∂p ⎠T
Codes:
ww

A B C D A B C D
(a) 3 4 2 1 (b) 4 1 3 2
(c) 3 4 1 2 (d) 4 1 2 3
70. Ans. (c)

82. Assertion (A): Specific heat at constant pressure for an ideal gas is always greater than the
specific heat at constant volume. [IES-2002]
Reason (R): Heat added at constant volume is not utilized for doing any external work.
82. Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

36. An insulated box containing 0.5 kg of a gas having Cv = 0.98 kJ/kgK falls from a
balloon 4 km above the earth's surface. What will be the temperature rise of the gas when
the box hits the ground? [IES-2004]
(a) 0 K (b) 20 K (c) 40 K (d) 60 K
36. Ans. (c) Potential energy will converted to heat energy.
gh 980 × 4000
mgh = mc v ΔT or ΔT = = = 40K
cv 980

m
83. As compared to air standard cycle, in actual working, the effect of variations in specific heats
is to [IES-1994]

co
(a) increase maximum pressure and maximum temperature.
(b) reduce maximum pressure and maximum temperature.
(c) increase maximum pressure and decrease maximum temperature.
(d) decrease maximum pressure and increase maximum temperature.
83. Ans. (b)

.
(a)
⎛ ∂v ⎞
T⎜ ⎟ (b)
⎛ ∂T ⎞
T⎜ ⎟
tas
95. The specific heat Cp is given by [IAS-2000]

(c)
⎛ ∂s ⎞
T⎜ ⎟ (d)
⎛ ∂T ⎞
T⎜ ⎟
⎝ ∂T ⎠ p ⎝ ∂s ⎠ p ⎝ ∂T ⎠ p ⎝ ∂v ⎠ p
dQp ⎛ ∂s ⎞
95. Ans. (c) Cp = = T⎜ ⎟ [∵ dQ = TdS]
lda
∂T ⎝ ∂T ⎠p

29. The number of degrees of freedom for a diatomic molecule [IES-1992]

(a) 2 (b) 3 (c) 4 (d) 5
29. Ans. (d)
vi

A diatomic gas (such as that of oxygen) has six degrees of freedom in all-three corresponding to translator
motion, two corresponding to rotatory motion and one corresponding to vibratory motion. Experiments have
shown that at ordinary temperatures, the vibratory motion does not occur. Hence, at 27°C, an oxygen molecule
has just five degrees of freedom.
Ci

Cp
32. The ratio for a gas with n degrees of freedom is equal to: [IES-1992]
Cv
2 2
w.

(a) n + I (b) n – I (c) −1 (d) 1 +

n n
32. Ans. (d)
ww

12. The specific heats of an ideal gas depends on its [GATE-1996]

(a) Temperature (b) pressure (c) volume (d) molecular weight and structure.
Ans. (d)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

⎡ ⎛ ∂s ⎞ ⎛ ∂s ⎞ ⎤
1.9 For an ideal gas the expression ⎢T ⎜ ⎟ −T ⎜ ⎟ ⎥ is always equal to
⎢⎣ ⎝ ∂T ⎠ p ⎝ ∂T ⎠v ⎥⎦
[IAS-2003, GATE-1997]
cp
(a ) zero (b) (c ) R (d ) RT
cv
1.9 Ans. (c)
⎛ ∂S ⎞ ⎛ T∂S ⎞

m
⎛ dQ ⎞
T⎜ ⎟ =⎜ ⎟ =⎜ ⎟ = CP
⎝ ∂T ⎠P ⎝ ∂T ⎠P ⎝ ∂T ⎠P
⎛ ∂S ⎞ ⎛ T∂S ⎞ ⎛ dQ ⎞
T⎜ ⎟ =⎜ ⎟ =⎜ ⎟ = CV
⎝ ∂T ⎠ V ⎝ ∂T ⎠ V ⎝ ∂T ⎠ V

co
⎛ ∂S ⎞ ⎛ ∂S ⎞
∴T⎜ ⎟ − T ⎜ ∂T ⎟ = CP − CV = R
⎝ ∂T ⎠P ⎝ ⎠V

.
14. Assertion (A): Specific heat at constant pressure for an ideal gas is always greater
tas
than the specific heat at constant volume.
Reason (R): Heat added at constant volume is not utilized for doing any external
work. [IAS-2000, IES-2002]
Ans. (a) Both A and R correct and R is the correct explanation of A
lda
16. Assertion (A): Ratio of specific heats C p decreases with increase in temperature.
C v

Reason (R): With increase in temperature, C p decreases at a higher rate than C v.

[IES-1996]
vi

Ans. (C): A is correct but R is false.

We know that C p =a+KT+K1T2+K2T3
C v =b+ KT+K1T2+K2T3
See Cp and C v both increase with temperature and by same amount. As Cp>C v then
Ci

percentage increase of Cp is less than C v. So C p decreases with temperature.

Cv

17. It can be shown that for a simple compressible substance, the relationship
w.

2
⎛ ∂V ⎞ ⎛ ∂P ⎞
Cp-C v= -T ⎜ ⎟ ⎜ ⎟ exists. Where C p and C v are specific heats at
⎝ ∂T ⎠ p ⎝ ∂T ⎠ T
constant pressure and constant volume respectively. T is the temperature V is volume
ww

and P is pressure. [IES-1998]

Which one of the following statements is NOT true?
(a) C p is always greater than C v.
(b) The right side of the equation reduces to R for ideal gas.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

2
⎛ ∂P ⎞ ⎛ ∂V ⎞
(c) Since ⎜ ⎟ can be either positive or negative, and ⎜ ⎟ must be positive, T
⎝ ∂T ⎠ T ⎝ ∂T ⎠ p
⎛ ∂P ⎞
must have a sign that is opposite to that of ⎜ ⎟
⎝ ∂T ⎠ T
(d) Is very nearly equal to for liquid water.
⎛ ∂P ⎞
Ans. (c) Sign of T must be positive. ⎜ ⎟ is always negative.

m
⎝ ∂T ⎠ T

18 Match List-I with List-II and select the correct answers using the codes given below
the lists.

co
[IAS-2002]
List-I List-II
A. Joule Thomson co-efficient 1. 5 R
2

.
B. Cp for monatomic gas 2. Cv
C. Cp - Cv for diatomic gas
D. ⎛ ∂U ⎞
⎜ ⎟
tas 3. R
4. ⎛⎜ ∂T ⎞⎟
⎝ ∂T ⎠v ⎝ ∂P ⎠ h
lda
Codes: A B C D A B C D
(a) 3 2 4 1 (b) 4 1 3 2
(c) 3 1 4 2 (d) 4 2 3 1
Ans. (b) Cp-Cv for all ideal gas is R, So C-3, (a) & (c) out. A automatically match 4,and
vi

Cp= γ R for monatomic gas γ= 5 so =5R

γ −1 3 2
Ci

108. Ratio of specific heats for an ideal gas is given by (symbols have the usual
meanings) [IAS-1999]
1 1 1 1
(a) (b) (c) (d)
R C C R
1− 1− p 1+ p 1+
w.

Cp R R Cp
Cp Cp 1
108. Ans. (a) Cp − Cv = R and γ = = =
Cv Cp − R R
1−
Cp
ww

14. A 2 kW, 40 litre water heater is switched on for 20 minutes. The heat capacity Cp for water is
4.2 kJ/kg K. Assuming all the electrical energy has gone into heating the water, increase of the
water temperature in degree centigrade is [GATE-2003]
(a) 2.7 (b) 4.0 (c) 14.3 (d) 25.25
14. Ans. (c)
Heat absorbed by water = Heat supplied by heater
m w c pw ( ΔT )w = P × t or 40 × 4.2 × ( ΔT ) w = 2 × 20 × 60 or ( ΔT )w = 14.3o C

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Energy Equation

Joule-Kelvin Effect or Joule-Thomson coefficient

44. Joule-Thomson coefficient is defined as [IES-1995]
⎛ ∂T ⎞ ⎛ ∂h ⎞ ⎛ ∂h ⎞ ⎛ ∂p ⎞
(a) ⎜ ⎟ (b) ⎜ ⎟ (c) ⎜ ⎟ (d) ⎜ ⎟
⎝ ∂p ⎠ h ⎝ ∂p ⎠T ⎝ ∂T ⎠ p ⎝ ∂T ⎠h

m
44. Ans. (a)

58. Which combination of the following statements is correct? [GATE-2007]

co
P: A gas cools upon expansion only when its Joule-Thomson coefficient is positive in the
temperature range of expansion.
Q: For a system undergoing a process, its entropy remains constant only when the process is
reversible.
R: The work done by a closed system in an adiabatic process is a point function.

.
S: A liquid expands upon freezing when the slop of its fusion curve on Pressure Temperature
diagram is negative.
(a) R and S
58. Ans. (b)
(b) P and Q
tas (c) Q, R and S (d) P, Q and R

1.19 A positive value to Joule-Thomson coefficient of a fluid means [GATE-2002]

(a) temperature drops during throttling (b) temperature remains constant during throttling
lda
(c) temperature rises during throttling (d) none of these
⎛ ∂T ⎞
1.19 Ans. (a) μ = ⎜ ⎟ i,e. μ > o, ∂P is ( −ive ) so ∂T must be − ive.
⎝ ∂P ⎠h
1.14 A gas having a negative Joule-Thompson coefficient (µ< 0), when throttled, will
(a) become cooler (b) become warmer [GATE-2001]
vi

(c) remain at the same temperature (d) either be cooler or warmer depending on the type of gas
∂T ∂T
1.14 Ans. (b) Joule-Thomson co-efficient ⎛⎜ ⎞⎟ Here ∂p, − ive and ⎛⎜ ⎞⎟ , -ive so ∂T
⎝ ∂P ⎠h ⎝ ∂P ⎠h
must be +ive so gas will be warmer
Ci

4.3 Match 4 correct pairs between list I and List II for the questions
For a perfect gas:
List I List II [GATE-1994]
w.

(a) Isobaric thermal expansion coefficient 1. 0

(b) Isothermal compressibility 2. ∞
(c) Isentropic compressibility 3. 1/v
(d) Joule - Thomson coefficient 4. 1/T
5. 1/p
6. 1/ γ p
ww

4.3 Ans. (a) – 4, (b) - 5, (c) - 6, (d) – 1

26. The throttling of certain gasses may be used for getting the refrigeration effect. What
is the value of Joule – Thomson coefficient (µ) for such a throttling process?
(a) µ = 0 (b) µ = 1
(c) µ < 1 (d) µ > 1 [IES 2007]
Ans. (d) Actually Joule-Thomson coefficient will be positive.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

81. Which one of the following is correct?

When a real gas undergoes Joule-Thomson expansion, the temperature
(a) may remain constant. (b) always increases.
(c) may increase or decrease. (d) always decreases. [IES 2007]
Ans. (c) For ideal gas μ = 0 and for real gas μ may be positive (N2, O2, CO2 etc.) or
negative (H2)

36. Assertion (A): Throttling process for real gases at initial temperature higher than

m
maximum inversion temperature is accompanied by decrease in temperature of the gas.
Reason (R): Joule-Kelvin coefficient μj is given ( ∂T / ∂p )h and should have a positive
value for decrease in temperature during throttling process. [IES-2003]

co
36. Ans. (a)
18. Match List-I (Name of entity) with List-II (Definition) and select the correct answer using the
codes given below the lists:

.
List-I (Name of entity) List-II (Definition) [IES-2001]
tas 1 ⎛ ∂v ⎞
A. Compressibility factor 1. − ⎜ ⎟
v ⎝ ∂T ⎠ p
⎛ ∂h ⎞
B. Joule- Thomson coefficient 2. ⎜ ⎟
⎝ ∂T ⎠ p
lda
⎛ ∂T ⎞
C. Constant pressure specific heat 3. ⎜ ⎟
⎝ ∂p ⎠h
⎛ pv ⎞
D. Isothermal compressibility 4. ⎜ ⎟
⎝ RT ⎠
vi

Codes: A B C D A B C D
(a) 2 1 4 3 (b) 4 3 2 1
(c) 2 3 4 1 (d) 4 1 2 3
18. Ans. (b)
Ci

54. Joule-Thomson coefficient is the ratio of [IES-1999]

(a) pressure change to temperature change occurring when a gas undergoes the process of
adiabatic throttling
(b) temperature change to pressure change occurring when a gas undergoes the process of
w.

adiabatic throttling
(c) temperature change to pressure change occurring when a gas undergoes the process of
adiabatic compression
(d) pressure change to temperature change occurring when a gas undergoes the process of
adiabatic compression
ww

54. Ans. (b) Joule Thomson coefficient is the ratio of temperature change to pressure change
when a gas undergoes adiabatic throttling.

78. The Joule-Thomson coefficient is the [IES-1996]

⎛ ∂T ⎞ ⎛ ∂T ⎞
(a) ⎜ ⎟ of pressure-temp curve of real gases (b) ⎜ ⎟ of temp.-entropy curve of real gases
⎝ ∂p ⎠h ⎝ ∂s ⎠v
⎛ ∂h ⎞
(c) ⎜ ⎟ of enthalpy-entropy curve of real gases
⎝ ∂s ⎠T

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

⎛ ∂V ⎞
(d) ⎜ ⎟ of pressure-volume curve of real gases.
⎝ ∂T ⎠ p
78. Ans. (a) The slope of the isenthalpic curve at any point is know!) as Joule-Thomson
⎛ ∂T ⎞
coefficient and is expressed as μ =⎜ ⎟
⎝ ∂p ⎠h

69. Match the following

m
List I List II
A. Work 1. Point function
B. Heat 2. ∫ Tds

co
⎛ ∂u ⎞
C. Internal energy 3. ⎜ ⎟
⎝ ∂T ⎠h
D. Joule Thomson Coefficient 4. ∫ pdv [IES-1992]

.
Code: A B C D A B C D
(a)
(c)
4
4
69. Ans. (a)
2
1
1
2
3
3
tas (b)
(d)
1
2
2
1
4
4
3
3

85. Which one of the following properties remains unchanged for a real gas during Joule-
Thomson process? [IAS-2000]
lda
(a) Temperature (b) Enthalpy (c) Entropy (d) Pressure
85. Ans. (b)

Clausius-Clapeyron Equation
53. Consider the following statements in respect of the Clausius – Clapeyron equation:
1. It points to one possible way of measuring thermodynamic temperature.
vi

2. It permits latent heat of vaporization to be estimated from measurements of specific

volumes of saturated liquid, saturated vapour and the saturation temperatures at two
nearby pressures.
Ci

3. It does not apply to changes from solid to the liquid phase and from solid to the
Vapour phase.
Which of the statements given above are correct?
(a) 1,2 and 3 (b) 1 and 2 only
w.

(c) 2 and 3 only (d) 1 and 3 only [IES 2007]

Ans. (b)

32. The equation relating the following measurable properties: [IES-2005]

ww

(i) the slope of saturation pressure temperature line

(ii) the latent heat, and
(iii) the change in volume during phase transformation; is known as :
(a) Maxwell relation (b) Joules equation
(c) Clapeyron equation (d) None of the above
32. Ans. (c)

43. The variation of saturation pressure with saturation temperature for a liquid is 0.1
bar/K at 400 K. The specific volume of saturated liquid and dry saturated vapour at 400

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

K are 0.251 and 0.001 m3/kg What will be the value of latent heat of vaporization using
Clausius Clapeyron equation ? [IES-2004]
(a) 16000 kJ/kg (b) 1600 kJ/kg (c) 1000 kJ/kg (d) 160 kJ/kg
hfg
43. Ans. (c) ⎛⎜ ⎞⎟
dP
=
⎝ dT ⎠sat T ( Vg − Vf )
⎛ dP ⎞
or hfg = T ( Vg − Vf ) × ⎜ ⎟ = 400 × ( 0251 − 0.001) × 0.1× 10 J / kg = 1000kJ / kg
5

⎝ dT ⎠sat

m
74. If h, p, T and v refer to enthalpy, pressure, temperature and specific volume respectively and
subscripts g and f refer to saturation conditions of vapour and liquid respectively then Clausius-
Clapeyron equation applied to change of phase from liquid to vapour states is
dp ( hg − h f ) dp (hg − h f )
= =

co
(a) (b) [IAS-2003, IES-1996, 2006]
dt (vg − v f ) dt T (vg − v f )
dp (hg − h f ) dp (vg − v f )T
(c) = (d) =
dt T dt (hg − h f )

.
74. Ans. (b)

change of phase of a pure substance?

tas
39. Which one of the following functions represents the Clapeyron equation pertaining to the
[IES-2002]
(a) f (T, p, hfg) (b) f (T, p, hfg, vfg) (c) f (T, p, hfg, sfg) (d) f (T, p, hfg, sfg, vfg)
39. Ans. (b)
lda
43. The Clapeyron equation with usual notations is given by [IES-2000]
⎛ dT ⎞ h fg ⎛ dP ⎞ h fg ⎛ dT ⎞ Th fg ⎛ dP ⎞ Th fg
(a) ⎜ ⎟ = (b) ⎜ ⎟ = (c ) ⎜ ⎟ = (d ) ⎜ ⎟ =
⎝ dP ⎠ sat Tv fg ⎝ dT ⎠ sat Tv fg ⎝ dP ⎠ sat v fg ⎝ dT ⎠ sat v fg
43. Ans. (b)
vi

59. Clausius-Clapeyron equation gives the 'slope' of a curve in [IES-1999]

(a) p-v diagram (b) p-h diagram (c) p – T diagram (d) T-S diagram
59. Ans. (c)
Ci

34. The thermodynamic parameters are: [IES-1997]

I. Temperature II. Specific Volume III. Pressure IV. Enthalpy V. Entropy
The Clapeyron Equation of state provides relationship between:
(a) I and II (b) II, III and V (c) III, IV and V (d) I, II, III and TV
w.

34. Ans. (d) Clapeyron equation state provides relationship between temperature, specific
volume, pressure and enthalpy.

52. Which one of the following is the correct statement? [IAS-2007]

ww

Clapeyron equation is used for

(a) finding specific volume of vapour (b) finding specific volume of liquid
(c) finding latent heat of vaporization (c) finding sensible heat
52. Ans. (c)

118. Assertion (A): Water will freeze at a higher temperature if the pressure is increased.
Reason (R): Water expands on freezing which by Clapeyron's equation gives negative slope for
the melting curve. [IAS-2003]
118. Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

36. Match List I with List II and select the correct answer using the codes given below the lists
List I List II [IAS 1994]
A. Mechanical work 1. Clausius-Clapeyron equation
dQ
B. ∫ T
≤0 2. Gibb's equation

C. Zeroth Law 3. High grade energy

D. H-TS 4. Concept of temperature

m
Codes: A B C D A B C D
(a) 1 3 2 4 (b) 3 - 2 4
(c) - 2 3 1 (d) 3 - 4 2
Ans. (d)

co
Mixtures of Variable Composition

.
tas
Conditions of Equilibrium of a Heterogeneous System

Gibbs Phase Rule

21. Number of components (C), phase (P) and degrees of freedom (F) are related by Gibbs-
phase rule as [IES-2001]
lda
(a) C – P – F = 2 (b) F – C – P = 2 (c) C + F – P = 2 (d) P + F – C = 2
21. Ans. (d)

3. As per Gibb's phase rule, if number of components is equal to 2 then the number of
phases will be [IES-2002]
(a) ≤ 2 (b) ≤ 3 (c) ≤ 4 (d) ≤ 5
vi

3. Ans. (c)

4. Gibb's phase rule is given by [IES-1999]

(F = number of degrees of freedom; C = number of components; P = number of phases)
Ci

(a) F= C+ P (b) F=C+P-2 (c) F= C-P-2 (d) F= C-P+ 2

4. Ans. (d) F = C - P + 2

55. Gibb's free energy 'c' is defined as [IES-1999]

w.

(a) G = H - TS (b) G = U - TS (c) G = U + pV (d) G = H + TS

55. Ans. (a) Gibb's free energy 'G' is defined as G = H - TS.

88. Which one of the following relationships defines Gibb's free energy G? [IAS-
2007]
ww

(a) G=H + TS (b) G=H-TS (c) G= U+ TS (d) G= U-TS

88. Ans. (b)

52. Which one of the following relationships defines the Helmholtz function F?
(a) F = H + TS (b) F = H – TS
(c) F = U – TS (d) F = U +TS [IES 2007]
Ans. (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

20. The Gibbs free-energy function is a property comprising [IAS-1998]

(a) pressure, volume and temperature (b) ethalpy, temperature and entropy
(c) temperature, pressure and ethalpy (d) volume, ethalpy and entropy
20. Ans. (b)

53. Assertion (A): For a mixture of solid, liquid and vapour phases of a pure substance in
equilibrium, the number of independent intrinsic properties needed is equal to one.
Reason(R): Three phases can coexist only at one particular pressure. [IES-2005]

m
53. Ans. (d) F = C – P + 2
C = 1, P = 3 or F = 1 – 3 + 2 = 0
112. Consider the following statements: [IES-2000]

co
1. Azeotropes are the mixtures of refrigerants and behave like pure substances.
2. Isomers refrigerants are compounds with the same chemical formula but have different
molecular structures.
3. The formula n + p + q = 2m is used for unsaturated chlorofluorocarbon compounds
(m, n, p and q are the numbers atoms of carbon, hydrogen, fluorine and chlorine respectively)

.
Which of these statements are correct?
(a) 1 and 3 (b) 2 and 3
tas (c) 1 and 2 (d) 1, 2 and 3
112. Ans. (a)

Types of Equilibrium
lda
Local Equilibrium Conditions

Conditions of Stability
vi
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

7. PURE SUBSTANCES
Common data for Question 71, 72 and 73

In the figure shown, the system is a pure substance kept in a piston-cylinder arrangement. The
system is initially a two-phase mixture containing 1 kg of liquid and 0.03 kg of vapour at a
pressure of 100 kPa. Initially, the piston rests on a set of stops, as shown in the figure. A pressure
of 200 kPa is required to exactly balance the weight of the piston and the outside atmospheric

m
pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat
transfer to the system occurs in such a manner that the piston, when allowed to move, does so in
a very slow (quasi-static I quasi-equilibrium) process. The thermal reservoir from which heat is
transferred to the system has a temperature of 400°C. Average temperature of the system

co
boundary can be taken as 170C. The heat transfer to the system is I kJ, during which its entropy
increases by 10 J/K.
Atmospheric pressure

.
tas
vi lda

Specific volumes of liquid (vf) and vapour (vg) phases, as well as values of saturation
temperatures, are given in the table below.
Pressure (kPa) Saturation vf(m3/kg) vg(m3/kg)
o
temperature, Tsat ( C)
Ci

100 100 0.001 0.1

200 200 0.0015 0.002

71. At the end of the process, which one of the following situations will be true? [GATE-2008]
w.

(A) superheated vapour will be left in the system

(B) no vapour will be left in the system
(C) a liquid + vapour mixture will be left in the system
(D) the mixture will exist at a dry saturated vapour state
71. Ans. (A) Initial Volume (V1) = 0.001 + 0.03 × 0.1 m3 = 0.004 m3
ww

let dryness fraction = x

Therefore 0.004 × 1.5 = (1-x) × 0.0015 × 1.03+x × 0.002 × 1.03
That gives an absurd value of x =8.65 ( It must be less than equal to unity)
So vapour is superheated.

72. The work done by the system during the process is [GATE-2008]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(A) 0.1 kJ (B) 0.2 kJ (C) 0.3 kJ (D) 0.4kJ

72. Ans. (D) Work done = first constant volume heating + ∫ pdv
= 0 + P (V2-V1) = 200 × (0.006-0.004) = 0.4 kJ
73. The net entropy generation (considering the system and the thermal reservoir together)
during the process is closest to [GATE-2008]
(A) 7.5 J/K (B) 7.7 J/K (C) 8.5 J/K (D) 10 J/K
73. Ans. (C)

m
1000
( ΔS = (ΔS ) syatem + (ΔS ) surroundings = 10 – =8.51 J/K
(273 + 400)

38. Assertion (A): Water is not a pure substance. [IES-1999]

co
Reason (R): The term pure substance designates a substance which is homogeneous and has
the same chemical composition in all phases.
38. Ans. (d) Water for all practical purpose can be considered as pure substance because it is
homogeneous and has same chemical composition under all phases.

.
tas
71. The given diagram shows an isometric cooling
process 1-2 of a pure substance. The ordinate and
abscissa are respectively
(a) pressure and volume
(b) enthalpy and entropy
(c) temperature and entropy
lda
(d) pressure and enthalpy
[IES-1998]

71. Ans. (c)

vi

31. The ordinate and abscissa in the given figure

showing the saturated liquid and vapour regions of a
pure substance represent:
Ci

(a) temperature and pressure

(b) enthalpy and entropy
(c) pressure and volume
(d) pressure and enthalpy
[IES-1997]
w.

31. Ans. (d) The ordinate and abscissa in given figure are pressure and enthalpy. Such diagram is
common in vapour compression refrigeration systems.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

49. The given diagram shows the throttling process of a pure

substance. [IES-1995]
The ordinate and abscissa are respectively
(a) pressure and volume (b) enthalpy and entropy
(c) temperature and entropy (d) pressure and enthalpy.

m
49. Ans. (d) The throttling process given in figure is on pressure-enthalpy diagram.

co
44. Which one of the following systems can be considered to be containing a pure substance?
[IES-1993, IAS 1998]

.
tas
vi lda
Ci
w.
ww

44. Ans. (a) Air and liquid air can be considered to be containing a pure substance, because air is
also considered to be a perfect gas. All other mixtures are not pure substances.

64. Assertion (A): On the enthalpy-entropy diagram of a pure substance the constant
dryness fraction lines start from the critical point. [IAS-2001]
Reason (R): All the three phases co-exist at the critical point.
64. Ans. (c) Only two phase liquid-vapour is co-exists at the critical point, but at triple point-all
three phase are co-exists.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

26. Assertion (A): Air, a mixture of O2 and N2, is a pure substance. [IAS-2000]
Reason(R): Air is homogeneous in composition and uniform in chemical aggregation.
26. Ans. (a) A pure substance is a substance of constant chemical composition throughout its
mass.

16. If a pure substance contained in a rigid vessel passes through the critical state on heating, its
initial state should be [IAS-1998]

m
(a) subcooled water (b) saturated water (c) wet steam (d) saturated steam
16. Ans. (c)

. co
tas
80. Assertion (A): Air is a pure substance but a mixture of air and liquid air in a cylinder is not a
pure substance.
lda
Reason (R): Air is homogeneous in composition but a mixture of air and liquid air is
heterogeneous. [IAS-1996]
80. Ans. (a)

114. Assertion (A):Temperature and pressure are sufficient to fix the state of a two phase system.
Reason(R): Two independent and intensive properties are required to be known to define the
vi

state of a pure substance. [IAS-1995]

114. Ans. (d) A is false but R is true

53. Assertion (A): At a given temperature, the enthalpy of super-heated steam is the same as that
Ci

of saturated steam. [IES-1998]

Reason (R): The enthalpy of vapour at lower pressures is dependent on temperature alone.
53. Ans. (d)
w.

p-v Diagram for a Pure Substance

50. Which p-v diagram for steam illustrates correctly the isothermal process undergone
by wet steam till it becomes superheated?
ww

(a) (b)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
(c) (d) [IES 1995, 2007]

Ans. (c) Up to saturation point pressure must be constant. After saturation its slope will
p

co
be –ive, as pv=RT or pv=const. or vdp+pdv=0 or dpdv
= −
v

114. Two-phase regions in the given

pressure-volume diagram of a pure

.
substance are represented by tas
(a) A, E and F
(b) B, C and D
(c) B, D and F
(d) A, C and E
lda
[IAS-1999]
114. Ans. (c)

23. A cyclic process ABC is shown on a V- T [IAS-1996]

vi

diagram in fig.

The same process on a P-V diagram will be

represent as
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
. co
23. Ans. (d) tas
57. The network done for the closed shown in the
given pressure-volume diagram, is
(a) 600kN-m (b) 700kN-m
(c) 900kN-m (d) 1000kN-m
vi lda

[IAS-1995]
57. Ans. (d)
Ci
w.
ww

Triple point
89. Triple point temperature of water is [IAS-2000]
(a) 273 K (b) 273.14 K (c) 273.15K (d) 273.16 K
89. Ans. (d) Remember: Triple point temperature of water = 273.16 K = 0.01°C

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

p-T Diagram for a Pure Substance

17. In the following P-T diagram of water showing phase
equilibrium lines, the sublimation line is
(a) p
[IAS-1998]

m
co
17. Ans. (a)

20. Consider the phase diagram of a certain

substance as shown in the given figure. Match List-I
(Process) with List-II (Curves/lines) and select the

.
correct answer using the codes given below the lists:

List-I (Process
A. Vaporization
tas
List-II (Curves/lines)
1. EF
B. Fusion 2. EG
C. Sublimation 3. ED
A B
lda
(a) 1 3
(c) 3 2
[IES-2001]
vi

20. Ans. (a)

Ci

p-v-T Surface
58. The p-v-T surface of a pure substance is
shown in the given p figure. The two-phase
regions are labelled as
w.

(a) R, T and X
(b) S, U and W
(c) S, W and V
(d) R, T and V
[IES-1999]
ww

58. Ans. (c)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

T-s Diagram for a Pure Substance

76. The conversion of water from 40°C to steam at 200°C pressure of 1 bar is best represented as
[IES-1994]

m
. co
tas
lda

76. Ans. (a)

vi

77. The following figure shows the T-s diagram for

steam. With respect to this figure, match List I with
List II and select the correct answer using the codes
given below the Lists:
Ci

List I
A. Curve I 1. Saturated liquid line
B. Curve II 2. Saturated vapour line
C. Curve III 3. Constant pressure line
D. Curve IV 4. Constant volume line
w.

[IES-1994]
Codes: A B C D A B C D
(a) 2 1 4 3 (b) 2 1 3 4
(c) 1 2 3 4 (d) 1 2 4 3
ww

77. Ans. (c)

86. Entropy of a saturated liquid at 2270 C is 2.6 kJ/kgK. Its latent heat of vaporization is 1800
kJ/kg; then the entropy of saturated vapour at 2270C would be [IAS-2001]
(a) 2.88 kJ/kg K (b) 6.2 kJ/kg K (c) 7.93 kJ/kg K (d) 10.53 kJ/kg K
h fg 1800
86. Sg = S f + = 2.6 + = 6.2 kJ / kgK
Tsat 500

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

115. Two heat engine cycles (l - 2 - 3- 1

and l' - 2' - 3' - l’) are shown on T-s co-
ordinates in
[IAS-1999]

m
. co
tas
lda
115. Ans. (d)

117. The mean effective pressure of the

thermodynamic cycle shown in the given
pressure-volume diagram is
vi

(a) 3.0 bar

(c) 4.0 bar
Ci

[IAS-1999]

1
117. Ans. (a) Work (W) = ( 0.03 − 0.01) × ( 400 − 200 ) + × ( 600 − 400 ) × ( 0.03 − 0.01) = 6kJ
w.

2
W 6
W = pm × ΔV or pm = = kPa = 3bar
ΔV ( 0.03 − 0.01)
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

25. The given figure shows a thermodynamic

cycle on T-s diagram. All the processes are
straight times. The efficiency of the cycle is
given by

(a) (0.5 Th- Te)/ Th

(b) 0.5 (Th – Te)/ Th
(c) (Th – Te)/ 0.5 Th
(d) (Th - 0.5 Te)/ Th

m
[IAS-1996]

co
1
25. Ans. (b) Work output = Area 123 = × ( Th − Tc ) × ( S2 − S1 )
2
Heat added = Area under 1 − 2 = Th ( S2 − S1 )

.
1
( Th − Tc )( S2 − S1 )
tas
∴η = 2 = 0.5 ( Th − Tc ) / Th
Th ( S2 − S1 )

Common Data for Questions 59-60

A thermodynamic cycle with an ideal gas as working fluid is shown below.
vi lda
Ci
w.

59. The above cycle is represented on T-S plane by [GATE-2007]

ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
. co
tas
vi lda
Ci

59. Ans. (c)

60. If the specific heats of the working fluid are constant and the value of specific heat ratio γ is
1.4, the thermal efficiency (%) of the cycle is [GATE-2007]
w.

(a) 21 (b) 40.9 (c) 42.6 (d) 59.7

60. Ans. (b)

2.12. The slopes of constant volume and constant pressure lines in the T- s diagram are…..
and….. respectively. [GATE-1994]
ww

2.12 Ans. Higher, Lower

Critical Point
100. Which one of the following statements is correct when saturation pressure of a
vapour increases?
(a) Saturation temperature decreases (b) Enthalpy of evaporation decreases
(c) Enthalpy of evaporation increases

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(d) Specific volume change of phase increases [IES 2007]

Ans. (b)

64. Match List I with List II and select the correct answer using the code given below the
Lists:
List I List II [IES-2005]
A Critical point 1. All the three phases - solid, liquid and vapour co-exists in
equilibrium

m
B. Sublimation 2. Phase change form solid to liquid
C. Triple point 3. Properties of saturated liquid and saturated vapour are identical
D. Melting 4. Heating process where solid gets directly transformed to
gaseous phase

co
A B C D A B C D
(a) 2 1 4 3 (b) 3 4 1 2
(c) 2 4 1 3 (d) 3 1 4 2
64. Ans. (b)

.
(a) remains same
33. Ans. (c)
tas
33. With increase of pressure, the latent heat of steam
(b) increases (c) decreases
[IES-2002]
(d) behaves unpredictably

45. Consider the following statements about critical point of water:

lda
[IES-1993]
1. The latent heat is zero.
2. The liquid is denser than its vapour.
3. Steam generators can operate above this point.
Of these statements
(a) 1,2 and 3 are correct (b) 1 and 2 are correct
(c) 2 and 3 are correct (d) 1 and 3 are correct
vi

45. Ans. (d) At critical point, the latent heat in zero and steam generators can operate above this
point as in the case of once through boilers.
The density of liquid and its vapour is however same and thus statement 2 is wrong.
Ci

48. List I gives some processes of steam whereas List II gives the effects due to the processes.
Match List I with List II, and select the correct answer using the codes given below the lists:
List I List II [IES-1995]
A. As saturation pressure increases 1. Entropy increases.
w.

B. As saturation temperature increases 2. Specific volume increases.

C. As saturation pressure decreases 3. Enthalpy of evaporation decreases.
D. As dryness fraction increases 4. Saturation temperature increases.
Code: A B C D A B C D
ww

(a) 1 3 2 4 (b) 4 3 2 1
(c) 4 3 1 2 (d) 2 4 3 1
48. Ans. (c)

h-s Diagram or Mollier Diagram for a Pure Substance

1.12. Constant pressure lines in the superheated region of the Mollier diagram will have
[GATE-1995]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

(a) a positive slope (b) a negative slope (c) zero slope (d) both positive and negative slope
1.12 Ans. (a)

86. Constant pressure lines in the superheated region of the Mollier diagram have what
type of slope? [IAS-2007]
(a) A positive slope (b) A negative slope
(c) Zero slope (d) May have either positive or negative slopes
86. Ans. (a) Mollier diagram is a h-s plot.

m
⎛ ∂h ⎞
= T = slope
Tds= dh - υ dp or ⎜⎝ ∂s ⎟⎠ P

co
T is always + ive so slope always +ive. Not only this if T ↑ then slope ↑

113. Assertion (A): In Mollier chart for steam, the constant pressure lines are straight lines in wet
region.

.
Reason (R): The slope of constant pressure lines in wet region is equal to T. [IAS-1995]
113. Ans. (a) Both A and R are true and R is the correct explanation of A
tas
44. Which one of the following represents the condensation of a mixture of saturated
liquid and saturated vapour on the enthalpy-entropy diagram? [IES-2004]
(a) A horizontal line (b) An inclined line of constant slope
(c) A vertical line (d) A curved line
lda
44. Ans. (b) Tds = dh – Vdp Or ⎛ ∂h ⎞
⎜ ∂s ⎟ = T
⎝ ⎠P
The slope of the isobar on the h-s diagram
is equal to the absolute temp, for
vi

condensation T is cost so slope is const, but

not zero so it is inclined line.
Ci
w.

Quality or Dryness Fraction

84. Dryness fraction of steam means the mass ratio of [IAS-2001]
(a) wet steam, to dry steam (b) dry steam to water particles in steam
(c) water particles to total steam (d) dry steam to total steam
ww

84. Ans. (d)

3.6 Consider a Rankine cycle with superheat. If the maximum pressure in tile cycle is increased
without changing the maximum temperature and the minimum pressure, the dryness fraction of
steam after the isentropic expansion will increase. [GATE-1995]
3.6 Ans. False

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
co
Steam Tables

.
Charts of Thermodynamic Properties

Measurement of Steam Quality

tas
69. Saturated liquid at a high pressure P1 having enthalpy of saturated liquid 1000 kJ/kg
is throttled to a lower pressure P2. At pressure p2 enthalpy of saturated liquid and that of
the saturated vapour are 800 and 2800 kJ/kg respectively. The dryness fraction of vapour
lda
after throttling process is [IES-2003]
(a) 0.1 (b) 0.5 (c) 18/28 (d) 0.8
69. Ans. (a)
For throttling process (1- 2), h1 = h 2
h1 =h f = 1000 kJ/kg at pressure P1
vi

h 2 = h f + x (h g - h f ) at pressure P2
∴1000 = 800 + x (2800 - 800)
Ci

or x = 0.1

34. Consider the following statements regarding the throttling process of wet steam: [IES-2002]
1. The steam pressure and temperature decrease but enthalpy remains constant.
2. The steam pressure decreases, the temperature increases but enthalpy remains constant.
w.

3. The entropy, specific volume, and dryness fraction increase.

4. The entropy increases but the volume and dryness fraction decrease.
Which of the above statements are correct?
(a) 1 and 4 (b) 2 and 3 (c) 1 and 3 (d) 2 and 4
34. Ans. (c)
ww

79. Match List - I with List - II and select the correct answer using the code given below
the Lists: [IES-2006]
List - I (Apparatus) List - II (Thermodynamic process)
A. Separating calorimeter 1. Adiabatic process
B. Throttling calorimeter 2. Isobaric process
C. Sling psychrometer 3. Isochoric process
D. Gas thermometer 4. Isenthalpic process

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

A B C D A B C D
(a) 1 3 2 4 (b) 2 4 1 3
(c) 1 4 2 3 (d) 2 3 1 4
79. Ans. (c)
79. Select the correct answer using the codes given below the Lists:
List-I List-II [IES-1998]
A. Bomb calorimeter 1. Pressure

m
B. Exhaust gas calorimeter 2. Enthalpy
C. Junker gas calorimeter 3. Volume
D. Throttling calorimeter 4. Specific heats
Code: A B C D A B C D
(a) 3 4 1 2 (b) 2 4 1 3

co
(c) 3 1 4 2 (d) 4 3 2 1
79. Ans. (a)

Throttling

.
36. Consider the following statements: [IES-2000]
tas
When dry saturated steam is throttled from a higher pressure to a lower pressure, the
1. pressure decreases and the volume increases
2. temperature decreases and the steam becomes superheated
3. temperature and the dryness fraction increase
4. entropy increases without any change in enthalpy
Which of these statements are correct?
lda
(a) 1and 4 (b) 1, 2 and 4 (c) 1 and 3 (d) 2 and 4
36. Ans. (b)

34. The process 1-2 for steam shown in the given figure
is
vi

(a) isobaric
(b) isentropic
(c) isenthalpic
(d) isothermal
Ci

[IES-2000]
w.

34. Ans. (c)

A fluid flowing along a pipe line undergoes a throttling process from 10 bar to 1
ww

Bar in passing through a partially open valve. Before throttling, the specific
volume of the fluid is 0.5 m3 /kg and after throttling is 2.0 m3 /kg. What is the
Change in specific internal energy during the throttling process?
(a) Zero (b) 100 kJ/kg
(c) 200 kJ/kg (d) 300 kJ/kg [IES 2007]
Ans. (d) Throttling is a isenthalpic process
h1=h2 or u1+p1v1=u2+p2v2 or u2-u1=p1v1-p2v2 = 1000x0.5 – 100x2 = 300 kJ/kg

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

84b. When saturated liquid at 40°C is throttled to -20°C, the quality at exit will be [GATE-2005]
(a) 0.189 (b) 0.212 (c) 0.231 (d) 0.788
84b. Ans. (b)
h 40 = h−20 = (1 − x ) hf − 20 + xhg
or 371.43 = (1 − x ) 89.05 + x × 1418.0 or x = 0.212
1.17 When wet steam flows through a throttle valve and remains wet at exit [GATE-1996]
(a) its temperature and quality increases
(b) its temperature decreases but quality increases

m
(c) its temperature increases but quality decreases
(d) its temperature and quality decreases
1.17 Ans. (b)

co
2.7 When an ideal gas with constant specific heats is throttled adiabatically, with negligible
changes in kinetic and potential energies [GATE-2000]
(a ) Δh = 0, ΔT = 0 (b) Δh > 0, ΔT = 0 (c) Δh > 0, ΔS > 0 (d ) Δh = 0, ΔS > 0
Where h, T and S represent respectively, enthalpy, temperature and entropy, temperature and
entropy

.
2.7 Ans. (d)
Δh = o
Δs > 0
tas
ΔT < 0
vi lda

5.9 One kilomole of an ideal gas is throttled from an initial pressure of 0.5 MPa to 0.1 MPa. The
Ci

initial temperature is 300 K. The entropy change of the universe is [GATE-1995]

(a) 13.38 kJ/K (b)401.3 kJ/K (c) 0.0446 kJ/K (d) -0.0446 kJ/K
5.9 Ans. (a)
w.
ww

74. The throttling process undergone by a gas

across an orifice is shown by its states in the
following figure:

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

[IES-
1996]

m
74. Ans. (c) The throttling process takes places with enthalpy remaining constant. This process
on T-S diagram is represented by a line starting diagonally from top to bottom.

co
26. In the figure shown, throttling process is
represented by
(a) a e
(b) a d
(c) a c

.
(d) a b tas [IES-1
lda
26. Ans. (b)

20. Assertion (A): Throttle governing is thermodynamically more efficient than nozzle control
governing for steam turbines. [IAS-2000]
Reason (R): Throttling process conserves the total enthalpy.
vi

20. Ans. (d) If throttle governing is done at low loads, the turbine efficiency is considerably
reduced. The nozzle control may then be a better method of governing.
Ci
w.
ww

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

8. PROPERTIES OF GASSES AND GAS MIXTURE

HIGHLIGHTS
1. The functional relationship among the independent properties, pressure P, molar
or specific volume v, and temperature T, is known as ‘Equation of state’ i.e.
PV=RT for gases.

m
2. A hypothetical gas which obeys the law PV = RT at all temperatures and
pressures is called an ‘ideal gas’
• An ‘ideal gas’ has no forces of intermolecular attraction.

co
• The specific heat capacities are constant.
3. ‘Real gas’ does not conform to equation of state with complete accuracy. As PÆ0
or T Æ∞ , the real gas approaches the ideal gas behavior.

.
tas
4. Joule’s law states that the specific internal energy of a gas depends only on the
temperature of the gas and is independent of both pressure and volume.

5. Reversible adiabatic process for a gas

(i) PVγ =c (ii) T 2 = ⎛⎜ P 2 ⎞⎟ (γ-1)/γ = ⎛⎜ V 1
⎞
⎟
(γ-1)
lda
T 1 ⎜⎝ P 1 ⎟⎠ ⎜
⎝V 2
⎟
⎠
6. For Isentropic processes
2
P1V 1 − P2 V 2 (note p v terms comes first)
(i) For closed system ∫ pdv =
1 γ −1
1 1

γ
vi

2
(ii) For flow work or steady flow - = (P1V1-P2V2)
∫
1
vdp
γ −1
7. Polytropic process
It is not adiabatic, but it can be reversible.
Ci

For reversible polytropic process [all n]

2
P1V1 − P2V 2
For closed system, W= ∫ pdv =
1 n −1
w.

2
For Open system, W= - ∫ vdp = n [P V - P V ]
1 1 2 2
1 n −1
If the process is polytropic but we don’t know it is reversible or not then
use [mix. of n & γ ]
ww

n −1

For closed system, First Law W-Q= P1V 1 − P2 V 2 = P1V 1 [ ⎛⎜ P2 ⎞⎟ n

-1]
γ −1 γ −1 ⎜ P ⎟
⎝ 1 ⎠
v2 γ
For steady flow, SFEE W-Q + Δ( + gz ) = (P1V1-P2V2)
2 γ −1

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

γ ⎡ (n −1 )
⎤
= P1V1 ⎢ ⎛⎜ P2 ⎞⎟
− 1⎥
n

γ −1 ⎢⎜ P ⎟ ⎥
⎢⎣ ⎝ 1 ⎠ ⎥⎦

8. a. Isobaric process (P=C), i.e. n=0 (PV0=C)

b. Isothermal process (T=C), i.e. n=1 (PV=RT)
c. Isentropic process (S=C), i.e. n=γ (PVγ=C)

m
1 1

d. Isomeric or Isochoric process (V=C) i.e. n=∞ ( P γ V = C γ if γ Æ∞, V = C)

9. For minimum work in multistage compression P2 = √P1P3

co
a. Equal pressure ratio i.e. P2 P3
=
P1 P2
b. Equal discharge temperature i.e. T2=T3

.
c. Equal work required for both the stages.

a. Van der waals equation

tas
10. Equation of states for real gas
a
(P+)(v-b)=RT
v2
The coefficient a is introduced to account for the existence of mutual attraction
lda
between the molecules. The term a/v2 is called the force of cohesion. The coefficient b
is introduced to account for the volumes of the molecules, and is known as co-volume.
vi

b. Beattie Bridgeman equation

RT (1 − e)(v + B ) A a
P= - v2 Where A=A0 (1- )
Ci

v2 v2
b
B=B0 (1- )
v
c
w.

e=
vT 3
This equation does not give satisfactory results in the critical point region.

11. The ratio PV is called the compressibility factor.

ww

RT
Value of compressibility factor (Z) at critical point is 0.375 for Van der waals gas.
For ideal gas z = 1

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

12. Critical Properties

Vc 8 PcVc
a =3PcVc2, b= , and R=
3 3 Tc
Where Pc, Vc and Tc are critical point pressure, volume
and temperature respectively.
At Critical Point
(i) Three real roots of Vander Waal equation coincide.

m
⎛ ∂p ⎞
(ii) ⎜ ⎟ = 0 i.e. Slope of p-v diagram is zero.
⎝ ∂v ⎠Tc
⎛ ∂2 p ⎞

co
(iii) ⎜ 2 ⎟ = 0 i.e. Change of slope also zero. Fig. Critical properties on p-v diagram
⎝ ∂v ⎠Tc
⎛ ∂3 p ⎞
(iv) ⎜ 3 ⎟ < 0 i.e. negative, and equal to -9pc
⎝ ∂v ⎠Tc

.
13. Boyle’s Temperature (TB) =
tas
a
bR
Boyle’s Law is obeyed fairly accurately up to a moderate pressure and the corresponding
temperature is called the Boyle’s Temperature.
lda
14. Dalton’s Law
a. The pressure of a mixture of gases is equal to the sum of the partial pressures of
the constituents.
b. The partial pressure of each constituent is that pressure which the gas would
exert if it occupied alone that volume occupied by the mixture at the same
temperature.
vi

15. Gibbs-Dalton Law

a. The internal energy, enthalpy and entropy of a gaseous mixture are respectively
Ci

equal to the sum of the internal energies, enthalpies, entropies of the constituents.
b. Each Constituent has that internal energy, enthalpy and entropy, which it would
have if it occupied alone that volume occupied by the mixture at the same
temperature.
w.

16. Equivalent molecular weight (Me) = x1M1+x2M2+------------+ xnMn

Equivalent gas constant (Re) = x1R1+x2R2+-------------+xnRn
Equivalent constant volume specific heat (Cve) = x1Cv1+x2Cv2+------------+xnCnv
Equivalent constant pressure specific heat (Cpe) = x1Cp1+x2Cp2+------------+xnCpn
ww

m
xi = i = mass fraction of a constituent
m
17. The value of Universal Gas constant R = 8.3143 KJ/Kg mole K

Avogadro's Law
6. Assertion (A): The mass flow rate through a compressor for various refrigerants at
same temperature and pressure, is proportional to their molecular weights.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Reason (R): According to Avogardo’s Law all gases have same number of moles in a
given volume of same pressure and temperature. [IES-2002]
Ans. (a) Both A and R correct and R is the correct explanation of A

Ideal Gas
1. Variation of pressure and volume at constant temperature are correlated through
(a) Charles law (b) Boyle’s law (c) Joule’s Law (d) Gay Lussac’s Law

m
[IAS-2002]
Ans. (b) Boyle’s law: It states that volume of a given mass of a perfect gas varies
inversely as the absolute pressure when temperature is constant.

co
30. Assertion (A): A perfect gas is one that satisfies the equation of state and whose
specific heats are constant. [IES-1993]
Reason (R): The enthalpy and internal energy of a perfect gas are functions of

.
temperature only.
30. Ans. (b) For perfect gas, both the assertion A and reason R are true. However R is
tas
not the explanation for A. A provides definition of perfect gas. R provides further
relationship for enthalpy and internal energy but can't be reason for definition of perfect
gas.

30. Consider an ideal gas contained in vessel. If intermolecular interaction suddenly

lda
begins to act, which of the following happens? [IES-1992]
(a) The pressure increase (b) The pressure remains unchanged
(c) The pressure increase (d) The gas collapses
30. Ans. (a)

52. Which of the following statement is correct?

vi

[IES-1992]
(a) Boilers are occasionally scrubbed by rapidly and artificially circulating water inside
them to remove any thin water film may have formed on their inside
(b) A sphere, a cube and a thin circular plate of the same mass are made of the same
Ci

material. If all of them are heated to the same high temperature, the rate of cooling is
maximum for the plate and minimum for the sphere.
(c) One mole of a monoatomic ideal gas is mixed with one mole of diatomic ideal gas.
The molar specific heat of the mixture a constant volume is 2R, where R is the molar gas
constant.
w.

(d) The average kinetic energy of 1 kg of all ideal gases, at the same temperature, is the
same.
52. Ans. (d)
(a) True. A water film, if formed, will act as a very poor conductor of heat and will not
easily let the heat of
ww

the furnace pass into the boiler. An oil film if present, is even worse than water film and
the formation of
such films inside the boiler must be avoided.
(b) Since the mass and material are the same, the volumes must also be the same. For
the same volume, the
surface area of the plate is the greatest and that of the sphere is the least. The rate of
loss of heat by

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

radiation being proportional to the surface area, the plate cools the fastest and the
sphere the slowest.

m
84. Assertion (A): For a perfect gas, hyperbolic expansion is an isothermal expansion.
Pv
Reason (R): For a perfect gas, = constant. [IAS-2007]

co
T
84. Ans. (a)

57. Variation of pressure and volume at constant temperature are correlated through
(a) Charle’s law (b) Boyle’s law (c) Joule’s law (d) Gay Lussac’s law

.
57. Ans. (b) [IAS-2002]
tas
83. An ideal gas with initial volume, pressure and temperature of 0.1 m3, 1bar and 270C
respectively is compressed in a cylinder by a piston such that its final volume and
pressure are 0.04m3 and 5bars respectively, then its final temperature will be [IAS-
2001]
(a) - 1230 C (b) 540 C (c) 3270 C (d) 6000 C
lda
PV PV PV 5 × 0.04
83. Ans. (c) 1 1
= 2 2 or T2 = 2 2
× T1 = × (300) = 600 K = 3270 C
T1 T2 PV
1 1 1 × 0.1

5. Consider the following statements:

A real gas obeys perfect gas law at a very
vi

1. High temperature 2. High-pressure 3. Low pressure

Which of the following statements is/are correct?
(a) 1 alone (b) 1 and 3 (c) 2 alone (d) 3 alone [IES-2000]
Ci

Ans (b) In Perfect gas intermolecular attraction is zero. It will be only possible when
intermolecular distance will be too high. High temperature or low pressure or both cause
high intermolecular distance so choice 1 and 3.
w.

Equation of State of a Gas

46. The correct sequence of the decreasing order of the value of characteristic gas
constants of the given gases is [IES-1995]
(a) hydrogen, nitrogen, air, carbon dioxide (b) carbon dioxide, hydrogen, nitrogen, air.
ww

(c) air, nitrogen, carbon dioxide, hydrogen (d) nitrogen, air, hydrogen, carbon dioxide.
46. Ans. (a) The correct sequence for decreasing order of the value of characteristic gas
constants is hydrogen, nitrogen, air and carbon dioxide.

63. If a real gas obeys the Clausius equation of state p(v - b) = RT then, [IES-1992]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞
(a) ⎜ ⎟ ≠0 (b) ⎜ ⎟ =0 (c) ⎜ ⎟ = 1 (d)
⎝ ∂v ⎠T ⎝ ∂v ⎠T ⎝ ∂v ⎠T
⎛ ∂u ⎞ 1
⎜ ⎟ =
⎝ ∂v ⎠T p
63. Ans. (b)

87. The volumetric air content of a tyre at 27°C and at 2 bars is 30 litres. If one morning,

m
the temperature dips to -3oC then the air pressure in the tyre would be [IAS-2000]
(a) 1.8 bars (b) 1.1 bars (c) 0.8 bars (d) the same as at 27°C
87. Ans. (a) Apply equation of states
P1V1 P2 V2 T2 ( 273 − 3 )

co
= [∵V1 = V2 ] or P2 = P1 × = 2× = 1.8bar
T1 T2 T1 ( 273 + 27 )
8.An Ideal gas with initial volume, pressure and temperature of 0.1m3, 1 bar and 270C
respectively is compressed in a cylinder by piston such that its final volume and pressure
0.04 m3 and 5bar respectively, then its final temperature will be

.
(a) -1230C (b) 540C tas (c) 3270C (d) 6000C [IAS-2001]
P1V1 P2V2 PV
Ans. (c ) : Apply equation of states = or T2 = 2 2 xT1
T1 T2 P1V1
5 0.04
∴T2 = ( )x( ) x (273+27) = 600K = 3270C
1 0.1
lda
10. Pressure reaches a value of absolute zero
(a) at a temperature of -273K
(b) under vacuum condition
(c) at the earth’s centre
(d) when molecular momentum of system becomes zero. [IES-
2002]
vi

1
Ans. (d) we know that P= ρ C 2 If momentum is zero then C must be zero. Hence P
3
Ci

would be zero. That will occur at absolute zero temperature. But note here
choice (a) has in defined temp. –273K which is imaginary temp.

62. Which one of the following PV-T diagrams correctly represents the properties of an
w.

ideal gas?
ww

[IAS-1995]

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

62. Ans. (c)

m
For an ideal gas PV = MRT i.e. P and T follow direct straight line relationship, which is
depicted in figure (c).

co
Van der Waals equation
85. Which one of the following is the characteristic equation of a real gas? [IES-2006]

.
⎛ a⎞ ⎛ a⎞
(a) ⎜ p + 2 ⎟ ( v − b ) = RT tas (b) ⎜ p − 2 ⎟ ( v + b ) = RT
⎝ v ⎠ ⎝ v ⎠
(c) pv = RT (d) pv = nRT
85. Ans. (a)

41. Which of the following statement about Van der waal's equation i valid?
lda
(a) It is valid for all pressure and temperatures [IES-1992]
(b) It represents a straight line on pv versus v plot
(c) It has three roots of identical value at the critical point
(d) The equation is valid for diatomic gases only.
41. Ans. (c)
vi

RTc
75. If a gas obeys van der Waals' equation at the critical point, then is equal to
pc vc
which one of the following? [IAS-2004; 2007]
Ci

(a) 0 (b) 1 (c) 1·5 (d) 2·67

V 8 PV
75. Ans. (d) a=3 pc Vc2 , b= c , R = c c

3 3 Tc
w.

28 In Van der Waal’s gas equation

⎛ a ⎞⎟
⎜P + (v − b ) = RT (R = Universal gas constant)
⎜
v ⎟⎠
2
ww

the unit of ‘b’ is [IAS-1997]

(a) liter/mole0C (b) m3/mole (c) Kg-liter/mole (d) dimensionless

Ans. (b): According to dimensional homogeneity law unit of molar-volume and ‘b’ must
be same. i.e. m3/mole

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

29 Nitrogen at an initial stage of 10 bar, 1 m3 and 300K is expanded isothermally to a

final volume of 2 m3. The P-V-T relation is ⎛⎜ P + a ⎞

⎟ v = RT , where a>0. The
⎜
v ⎟⎠
2
⎝
final pressure will be [GATE-2005]
(a) slightly less than 5 bar (b) slightly more than 5 bar

m
(c) exactly 5 bar (d) cannot be ascertained.

co
Ans. (b): Let no of mole = n
Initial P1 = 10 bar Final P2 =?
1 2
V1 = ( ) m3/mole V2 = ( ) m3/mole

.
n n
T1 = 300K
∴ (P1+a/v12) v1 =(P2+a/v22) v2
tas T2 = 300K = T1=T (say)

⇒ (10 + an2) x (1/n)= (P2 + an2/4) x (2/n)

⇒ 2P2 = 10 + an2-an2/2 = 10 + an2/2 ⇒ P2 = 5 + an2/4
lda
as a>0 ∴P2 is slightly more than 5 bar.

30 A higher value of Van der waal’s constant for a gas indicates that the
vi

(a) Molecules of the gas have smaller diameter. [IAS-2003]

(b) Gas can be easily liquefied.
(c) Gas has higher molecular weight.
Ci

(d) Gas has lower molecular weight.

Ans. (b)
w.

31. The internal energy of a gas obeying Van der Waal’s equation
⎛ a ⎞⎟
ww

⎜P +
⎜ 2
⎟
(v − b ) = RT , depends on [IES-2000]
⎝ v⎠
(a) temperature (b) temperature and pressure
(c) temperature and specific volume (d) pressure and specific volume

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Ans. (b): Joule’s law states that for an Ideal gas internal energy is a function of
temperature only. u = ƒ(T). But this is not Ideal gas it is real gas.

32 Van der Waal’s equation of state is given by ⎛⎜ P + a ⎞⎟

(v − b ) = RT
⎜
v ⎟⎠
2
⎝

The constant ‘b’ in the equation in terms of specific volume at critical point Vc is
equal to [IES-2003]

m
8a
(a) Vc/3 (b) 2 Vc (c) 3 Vc(d)
27VcR

co
Ans. (a): We know that at critical point
8PcVc
a = 3PcVc2 ; b = Vc/3 and R =
3Tc

.
Beattie-Bridgeman equation tas
Virial Expansions
lda
Compressibility
51. Consider the following statements:
1. A gas with a compressibility factor more than 1 is more compressible than a perfect
gas.
2. The x and y axes of the compressibility chart are compressibility factor on y-axis and
vi

reduced pressure on x-axis.

3. The first and second derivatives of the pressure with respect to volume at critical points
are zero.
Which of the statements given above is/are correct?
Ci

[IES 2007]
(a) 2 and 3 only (b) 1 and 3 only
(c) 1 and 2 only (d) 1, 2 and 3
Ans. (a) 1 is false. At very low pressure, all the gases shown have z ≈ 1 and behave
nearly perfectly. At high pressure all the gases have z>1, signifying that they are more
w.

difficult to compress than a perfect gas (for a given molar volume, the product pv is
greater than RT). Repulsive forces are now dominant. At intermediate pressure, must
gasses have Z < 1, including that the attractive forces are dominant and favour
compression.
ww

60. Which one of the following statements is correct?

(a) Compressibility factor is unity for ideal gases
(b) Compressibility factor is zero for ideal gases [IES 2007]
(c) Compressibility factor is lesser than unity for ideal gases
(d) Compressibility factor is more than unity for ideal gases
Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

64. Assertion (A): At very high densities, compressibility of a real gas is less than one.
[IES-2006]
Reason (R): As the temperature is considerably reduced, the molecules are brought closer
together and thermonuclear attractive forces become greater at pressures around 4 MPa.
64. Ans. (d)

38. The value of compressibility factor for an ideal gas may be: [IES-2002]
1. less or more than one 2. equal to one 3. zero 4. less than zero

m
The correct value(s) is/are given by
(a) 1 and 2 (b) 1 and 4 (c) 2 only (d) 1 only
38. Ans. (c)

co
88. Assertion (A): The value of compressibility factor, Z approaches zero of all isotherms as
pressure p approaches zero. [IES-1992]
Reason (R): The value of Z at the critical points is about 0.29.
88. Ans. (d)

.
Critical Properties
tas
113. The mathematical conditions at the critical point for a pure substance are represented
by [IAS-1999]
lda
δp δ2p δ3p δp δ2p δ3p
(a) < 0, 2 = 0 and 3 = 0 (b) = 0, 2 < 0 and 3 = 0
δv δv δv δv δv δv
δp δ p
2
δ p
3
δp δ p
2
δ3p
(c) = 0, 2 = 0 and 3 < 0 (d) = 0, 2 = 0 and 3 = 0
δv δv δv δv δv δv
113. Ans. (c)
vi

90. In the above figure, yc

Ci

corresponds to the critical point

of a pure substance under study.
Which of the following
mathematical conditions
w.

applies/apply at the critical

point?
⎛ ∂P ⎞
(a) ⎜ ⎟ =0
⎝ ∂v ⎠Tc
ww

⎛ ∂2P ⎞
(b) ⎜ 2 ⎟ = 0
⎝ ∂v ⎠Tc
⎛ ∂3 P ⎞
(c) ⎜ 3 ⎟ < 0 [IAS-
⎝ ∂v ⎠Tc
2007]
(d) All of the above

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

90. Ans. (d) Van der Waals equation

⎛ a ⎞ RT a
⎜ P + 2 ⎟ (υ − b ) = RT or P = − 2
⎝ υ ⎠ υ −b υ
V 8 PV
At critical point a= 3pcVc2, b= c , R = c c

3 3 Tc
⎛ ∂P ⎞ − RTc 2a
⎜ ⎟ = + 3 =0
⎝ ∂V ⎠T =Tc (Vc − b ) Vc
2

m
⎛ ∂2 P ⎞ 2.RTc 6a
⎜ 2⎟ = − 4 =0

co
⎝ ∂V ⎠T =Tc (Vc − b ) Vc
3

⎛ ∂3 P ⎞ 6 RTc 24a
& ⎜ 3⎟ =− − 5 = −9 pc i.e.-ive
( vc − b ) vc

.
⎝ ∂V ⎠T =Tc

Boyle temperature
tas
lda
Law of Corresponding States

Adiabatic process
55. Assertion (A): An adiabatic process is always a constant entropy process.
vi

Reason(R): In an adiabatic process there is no heat transfer. [IES-2005]

55. Ans. (d)
Ci

69. A control mass undergoes a process

from state 1 to state 2 as shown in the given
figure. During this process, the heat transfer
to state 2 to state 1 by another process,
then the work interaction during the return
w.

process (in kNm) would be

(a) -400 (b) -200
(c) 200 (d) 400
[IE
ww

69. Ans. (b) During adiabatic process, work done = change in internal energy.

30. A gas expands from pressure P1 to pressure P2 (P2 = p1/10). If the process of expansion is
isothermal, the volume at the end of expansion is 0.55 m3. If the process of expansion is
adiabatic, the volume at the end of expansion will be closer to [IES-1997]
(a) 0.45 m3 (b) 0.55 m3 (c) 0.65 m3 (d) 0.75 m3
p1
30. Ans. (a) For isothermal process, p1v1 = p2 v2 , or p1v1 = × 0.55, v1 = 0.055 m3
10

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

For adiabatic process

p1 1.4
2 , or p1 ( 0.055 )
1.4
p1v11.4 = p2 v1.4 = × v2 or v2 = 0.0551.4 10 = 0.45 m3
10

28. Consider the following statements: [IAS-2007]

1. During a reversible non-flow process, for the same expansion ratio, work done by
a gas diminishes as the value of n in pvn = C increases.
2. Adiabatic mixing process is a reversible process.

m
Which of the statements given above is/are correct?
(a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2
28. Ans. (a)

co
In adiabatic mixing there is always increase in
entropy so large amount of irreversibility is these.

.
tas
lda
Statement for Linked Answer Questions 80 & 81:
A football was inflated to a gauge pressure of 1 bar when the ambient temperature was
15°C. When the game started next day, the air temperature at the stadium was 5oC.
Assume that the volume of the football remains constant at 2500 cm3.

80. The amount of heat lost by the air in the football and the gauge pressure of air in the
vi

football at the stadium respectively equal [GATE-2006]

(a) 30.6 J, 1.94 bar (b) 21.8 J, 0.93 bar (c) 61.1 J, 1.94 bar (d) 43.7 J, 0.93 bar
80. Ans. (d)
Ci

81. Gauge pressure of air to which the ball must have been originally inflated so that it
would equal 1 bar gauge at the stadium is [GATE-2006]
(a) 2.23 bar (b) 1.94 bar (c) 1.07 bar (d) 1.00 bar
w.

81. Ans. (c)

48. A 100 W electric bulb was switched on in a 2.5 m x 3 m x 3 m size thermally

insulated room having a temperature of 20°C. The room temperature at the end of 24
hours will be [GATE-2006]
ww

(a) 321°C (b) 341°C (c) 450°C (d) 470°C

48. Ans. (c) Heat produced by electric bulb in 24 hr. = 100 × 24 × 60 × 60 J = 8640kJ
Volume of air = 2.5 × 3 × 3 = 22.5m3
Density (ρ) = 1.24 kg/m3
ΔQ 8640
ΔQ = mCv Δt or Δt = = = 430 o C ∴t = 430 + 20 = 450 o C
mCv 22.5 × 1.24 × 0.716

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

Isothermal Process
34. An ideal gas undergoes an
isothermal expansion from state R to
state S in a turbine as shown in the
diagram given below:
The area of shaded region is 1000
Nm. What is the amount is turbine
work done during the process?

m
(a) 14,000 Nm (b) 12,000 Nm
(c) 11,000Nm
[IES-2004]

co
34. Ans. (c) Turbine work = area under curve R-S

= ∫ P dv

.
= 1 bar × ( 0.2 − 0.1) m3 + 1000 Nm
tas
= 105 × ( 0.2 − 0.1) Nm + 1000Nm
= 11000Nm

35. The work done in compressing a gas isothermally is given by: [IES-1997]
lda
γ −1
⎡ ⎤
γ ⎛p ⎞ γ ⎛p ⎞
(a) p1v1 ⎢⎜ 2 ⎟ − 1⎥ (b) mRT1 log e ⎜ 2 ⎟
γ −1 ⎢⎝ p1 ⎠ ⎥ ⎝ p1 ⎠
⎢⎣ ⎥⎦
⎛ T ⎞
(c) mc p (T2 − T1 ) (d ) mRT1 ⎜1 − 2 ⎟
vi

⎝ T1 ⎠
35. Ans. (b)
Ci

31. The slope of log P-log V graph for a gas for isothermal change is m1 and for adiabatic
changes is m2. If the gas is diatomic gas, then [IES-1992]
(a)m1<m2 (b)m1>m2 (c) m1 + m2 = 1.0 (d) m1 = m2
31. Ans. (a)
w.
ww

35. The work done during expansion of a gas is independent of pressure if the expansion takes
place. [IES-1992]
(a) isothermally (b) adiabatically (c) in both the above cases (d) in none of the above cases
35. Ans. (d)

70. Identify the process of change of a close system in which the work transfer is maximum.
(a) Isothermal (b) Isochoric (c) Isentropic (d) Polytropic [IAS-
2003]
70. Ans. (c) Aamar mone hoy (a) hobe

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

38. Three moles of an ideal gas are compressed to half the initial volume at a constant
temperature of 300k. The work done in the process is [IES-1992]
(a) 5188 J (b) 2500 J (c) -2500 J (d) -5188 J
38. Ans. (d)

m
co
91. In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3
to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is
the rate of work done during the process? [IAS-2007]

.
(a) 20 kW (b) 35 kW (c) 80 kW (d) 100 kW
91. Ans. (d) For reversible isothermal expansion heat supplied is equal to work done
tas ⎛v ⎞
during the process and equal to Q = W =mRT1 ln ⎜ 2 ⎟
⎝ v1 ⎠
∵ T em perature constant so no change in internal energy dQ = dU + dW ; dU =0
Therefore dQ = dW
lda
86. In respect of a closed system, when an ideal gas undergoes a reversible isothermal process,
the [IAS-2000]
(a) heat transfer is zero (b) change in internal energy is equal to work transfer
(c) work transfer is zero (d) heat transfer is equal to work transfer
86. Ans. (d) In reversible isothermal process temperature constant. No change in internal
vi

energy. So internal energy constant dQ = δ u + δ W as δ u = 0, dQ = dW

Ci

Polytropic process
31. Assertion (A): Though head is added during a polytropic expansion process for which
γ > n> 1, the temperature of the gas decreases during the process. [IES 2007]
Reason (R): The work done by the system exceeds the heat added to the system.
w.

Ans. (a)

⎛ γ − n ⎞ ⎧ p1v1 − p2 v2 ⎫
70. In a polytropic process, the term ⎜ ⎟⎨ ⎬ is equal to: [IES-2005]
⎝ γ − 1 ⎠ ⎩ (n − 1) ⎭
ww

(a) Heat absorbed or rejected (b) Change in internal energy

(c) Ratio of T1/T2 (d) Work done during polytropic expansion
70. Ans. (a)

31. The heat absorbed or rejected during a polytropic process is equal to [IES-2002]
1/2
⎛γ −n⎞ ⎛γ −n⎞
(a) ⎜ ⎟ x work done (b) ⎜ ⎟ x work done
⎝ γ −1 ⎠ ⎝ n −1 ⎠

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

2
⎛γ −n⎞ ⎛γ −n⎞
(c) ⎜ ⎟ x work done (d) ⎜ ⎟ x work done
⎝ γ −1 ⎠ ⎝ γ −1 ⎠
31. Ans. (c)

m
Constant Pressure or Isobaric Process
72. Change in enthalpy in a closed system is equal to the heat transferred, if the
reversible process takes place at [IES-2005]
(a) Temperature (b) Internal energy (c) Pressure (d) Entropy

co
72. Ans. (c) dQ = du + pdυ + υpd − υ dp = d ( u + pυ ) − υ dp = dh − υ dp
if dp = 0 or p = const. these for ( dQ )p = ( dh )p

.
64. Which one of the following phenomena occurs when gas in a piston-in-cylinder
tas
assembly expands reversibly at constant pressure? [IES-2003]
(a) Heat is added to the gas
(c) Gas does work from its own stored energy
(b) Heat is removed from the gas
(d) Gas undergoes adiabatic expansion
64. Ans. (b)
lda
32. A standard vapour is compressed to half its volume without changing its temperature. The
result is that: [IES-1997]
(a) All the vapour condenses to liquid
(b) Some of the liquid evaporates and the pressure does not change
(c) The pressure is double its initial value
(d) Some of the vapour condenses and the pressure does not change
vi

32. Ans. (d) By compressing a vapour, its vapours condense and pressure remains unchanged.

78. For a non-flow constant pressure process the heat exchange is equal to
(a) zero (b) the work done [IAS-2003]
Ci

(c) the change in internal energy (d) the change in enthalpy

78. Ans. (d)
w.

Constant volume or isochoric Process

35. Which one of the following thermodynamic processes approximates the steaming of
food in a pressure cooker?
(a) Isenthalpic (b) Isobaric
ww

(c) Isochoric (d) Isothermal [IES 2007]

Ans. (c) In a pressure cooker, the volume of the cooker is fixed so constant volume
process but for safety some of steam goes out to maintain a maximum pressure. But it
occurs after proper steaming.

57.

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

m
Consider the four processes A, B, C and D shown in the graph given above:
Match List 1 with List 2 and select the correct answer using the code given

co
below the lists:

List 1 List 2
(Processes shown in the graph) (Index ‘n’ in the equation pvn = Const)

.
A. A 1. 0
B. B
C. C
D. D
tas 2. 1
3. 1.4
4. ∞

Code: [IES 2007]

lda
A B C D A B C D
(a) 4 2 3 1 (b) 1 2 3 4
(c) 1 3 2 4 (d) 4 3 2 1
Ans. (b)

21 Match List-I (process) with List-II (index n in PVn = constant) and select the correct
vi

answers using the codes given below the lists. [IES-1999]

List-I List-II
A. Adiabatic 1. n = infinity
Ci

B. Isothermal 2. n = C p
Cv
C. Constant pressure 3. n = 1
D. Constant volume 4. n = C p -1
Cv
w.

5. n = zero
Codes: A B C D A B C D
(a) 2 3 5 4 (b) 3 2 1 5
(c) 2 3 5 1 (d) 2 5 3 1
Ans. (c)
ww

72. A system at a given state undergoes change through the following expansion processes to
reach the same final volume [IES-1994]
1. Isothermal 2. Isobaric 3. Adiabatic ( γ =1.4) 4. Polytropic (n =1.3).
The correct ascending order of the work output in these four processes is
(a) 3,4,1,2 (b) 1,4,3,2 (c) 4,1,3,2 (d) 4,1,2,3
72. Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

70. Match the curves in Diagram I with the curves in Diagram II and select the correct answer.
Diagram I (Process on p- V plane) Diagram II (Process on T-s plane)

m
co
Code: A B C D A B C D
(a) 3 2 4 5 (b) 2 3 4 5
(c) 2 3 4 1 (d) 1 4 2 3 [IES-1996]

.
70. Ans. (b)

70.
tas
vi lda
Ci

Four processes of a thermodynamic cycle are shown above in Fig. I on the T-s plane in the
sequence 1-2-3-4. The corresponding correct sequence of these processes in the p- V plane as
shown above in Fig. II will be [IES-1998]
(a) (C-D-A-B) (b)(D-A-B-C) (c)(A-B-C-D) (d)(B-C-D-A)
70. Ans. (d)
w.

24. An ideal gas is heated (i) at constant volume and (ii) at constant pressure from
the initial state 1. Which one of the following diagrams shows the two processes
correctly? [IAS-1996]
ww

24. Ans. (d)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

69. Match List I with List II and select the correct answer [IES-1996]
List I List II
A. Work done in a polytropic process 1. − ∫ vdp
B. Work done in a steady flow process 2. zero
p1V1 − p2V2
C. Heat transfer in a reversible adiabatic process 3.
γ −1

m
p V − p2V2
D. Work done in an isentropic process 4. 1 1
n −1
69. Ans. (c)

co
88. One kg of a perfect gas is compressed from pressure P1 to pressure P2 by
1. isothermal process 2. adiabatic process 3. the law pv1.4= constant [IAS-2000]
The correct sequence of these processes in increasing order of their work requirement is
(a)1, 2, 3 (b) 1, 3, 2 (c) 2, 3, 1 (d) 3, 1, 2

.
88. Ans. (b)
Work requirement
1. isothermal – area under 121B1A
2. adiabatic – area under 122B2A
tas
3. pv1.1= c – area under 123B3A
vi lda

39. A perfect gas at 27°C was heated until its volume was doubled using the following
Ci

three different processes separately: [IES-2004]

1. Constant pressure process 2 Isothermal process 3. Isentropic process
Which one of the following is the correct sequence in the order of increasing value of the
final temperature of the gas reached by using the above three different processes?
w.

(a) 1 - 2 - 3 (b) 2 - 3 – 1 (c) 3 - 2 - 1 (d) 3 - 1 – 2

39. Ans. (b) Heat addition:
Minm heat required for isothermal process.
Medium heat required for isentropic process.
ww

Maxm heat required for constant pressure process.

90. Match List-I with List-II and select the correct answer using the codes given below the Lists:
List-I List- II
dP P
=−
A. Constant volume process I. dV V [IAS-1997]
dP γP
=−
B. Constant pressure process 2. dV V

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

dT T
=−
C. Constant temperature process 3.
ds CV
dT T
=−
D. Constant entropy process 4.
ds CP
Codes: A B C D A B C D
(a) 3 2 1 4 (b) 2 4 3 1
(c) 3 4 1 2 (d) 1 3 4 2

m
90. Ans. (c)

54. A reversible thermodynamic cycle containing only three processes and producing work is to
be constructed. The constraints are

co
(i) there must be one isothermal process, [GATE-2005]
(ii) there must be one isentropic process,
(iii) the maximum and minimum cycle pressures and the clearance volume are fixed, and
(iv) polytropic processes are not allowed. Then the number of possible cycles are
(a) 1 (b) 2 (c) 3 (d) 4

.
54. Ans. (a) two possible cycle are given below.
tas
vi lda

1-2: Isentropic 1-2: Isentropic

2-3: Isothermal 2-3: Isothermal
Ci

3-1: Constant volume 3-1: Constant pressure

Properties of Mixtures of Gases

w.

74. If M1, M2, M3, be molecular weight of constituent gases and m1, m2, m3… their
[IAS-2007]corresponding mass fractions, then what is the molecular weight M of the
mixture equal to?
1
(a) m1M 1 + m2 M 2 + m3 M 3 + ...........
ww

(b)
m1M 1 + m2 M 2 + m3 M 3 + ...........
1 1 1 1
(c) + + + ........... (d)
m1M 1 m2 M 2 m3 M 3 ⎛ m1 ⎞ ⎛ m2 ⎞ ⎛ m3 ⎞
⎜ ⎟+⎜ ⎟+⎜ ⎟ + ...............
⎝ M1 ⎠ ⎝ M 2 ⎠ ⎝ M 3 ⎠
74. Ans. (a)

Visit : www.Civildatas.com
 Visit : www.Civildatas.com

19. The entropy of a mixture of pure gases is the sum of the entropies of constituents evaluated
at [IAS-1998]
(a) temperature and pressure for the mixture
(b) temperature of the mixture and the partical pressure of the constituents
(c) temperature and volume of the mixture
(d) pressure and volume of the mixture
19. Ans. (b)

m
13. 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing
chamber, so that the final total pressure and temperature of the mixture become same as those
of the individual constituents at their initial states. The universal gas constant is given as R. The
change in entropy due to mixing, per mole of oxygen, is given by [GATE-2008]

co
(A) –Rln2 (B) 0 (C) Rln2 (D)
Rln4
13. Ans. (B) Remember if we mix 2 mole of oxygen with another 2 mole of other gas the volume
Vtotal
will be doubled for first and second constituents ΔS = nR ln = 2 R ln 2 ∴ Total Entropy

.
Vinitial
tas
change = 4Rln2 So, Entropy change per mole=Rln2. And it is due to diffusion of one gas into
another.
vi lda
Ci
w.
ww

Visit : www.Civildatas.com