c) adiabatic energy

d) isothermal energy
Q 15 The enthalpies of all elements in their standard states are:
(a) Unity (b) zero (c) < 0 (d) different for each element
(2X10=20)
SECTION B (short answer questions- 2 Marks)
Q 16. Explain which of the following are extensive properties:
Temperature, internal energy, mass, pressure, volume and heat capacity
Q 17. Heat capacity (Cp) is the extensive property; however, the specific heat (c) is
the intensive property. What will be the relation between the Cp and c for 1
mol of the water?
Q 18. Is a closed system the same as an isolated system?
Q 19. In a process, 701 J of heat is absorbed by a system and 394 J of work is done
by the system. What is the change in the internal energy for the process?
Q 20. Predict the change in the internal energy for the isolated system at the constant
volume.
Q 21. Given the condition, ΔH = 0 for the mixing of the two gases. Explain if the
diffusion of these gases into each other in the closed container is a spontaneous
process or not.
Q 22. Why is the difference between ΔH and ΔU not significant for solids or liquids?
Q 23. What kind of system is the coffee held in a cup?
Q 24. How can we change the above system mentioned in q 23 into an isolated
system?
Q 25. Change in internal energy is a state function while work is not, why?
SECTION C ( 3 Marks) (3X10=30)
Q 26. If water vapour is assured to be perfect gas, molar enthalpy change for
vaporisation of 1 mol of water at 1 bar and 100°C is 41kJ/mol. Calculate the
internal energy change when 1 mol of water is vaporised at 1 bar pressure and
Q 27. Prove that Cp-Cv=R
Q 28. How is the internal energy of a system affected when
(a) Heat passes into or out of a system?
(b) Work is done by or on the system
Q 29. ΔG is net energy available to do useful work and is thus a measure of “free
energy”. Show mathematically that ΔG is a measure of free energy. Find the
unit of ΔG. If a reaction has positive enthalpy change and positive entropy
change, under what condition will the reaction be spontaneous?
Q30. The ΔH and ΔS for 2Ag2O(s)→4Ag(s)+O2(g) are given +61.17kJmol - 1and
+132Jk - 1mol - 1 respectively. Above what temperature will the reaction be
spontaneous?
Q31. At the temperature 298 K, the Kp of the reaction is given as
N2O4 (g) ⇄ 2NO2 (g) is 0.98. Predict when the reaction is spontaneous or not.
Q32. Calculate the number of kj of heat necessary to raise the temperature of 60 g
of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
Q33. (a)For an isolated system ∆U = 0; what will be ∆S?
(b)For a reaction at 298 K
2A + B————->C
∆H = 40Q kj mot1 and AS = 0.2 kj Kr-1 mol-1.
At what temperature will the reaction become spontaneous considering ∆H
and ∆S to be constant over the temperature range?
Q34. An increase In the enthalpy of the surroundings is the same as the decrease In
the enthalpy of the system. Will the temperature of the system and its
surroundings be equal if they are in thermal equilibrium?
Q35. The difference between the Cp and Cv can be derived as the empirical relation
H = U + PV. Calculate the difference between the Cp and Cv for the ten moles
of the ideal gas.
SECTION D ( 4 Marks ) (4X4=16)
Q 36 Derive the relationship between ΔH and ΔU for an ideal gas. Explain each
term Involved in the equation.
Q 37. Predict in which of the following the entropy increases or decreases:
(a) A liquid crystallizes into a solid
(b) Temperature of a crystalline solid is raised from 0K to 115K
(c) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2)(g)
(d) H2(g) → 2H(g)
Q 38. What do you understand by isothermal and free expansion of a gas?
Q 39. (a)Define entropy.
(b)Why is entropy a state function?
SECTION E ( 5 Marks) (5X4=20)
Q 40. Give reason for the following:
(a)Neither q nor w is a state function but q + w is a state function.
(b)A real crystal has more entropy than an ideal crystal.
Q41 (a)Two litres of an ideal gas at a pressure of 10 atm expands isothermally into
a vacuum until its total volume is 10 litres. How much heat is absorbed and
how much work is done in the expansion ?
(b) Consider the same expansion, but this time against a constant external
pressure of 1 atm.
(c) Consider the same expansion, to a final volume of 10 litres conducted
reversibly.
Q42 (a) Give the correct relation between equilibrium constant (K), standard
free energy (ΔG°) and temperature (T).
(b)Calculate the Standard Free Energy Change at 25℃ given the Equilibrium
constant of 1.3 × 104.
Q43 Derive the work done in
(a) an isothermally reversible process
(b) free expansion
Concepts of System and types of systems, surroundings, work, heat, energy,
extensive and intensive properties, state functions. First law of
thermodynamics -internal energy and enthalpy, heat capacity and Introduction
of entropy as a state function, Gibb's energy change for spontaneous and non-
spontaneous processes,

Answers:
Ans 15 SECTION A MCQS (1X15=15)
1. (d) 2. (c) 3. (c) 4. (b) 5. (b)
6. (a) 7. (b) 8. (b) 9. (c) 10. (b)
11. (b) 12. (a) 13. (a) 14. (b) 15.
SECTION B (2X10=20)
Ans 16 Mass, volume, heat capacity and internal energy are extensive properties as the
above mentioned values depends on quantity or size of matter present in the
system.
Ans 17 For the water, the heat capacity is = 18 × specific heat and Cp = 18 × c .Specific
heat is c = 4.18 Jg-1 K-1

Heat capacity is Cp = 18 × 4.18 JK-1 = 75.3 J K-1.

Ans 18. No. Systems can be classified as open, closed, or isolated. Open systems allow
energy and mass to pass across the system boundary. A closed system allows
energy but not mass across its system boundary. An isolated system allows
neither mass or energy to pass across the system boundary
Ans 19. In the given case, q =+701 J, w = -394 J. According to the first law of
thermodynamics, DeltaU = q + w = + 701 + (-394) = + 307 J Thus, internal
energy= 307J
Ans 20 There is no energy transfer as the heat or the work in the isolated system,Thus,
w=0 and q=0. According to the first law of thermodynamics-Δ U = q + w = 0 +
0 = 0 ΔU = 0
Ans 21. The diffusion will be a spontaneous process. As the change in the enthalpy is
zero, the change in the randomness or disorder that is ΔS increases. As a result,
for the equation ΔG = ΔH – TΔS, the term TΔS will be given as negative.
Thus, ΔG will be negative. Hence the process will be spontaneous.
Ans 22. The difference between ΔH and ΔU is not significant for solids or liquids
because systems made up entirely of solids and/or liquids do not experience
significant volume changes when heated, the difference between and is usually
insignificant.
Ans 23 Coffee held in a cup is an open system because it can exchange matter (water
vapour and energy (heat) with the surroundings.
Ans 24. Coffee held in a thermos flask is an example of an isolated system because it
can neither exchange energy nor matter with the surroundings.
Ans 25. In a process, the change in internal energy depends upon the initial and final
state of the system. Therefore, it is a state function. But work is dependent on
the path followed. Therefore, it is not a state function.
SECTION C (3X10=30)
Ans 26. The change H2O(l) → H2O(g)
∆H= ∆U +∆ngRT or
∆U= ∆H - ∆ngRT, substituting the values, we get
∆U=41.00kJmol-1 – 1 X8.3J mol-1 K-1 X 373K
= 41.00kJ mol-1 – 3.096 kJ mol-1
=37.904kJ mol-1

Ans 27. From the equation q = n C ∆T, we can say:

At constant pressure P, we have

qP = n CP∆T

This value is equal to the change in enthalpy, that is,

qP = n CP∆T = ∆H

Similarly, at constant volume V, we have

qV = n CV∆T

This value is equal to the change in internal energy, that is,

qV = n CV∆T = ∆U

We know that for one mole (n=1) of an ideal gas,

∆H = ∆U + ∆(pV ) = ∆U + ∆(RT) = ∆U + R ∆T

Therefore, ∆H = ∆U + R ∆T

Substituting the values of ∆H and ∆U from above in the former equation,

CP∆T = CV∆T + R ∆T
CP = CV + R
CP – CV = R
Ans 28. (a)When heat is transferred from the surroundings into the system there is a
rise in the temperature and the internal energy and when the heat is passed out
of the system into the surroundings the internal energy is decreased.

(b) If work is done by the system then Internal Energy decreases . As U

(Internal energy ) =Q(Heat) +W(Work done) . If work is done by the system so
work is negative and hence Internal energy decreases. On the other hand, when
the work is done on the system(e.g. compressing a gas), the internal energy
increases

Ans 29. ∆STotal = ∆Ssys + ∆Ssurr

∆STotal = ∆Ssys + ( – ∆Hsys/T)

T∆STotal = T∆Ssys – ∆Hsys

For spontaneous change, ∆STotal > 0

∴ T∆Ssys – ∆Hsys > 0

-(∆Hsys – T∆Ssys) > 0

But, ∆Hsys – T∆Ssys = ∆Gsys

∴ –∆Gsys > 0

∆Gsys = ∆Hsys – T∆Ssys < 0

∆Hsys = Enthalpy change of a reaction

T∆Ssys = Energy which is not available to do useful work

∆Gsys = Energy available for doing useful work

• If ∆G is negative (< 0), the process is spontaneous.

• If ∆G is positive (> 0), the process is non spontaneous.
Unit of ∆G is Joule.

The reaction will be spontaneous at high temperature.

Ans 30. From the above question we have ΔHand ΔS for 2Ag2O(s)→4Ag(s)+O2(g) are
given as +61.17kJmol - 1 and +132Jk - 1mol - 1 respectively.
As per the Gibbs Helmholtz equation, ΔG=ΔH−TΔS
Here, ΔG is the change in Gibbs free energy, ΔHis the change in the enthalpy, T
is the absolute temperature.
For the reaction 2Ag2O(s)→4Ag(s)+O2(g)
It is spontaneous when ΔGis negative.
Here ΔH,ΔSis positive. So ΔGis negative when ΔH<TΔS
T>ΔHΔS
T=61170mol - 1132Jk - 1mol - 1
T=463.4K
 Therefore, the process is spontaneous above a temperature of T=463.4K.
Ans 31. For the spontaneous reaction,

ΔrG° is given as -ve.

ΔrG° = – RT ln Kp = – RT ln (0.98)

In (0.98) is – 0.02

ΔrG° = – RT × –0.02

Δ will be positive.

Thus, the reaction will be considered as non-spontaneous.

Ans 32. No. of moles of Al (m) = (60g)/(27 g mol-1) = 2.22 mol
Molar heat capacity (C) = 24 J mol-1 K-1.
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol-1 K-1) x (2.22 mol) x (20 )
= 1065.6 J = 1.067 kj

Ans 33. (a)Change in internal energy (∆U) for an isolated system is zero for it does not
exchange any energy with the surroundings. But entropy tends to increase in
case of spontaneous reaction. Therefore, ∆S > 0 or positive.

(b) As per the Gibbs Helmholtz equation:

ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol-1)/(0.2 KJ K-1 mol-1) = 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be
spontaneous above this temperature.

Ans 34. Yes, if the system, as well as the surroundings, are in thermal equilibrium, their
temperatures are equal. Also, an increase in the enthalpy of the surroundings is
equal to a decrease in the enthalpy of the system.

Ans 35. For the 1 mole of the gas,

Cp – Cv = R

For n moles of the gas, the relation is given as

Cp – Cv = nR = 10 × 4.184 J

Cp – Cv = 41.84 J.
Section D (4X4=16)

Ans 36. (i) At constant volume-

By first law of thermodynamics: q = ΔU + (–w)

(–w) = pΔV
 ∴ q = ΔU + pΔV

ΔV = 0, since volume is constant.

∴ qV = ΔU + 0 ⇒ qV = ΔU = change in internal energy

(ii) At constant pressure qp = ΔU + pΔV

But, ΔU + pΔV = ΔH
∴ qp = ΔH = change in enthalpy.
So, at a constant volume and at constant pressure heat change is a state
function because it is equal to change in internal energy and change in enthalpy
respectively which are state functions.

Ans 37 Definition: Entropy is defined as a measure of randomness or disorder of a

system.
(a)A liquid crystallizing into a solid is a freezing process and after freezing, the
molecules attain an ordered state and therefore, entropy decreases.

(b) Temperature of a crystalline solid is raised from 0 K to 115 K. At 0 K, the

constituent particles are static, and entropy is minimum. If temperature is
raised to 115 K, these begin to move and oscillate about their equilibrium
positions in the lattice and system becomes more disordered, therefore entropy
increases.

(c) In the given reaction, reactant (NaHCO3) is a solid and it has low entropy.
Among products there are one solid and two gases. Therefore, the products
represent a condition of higher entropy.

(d) In the given reaction, here one molecule gives two atoms i.e., number of
particles increases leading to more disordered state. Two moles of H atoms
have higher entropy than one mole of dihydrogen molecule.

Ans 38. (a) An isothermal process is a change in the system such that the
temperature remains constant. In other words, in isothermal process ∆T
= 0. (b)Free expansion of a gas occurs when it is subjected to expansion
in a vacuum (pex=0).
Ans 39. (a)entropy: the degree of disorder or uncertainty in a system. 2. a. : the
degradation of the matter and energy in the universe to an ultimate state of
inert uniformity. Entropy is the general trend of the universe toward death and
disorder.

1. (b) The entropy is a state function because it depends on the final and
initial state of the process.
2. It does not depend on the path by which the process is completed.
3. For example, entropy change for a reversible process is the same when
the same process undergoes an irreversible manner but for work and
heat, it is not the same because entropy is a state function and work and
heat are path functions.
 SECTION E (5X4=20)

Ans 40. (a) q + w = ∆u

As ∆u is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its
structural arrangement whereas ideal crystal does not have any disorder.
Hence, a real crystal has more entropy than an ideal crystal.

Ans 41. (a) We have q = – w = pex (10 – 2) = 0(8) = 0 No work is done; no heat is
absorbed.

(b) We have q = – w = pex (8) = 8 litre-atm

(c) We have q = – w = 2.303 × 10 10 log 2 = 16.1 litre-atm

Ans 42. (a) The correct relation between equilibrium constant (K), standard free energy
(ΔG°) and temperature (T) can be given as-

Consider a reaction, A +B ⇄ C + D

ΔG = ΔGӨ + RT lnK

For equilibrium, ΔG = 0

0 = ΔGӨ + RT lnK

ΔGӨ = –RT lnK

ΔGӨ = –2.303RT logK

(b) ΔGӨ = –2.303 RTlogK

ΔGӨ = –2.303 RTlogK = –8.314 × 298 × log 1.3 × 104 = –23469 J = –23.4 KJ

Ans 43. (a) Derivation of work done in isothermal reversible expansion

dW=P×A×dl=P×dV

The amount of work done by isothermal reversible expansion of an ideal gas

from V1 to V2 is

W=∫V2V1PdV

From ideal gas equation, P=nRTV

Hence W=∫V2V1nRTVdV=nRT∫V2V1dVV

On integration we get
W=nRTlnV2V1

Since

P1V1=P2V2P1P2=V2V1W=nRTlnP1P2

(b) Expansion of a gas in vacuum (pex = 0) is called free expansion. No work

is done during free expansion of an ideal gas whether the process is reversible
or irreversible (equation 6.2 and 6.3). Now, we can write equation 6.1 in
number of ways depending on the type of processes. Let us substitute w = –
pex∆V (eq. 6.2) in equation 6.1, and we get ∆ = − ∆ U q p V ex If a process is
carried out at constant volume (∆V = 0), then ∆U = qV the subscript V in q V
denotes that heat is supplied at constant volume.