First Law of

thermodynamics
 SYSTEMS
❑ A system is defined as a quantity of matter or a
region in space chosen for study. The mass or
region outside the system is called the
surroundings.

❑ The real or imaginary surface that separates the

system from its surroundings is called the
boundary. The boundary of a system can be
fixed or movable
 Systems

Open Closed
System System

❑An open system allows heat and work to enter and/or leave the system, In
addition, an open system allows mass to enter and/or leave the system
❑ A closed system consists of a fixed amount of mass, and no mass can cross its
boundary. That is, no mass can enter or leave a closed system.
❑ Energy, in the form of heat or work, can cross the boundary
❑ The volume of a closed system does not have to be fixed.
❑ A special case, even energy is not allowed to cross the boundary, that system
is called an isolated system.
 STATE AND EQUILIBRIUM
❑ Consider a system not undergoing any change.
❑ At this point, all the properties can be measured or
calculated throughout the entire system, which gives us a
set of properties that completely describes the condition,
or the state, of the system.
❑ At a given state, all the properties of a system have fixed
values.
❑ If the value of even one property changes, the state will
change to a different one.
❑ In Fig. a system is shown at two different states
First law of thermodynamics
If a system undergoes a change
from an initial state to a final
state, then the change in the
internal energy ΔU is given by

𝑄 = 𝑊 + ∆𝑈
 ∆𝑈= U2-U1

More heat is added to system than More heat flows out of system than Heat added to system equals work
system does work: Internal energy work is done: Internal energy of done by system: Internal energy of
of system increases. system decreases. system unchanged
✓ The internal energy of an ideal gas depends only on its temperature,
not on its pressure or volume.
 WORK dW = F dx = PA dx = P dV
𝑉2
𝑊 = න 𝑃𝑑𝑣
𝑉1

the differential area dA is equal to P dV, which is

the differential work. The total area A under the
process curve 1–2 is obtained by integration
2 𝑉2
𝐴𝑟𝑒𝑎 = 𝐴 = න 𝑑𝐴 = 𝑊 = න 𝑃𝑑𝑣
1 𝑉1
The net work done by a system W depends on the
thermodynamic process (or the path) chosen
between its initial and final states
❑ A gas can follow several different paths as it
expands from state 1 to state 2. In general,
each path has a different area underneath it,
and since this area represents the magnitude of
the work, the work done is different for each
process
❑ The cycle shown in Fig. produces a net work
output because the work done by the system
during the expansion process (area under path
A) is greater than the work done on the system
during the compression part of the cycle (area
under path B), and the difference between these
two is the net work done during the cycle (the
colored area).
 Example 1
A gas expands from an initial volume of 0.4
m3 to a final volume of 0.62 m3 as the
pressure increases linearly from 110 kPa to
230 kPa. Find the work done by the gas.
 Problem 2 Problem 3

calculate (a) the work done on the gas A gas undergoes the three-part process,
and (b) the net energy transferred to connecting the states A and B. The three
the gas by heat in the process. parts of the process are labeled 1,2, and
3. Find the total work done by the gas
during this process
 Example 4
path (a) IAF
A gas expands from I to F along the three
paths indicated in Figure. Calculate the 𝑊 = 𝑊𝐼𝐴 + 𝑊𝐴𝐹
work done on the gas along paths (a)
IAF, (b) IF, and (c) IBF. 𝑊 = 4 4 − 2 + 0 = 8J

path (b) IF

1
𝑊 = 1 4 − 2 + (4 − 2)(4 − 1) = 5J
2
path (c) IBF

𝑊 = 𝑊𝐼𝐵 + 𝑊𝐵𝐹

𝑊 = 0 + 1 4 − 2 = 2J
 Kinds of Thermodynamic Processes P  T If V = constant
(1) Isochoric Process (Isovolumetric Process) P
= constant
P1 P
= 2
T T1 T2
➢ An isochoric process is a constant-volume process(the
piston is clamped to a fixed position to ensure an
isovolumetric process). When the volume of a
thermodynamic system is constant, it does no work on
its surroundings. Then W = 0 and 𝑄 = 𝑈2 − 𝑈1 = ∆ 𝑈
𝑄 = ∆ 𝑈 = 𝑚 𝑐𝑣 ∆𝑇 = 𝑛𝐶𝑣 ∆𝑇
CV is molar specific heat for constant volume
all the energy added as heat remains in the system kept
at constant volume, then all of the transferred energy
remains in the system as an increase in its internal
energy, and hence, temperature 𝑄
 Problem 5

An ideal gas is taken from a to b on the pV-

diagram shown in Fig. During this process,
700 J of heat is added and the pressure
doubles.
(a) How much work is done by or on the gas?
(a)
Explain.
(b)
(b) How does the temperature of the gas at a
compare to its temperature at b?
(c)
(c) How does the internal energy of the gas at a
compare to the internal energy at b?
(2) Isobaric Process V T If P = constant
An isobaric process is a constant-pressure V
process. In general, none of the three quantities = constant
T
ΔU, Q, and W is zero in an isobaric process,
V1 V
𝑊 = 𝑃(𝑉2 − 𝑉1) = 2
T1 T2
𝑄 = 𝑚 𝑐𝑝 ∆𝑇 = 𝑛𝐶𝑝 ∆𝑇
Cp is molar specific heat for constant pressure

❑ Assume that the piston is free to move in such a way that it is

always in equilibrium under the effect of the net force from a
gas pushing upward and the weight of the piston plus the force
due to atmospheric pressure pushing downward.
❑ Then, an isobaric process could be established by transferring
heat energy Q to or from the gas by any mechanism
 We can also express dW in terms of the temperature
(2) Relating Cp and CV for an Ideal Gas change dT by using the ideal gas equation of state PV =
nRT , Because p is constant, the change in V is
First consider the constant-volume process. We place n proportional to the change in T
moles of an ideal gas at temperature T in a constant-
dW = p dV = nR dT
volume container.
nC dT = dU + nR dT
p
dU = dQ = n CV dT
nCp dT = nCV dT + nR dT
The pressure increases during this process, but the gas
does no work (dW= 0) because the volume is constant. C=C +R
p V

The first law in differential form, is

dQ = dU + dW.
Since dW = 0, dQ = dU dU = n CV dT
Now consider a constant-pressure process with the same Molar Specific Heat for a Monatomic Ideal
temperature change dT.
Gas at Constant Volume
dQ = nCp dT
The work dW done by the gas in this constant-pressure Molar Specific Heat for a Monatomic Ideal
process is
dW = p dV Gas at Constant Pressure
 Problem 6
Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes
from 3.20 × 10-2 m3 to 4.50 × 10-2 m3 (a) Calculate the initial and final temperatures of the gas. (b) Calculate the
amount of work the gas does in expanding. (c) Calculate the amount of heat added to the gas. (d) Calculate the
change in internal energy of the gas (CV =12.47 J/mol ⋅K)
(a)

(b)

(c)

or
(d)
 Problem 7
One-half mole of an ideal gas is taken from state a to state c, (a) Calculate the final temperature
of the gas. (b) Calculate the work done on (or by) the gas as it moves from state a to state c. (c)
Does heat leave the system or enter the system during this process? How much heat?
A thermodynamic system is taken from state a to state c in Fig. along either Problem 8
path abc or path adc. Along path abc, the work W done by the system is 450
J. Along path adc, W is 120 J. The internal energies of each of the four
states shown in the figure are Ua = 150 J, Ub = 240J , Uc = 680J and Ud =
330J Calculate the heat flow Q for each of the four processes ab, bc, ad, and
dc. In each process, does the system absorb or liberate heat?

𝑊𝑎𝑏 = 0
𝑄 = 𝑈𝑏 − 𝑈𝑎 = 240 − 150 =90J
𝑄 = 𝑈𝑐 − 𝑈𝑏 + 𝑊𝑏𝑐
𝑄 = 𝑈𝑑 − 𝑈𝑎 + 𝑊𝑎𝑑 𝑄 = 330 − 150 + 120 = 300𝐽
𝑊𝑑𝑐 = 0 𝑄 = 𝑈𝑐 − 𝑈𝑑 + 𝑊𝑑𝑐 𝑄 = 680 − 300 + 0 = 350J
 Problem 9

The p-V diagram in Fig. shows two paths along which a sample
of gas can be taken from state a to state b, where Vb =3.0V1. Path
1 requires that energy equal to 5.0p1V1 be transferred to the gas
as heat. Path 2 requires that energy equal to 5.5p1V1 be
transferred to the gas as heat. What is the ratio p2/p1?

Since the combination “p1V1” appears frequently in this derivation we denote it as “x. Thus for process
1, the heat transferred is Q1 = 5x = ΔU1 + W1 , and for path 2 (which consists of two steps, one at
constant volume followed by an expansion accompanied by a linear pressure decrease) it is Q2 = 5.5x =
ΔU2 + W2 . If we subtract these two expressions and make use of the fact that internal energy is state
function (and thus has the same value for path 1 as for path 2) then we have 5.5x – 5x = W2 – W1 =
“area” inside the triangle = 1/2 (2 V1 )(p2 – p1) . Thus, dividing both sides by x (= p1V1), we find 0.5 =
(p2 / p1 ) -1, which leads immediately to the result: p2 / p1 = 1.5
 3 - Isothermal Process
❑ An isothermal process is a constant-
temperature process.
❑ For a process to be isothermal, any heat
flow into or out of the system must occur
slowly enough that thermal equilibrium is
maintained.
❑ In general, none of the quantities Q, or
W is zero in an isothermal process.
❑ For such systems, if the PV = constant P1V1 = P2V2
temperature is constant, the
internal energy is also constant
❑Any energy that enters the system by heat is transferred out of the system by
work

If the gas is compressed, V f

If the gas expands, the work
<V , and the work done is the
i
W equals the positive of the
negative of the area under the
shaded area under the PV
PV curve.
curve
work
❑ No change in the internal energy of the system occurs in an isothermal process
❑ Since T is constant and also n and R are constants
 Example 10
Consider the isothermal expansion of an ideal gas. The system contains n=4.0
moles of argon gas at room temperature(20oC). If the system then expands to
double its initial volume, what are (a) the work done by the system and (b) the
amount of heat that flows into the system from its surroundings?
(a)

The system is at room temperature (T =293 K), and the final volume is twice the
initial volume (Vf = 2Vi)

(b)
Example 11
A 1.0-mol sample of an ideal gas is kept at 0.0°C during an expansion from 3.0 L to 10.0 L. (A)
How much work is done on the gas during the expansion? (B) How much energy transfer by heat
occurs between the gas and its surroundings in this process? (C) If the gas is returned to the
original volume by means of an isobaric process, how much work is done on the gas?
(A)

(B)

(C)
Example 12
Sketch a P-V diagram showing the following processes in a cycle .
Process 1-2: isobaric work output of 10.5 J from an initial volume of
0.028 m3 and pressure 140 pa, Process 2-3: isothermal compression, and
Process 3-1: isochoric heat transfer to its original volume of 0.028 m3
and pressure 140 pa. Calculate (a) the maximum volume in the cycle, in
m3, (b) the isothermal work, in J, (c) the net work, in J.

❖ Process by process analysis,

❖ The isothermal work ❖ The net work

 4- Adiabatic Process

An adiabatic process is one that occurs

(1) so rapidly or occurs (2) in thermally
insulated systems during which no
transfer of heat energy enters or leaves
the system, i.e. Q = 0. With this
restriction and the application of the
first law of thermodynamics to an
adiabatic process, we get: Q = 0 and
ΔU = −W (Adiabatic process)
 4- Adiabatic Process

❑ No heat flows into or out of a system

❑ There are two ways that an adiabatic process

can come about
❑ First the gas can be expanded or
compressed very rapidly in what we
call an adiabatic expansion or an
adiabatic compression.
❑ In a rapid process there is essentially
no time for heat to be transferred
between the gas and the
environment
❑Second , a gas cylinder can be completely surrounded by thermal insulation,
such as thick pieces of Styrofoam

Expanding with no heat flow decreases P

Compressing with no heat flow increases P and T
and T

A gas that expands adiabatically gets

❑increases the thermal energy. Thus colder as its thermal energy decreases.
an adiabatic compression raises Thus an adiabatic expansion lowers the
the temperature of a gas temperature of a gas
 Adiabatic Ideal Gas: Relating V, T, and p
dU = n CV dT
the work done by the gas during the process is
given by

dW = p dV , dU = -dW
for an adiabatic process
n C dT = - p dV
V
Because γ is always greater than unity for a gas,
(γ -1) is always positive. This means that dV
and dT always have opposite signs. An
The coefficient R/CV can be expressed in terms adiabatic expansion of an ideal gas dV ˃ 0
of γ = Cp>CV We have always occurs with a drop in temperature dT < 0
and an adiabatic compression dV < 0 always
occurs with a rise in temperature dT ˃ 0 this
confirms our earlier prediction
 Adiabatic Ideal Gas: Relating V, T, and p

the work done by an ideal gas during an

adiabatic process. We know that Q = 0 and W
= - ΔU for any adiabatic process. For an ideal
gas, ΔU = n CV(T2 - T1). If the number of moles
n and the initial and final
Thus for an initial state (T1,V1) and a final state (T2,V2) temperatures T1, T2 and are known, we have
simply

using the ideal-gas equation in the form T= PV/nR

We may also use pV = nRT in above

equation to obtain
because n and R are constant

Thus for an initial state (P1,V1) and a final state (P2,V2)

Thermodynamic Processes and Their Characteristics
Problem 13
A container with an initial volume of 0.0607 m3 holds 2.45 moles of a monatomic
ideal gas at a temperature of 325 K. The gas is now compressed adiabatically to a
volume of 0.0377 m3. Find (a) the final pressure and (b) the final temperature of
the gas.
(a)

(b)
 Problem 14
An ideal monatomic gas is compressed starting at point A in the PV diagram, where PA = 100
kPa, VA = 1.0 m3 and TA = 300 K The gas is first compressed adiabatically to state B (PB = 200 kPa).
The gas is then further compressed from point B to point C (VC = 0.50 m3)
in an isothermal process. (a) Determine VB (b) Calculate the work done on the gas for the whole
process.
(a)

(b)
 100  10 3  (1) −  200  10 3  (0 .66 )
P v −P v W =     = − 48  10 3 J
W = A A B B AB 1 .67 − 1
AB  −1

The total work done on the gas is

 Problem 15
A sample of an ideal gas goes through the process shown in Figure. From
A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of
energy entering the system by heat; from C to D, the process is
isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the
system by heat. Determine the difference in internal energy between B,A.

ΔUint = ΔU AB + Δ U BC + Δ U CD + Δ U DA = 0
ΔU AB =− Δ U BC − Δ U CD − Δ U DA
 ‫ً‬
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