Chapter 15=Thermodynamics

Physics SF 016 Chapter 15
16

CHAPTER 15:
Thermodynamics

1

Remarks :
15.1 Learning Outcome

 Distinguish between
thermodynamic work done on
the system and work done by
the system.
 State and use first law of
thermodynamics,
Keypoint :

Q U W

15.1.1 Signs for heat, Q and work, W
(a) Surroundings
Sign convention for heat, Q : (environment)

Q = positive value
Heat flow into the system System
3 Q 0 W 0

Q = negative value (b) Surroundings
Heat flow out of the system (environment)

System
3 W 0

Sign convention for work, W: Surroundings
(environment)
W = positive value
Work done by the system
System
Q 0 W 0

Surroundings
W = negative value (environment)

Work done on the system

System
W 0
4

Surroundings
(environment)

Q = positive value
W = positive value
System
Q 0 W 0

Surroundings
(environment)
Q = negative value
W = negative value
System

5
W 0


Work done by gas (Expansion)
Motion of piston
When the air is expanded, the
molecule loses kinetic energy and
does positive work on piston. Air

Expansion

Work done on gas(Compression) Air
Motion of
Initially piston
When the air is compressed, the
molecule gains kinetic energy and
does negative work on piston.
Air
6
Compression


15.1.2 Work done in the thermodynamics system
Consider the infinitesimal work
A
done by the gas (system) during Initial
Gas
the small expansion, dx in a
cylinder with a movable piston as
shown in Figure 15.3. dx

F A
Suppose that the cylinder has a Final
cross sectional area, A and the
pressure exerted by the gas
(system) at the piston face is P.
Figure 15.3

7

15.1.3 First law of thermodynamics
It states that : “The heat (Q) supplied to a system is equal to the increase in the
internal energy ( U) of the system plus the work done (W) by the system on its
surroundings.”

Q U W (15.2)

and U U 2 U1
where Q : quantity of heat supplied
W : work done
U : change in the internal energy
U1 : initial internal energy
U 2 : final internal energy
For infinitesimal change in the energy,

dQ dU dW


•The first law of thermodynamics is P
a generalization of the principle of 1
conservation of energy to include P1 3
energy transfer through heat as well
as mechanical work.

P2 2
•The change in the internal energy 4
( U) of a system during any 0 V1 V2 V
thermodynamic process is
Figure 15.4
independent of path. For example a
thermodynamics system goes from
state 1 to state 2 as shown in Figure U1 2 U1 4 2 U1 3 2
16.5.

P(kPa )
Example 1 :
D C
300

150 B
Figure 15.6
A
0 2.0 4.0 V ( 10 2 m3 )
A vessel contains an ideal gas in condition A, as shown in Figure
16.6. When the condition of the gas changes from A to that of B,
the gas system undergoes a heat transfer of 10.5 kJ. When the gas
in condition B changes to condition C, there is a heat transfer of
3.2 kJ. Calculate
a. the work done in the process ABC,
b. the change in the internal energy of the gas in the process ABC,
c. the work done in the process ADC,
d. the total amount of heat transferred in the process ADC. 10

a. The work done in the process ABC is
given by :

Calculate:
WABC WAB WBC
but WBC 0
b. the change in the internal energy
of the gas in the process ABC, WABC PA VB VA
c. the work done in the process ADC WABC 150 103 4.0 10 2
2.0 10 2

,
d. the total amount of heat WABC 3000 J
transferred in the process ADC.

b. By applying the 1st law of
thermodynamics for ABC, thus

Calculate: QABC U ABC WABC

U ABC QAB QBC WABC
of the gas in the process ABC,
U ABC 10.5 103 3.2 103 3000
c. the work done in the process ADC
,
4
d. the total amount of heat U ABC 1.07 10 J

c. The work done in the process ADC is
given by

Calculate:
WADC WAD WDC
a. the work done in the process ABC, but WAD 0
of the gas in the process ABC, WADC PD VC VD
, WADC 300 103 4.0 10 2 2.0 10 2
d. the total amount of heat
WADC 6000 J

d. By applying the 1st law of
thermodynamics for ADC, thus

Calculate: QADC U ADC WADC
and U ADC U ABC
of the gas in the process ABC,
QADC U ABC WADC
,
d. the total amount of heat QADC 1.07 104 6000
4
QADC 1.67 10 J

Remarks :

Thermodynamics processes (1 hours)
 Define the following thermodynamics
processes:
i) Isothermal, ΔU= 0
ii) Isovolumetric, W = 0
iii) Isobaric, ΔP = 0
iv) Adiabatic, Q = 0
Keypoint :
 Sketch P V graph to distinguish
between isothermal process and
adiabatic process.

15


There are four specific kinds of thermodynamic
processes. It is :

 Isothermal process

 Isovolumetric @ Isochoric process

 Isobaric process

 Adiabatic process

15.2.1 Isothermal process
is defined as a process that occurs at Isothermal changes
constant temperature. When a gas expands or compresses
isothermally (constant temperature)
thus

U 0 PV constant (16.3)

Equation (16.3) can be expressed as
Thus,
PV1
1 P2V2
Q U W Q W If the gas expand isothermally, thus
V2>V1
W = positive

If the gas compress isothermally, thus
V2<V1
W = negative

15.2.2 Adiabatic Process
is defined as a process that occurs Notes :
without heat transfer into or out of a
system i.e.  For Adiabatic expansion
(V2>V1), W = positive value
Q 0 but U =negative value hence
the internal energy of the
thus Q U W system decreases.

U U 2 U1 W For Adiabatic compression
(V2<V1), W = negative value
but U =positive value hence
For example, the compression stroke in
an internal combustion engine is an the internal energy of the
approximately adiabatic process. system increases.

15.2.3 Isovolumetric @ 15.2.4 Isobaric
Isochoric is defined as a process that occurs at
is defined as a process that occurs at constant pressure i.e.
constant volume i.e.

W 0 thus P 0 and W P V

Q U W thus
Q U W
Q U U 2 U1
In an isochoric process, all the energy Q U P V
added as heat remains in the system
as an increase in the internal energy thus
the temperature of the system increases. For example, boiling water at constant
pressure is an isobaric process.
For example, heating a gas in a closed
constant volume container is an isochoric
process.

15.2.4 Pressure-Volume diagram (graph) for thermodynamic processes
Figure 15.5 shows a P V diagram for each
thermodynamic process for a constant T4 T3 T2 T1
amount of an ideal gas.
Path A B
Isothermal process (TB=TA)
P
Path A C
Adiabatic process (TC<TA)
A E
P1
Path A D
T4 Isochoric process (TD<TA)
P2 B T3
D C T2 Path A E
P3
T1 Isobaric process (TE>TA)
0 V1 V2 V3 V
Figure 16.8

Figure 15.5 shows a P V diagram for each From the Figure 15.5,
thermodynamic process for a constant For comparison between the
amount of an ideal gas. isothermal (A B) and adiabatic
expansions (A C):
The temperature fall (TC<TB) which
P accompanies the adiabatic
expansion results in a lower final
pressure than that produced by the
A E isothermal expansion (PC<PB).
P1
The area under the isothermal is
T4 greater than that under the
P2 B T3
D adiabatic, i.e. more work is done by
P3 C T2 the isothermal expansion than by
T1 the adiabatic expansion.
0 V
The adiabatic through any point is
steeper than the isothermal through
that point.


Example 3 :

Air is contained in a cylinder by a frictionless gas-tight piston.

a. Calculate the work done by the air as it expands from a
volume of 0.015 m3 to a volume of 0.027 m3 at a
constant pressure of 2.0 105 Pa.

b. Determine the final pressure of the air if it starts from the
same initial conditions as in (a) and expanding by the same
amount, the change occurs isothermally.

Example 3 : Solution :
a. Given
Air is contained in a cylinder by a
frictionless gas-tight piston. V1 0.015 m3 ; V2 0.027 m3 ;
P1 2.0 105 Pa
a. Calculate the work done by the air as
it expands from a volume of 0.015 m3 to The work done by the air is:
a volume of 0.027 m3 at a constant
pressure of 2.0 105 Pa. W PV V 1 2 1
5
b. Determine the final pressure of the air
W 2.0 10 0.027 0.015
if it starts from the same initial
conditions as in (a) and expanding by the
W 2400 J
same amount, the change occurs
isothermally

Example 3 :
b. The final pressure for the isothermal
Air is contained in a cylinder by a process is
frictionless gas-tight piston.
PV1
1 P2V2
a. Calculate the work done by the air as
it expands from a volume of 0.015 m3 to 2.0 105 0.015 P2 0.027
a volume of 0.027 m3 at a constant
pressure of 2.0 105 Pa. P2 1.11 105 Pa
b. Determine the final pressure of the air
if it starts from the same initial
conditions as in (a) and expanding by the
same amount, the change occurs
isothermally

Remarks :
Thermodynamics work (4 hour)
• Derive expression for work, W = pdV
• Determine work from the area under p-V graph.
• Derive the equation of work done in
isothermal, isovolumetric and isobaric processes.
• Calculate work done in :-
isothermal process and use Keypoint :
V2 P1
W nRT ln nRT ln
V1 P2

isobaric process, use

W PdV P V2 V1
isovolumetric process, use

25 W PdV 0

The work, dW done by the gas is
given by dW Fdx cos A
 Initial Gas
where 0 andF PA
dW PAdx
and Adx dV dx
dW PdV A
In a finite change of volume from Final 
V1 to V2, F
V2
Figure 15.6
dW PdV where
V1
V2 W : work done
W PdV (15.3) P : gas pressure
V1 V1 : initial volume of the gas
26
V2 : final volume of the gas

P
For a change in volume at constant
P1 1 2
pressure, P
W P V
Work done at
W P1 V2 V1 0
W P V2 V1 constant
pressure
0 V
P
For any process in the system which the P2 2
volume is constant (no change in
volume), the work done is
W 0
Work done at constant
W 0 volume P1
1
0 V
27 V1

When a gas is expanded from V1 to V2 P Area under graph
= work done by gas
Work done by gas, V2
W pdV P1
1
V1
V2 1
W nRT dV
V1 V
V
P2 2
nRT ln 2 W 0
V1 0 V1 V2 V
When a gas is compressed from V1=> V2 Expansion
P
Work done on gas, V2'
W pdV 2
V1' P2 Area under graph
V2' 1 = work done on gas
W nRT dV
V1' V
V2' P 1
nRT ln ' 1
V1 W 0
Since V2< V1 the value of work done is (-)
0 V2Compression V1 V

15.3.3 Work done in Isothermal Process
From the equation of state for an ideal
gas,
nRT V2
PV nRT then P W nRT ln V V1
V
Therefore the work done in the
isothermal process which change of W nRT ln V2 ln V1
volume from V1 to V2, is given
V2
V2
W PdV W nRT ln (15.9)
V1 V1
V2nRT
W dV
V1 V
V2 1
W nRT dV
V1 V

15.3.3 Work done in Isothermal Process

For isothermal process, the temperature By applying the 1st law of
of the system remains unchanged, thus Thermodynamics,thus

V2 P1 Q U W and U 0
PV1 P2V2
1
V1 P2 Q W
The equation (16.9) can be expressed as
V2 P1
Q nRT ln nRT ln
P V1 P2
W nRT ln 1 (15.10)
P2

15.3.3 Work done in isobaric process 15.3.3 Work done in isovolumetric
process
The work done during the isobaric
process which change of volume from V1
Since the volume of the system in
to V2 is given by
V2 isovolumetric process remains
W PdV unchanged, thus
V1
and P constant dV 0
V2
W P dV Therefore the work done in the
isovolumetric process is
V1

W P V2 V1
OR
W PdV 0 (15.11)

W P V (15.10)

Example 4 :
a. Process 3 is a process at constant
A quantity of ideal gas whose ratio of volume known as isovolumetric
molar heat capacities is 5/3 has a
(isochoric).
temperature of 300 K, volume of 64
10 3 m3 and pressure of 243 kPa. It is b. The graph of gas pressure (P) against
made to undergo the following three gas volume (V) for the changes
changes in order: described is shown in Figure 15.7.
1 : adiabatic compression to a volume P( 104 Pa)
27 10 3 m3, Figure 15.7
2 : isothermal expansion to 64 10 3 m3 ,
102 2
3 : a return to its original state.
Process 2
a. Describe the process 3.
b. Sketch and label a graph of pressure P3 3
Process 1 Process 3
against volume for the changes
described. 24.3 533 K
1 300 K
0 27 64 V ( 10 3 m3 )

Example 5 :
A vessel of volume 8.00 10 3 m3 contains an ideal gas at a pressure of 1.14 105
Pa. A stopcock in the vessel is opened and the gas expands adiabatically, expelling
some of its original mass until its pressure is equal to that outside the vessel (1.01
105 Pa). The stopcock is then closed and the vessel is allowed to stand until the
temperature returns to its original value. In this equilibrium state, the pressure is
1.06 105 Pa. Explain why there was a temperature change as a result of the
adiabatic expansion?

Solution :

Initial Final
P V1
1 P2 V2 P3 V3 V2
T1 T2 T3 T1
Adiabatic Isochoric
expansion process

Example 5 :
A vessel of volume 8.00 10 3 m3 contains an ideal gas at a pressure of 1.14 105
Pa. A stopcock in the vessel is opened and the gas expands adiabatically, expelling
some of its original mass until its pressure is equal to that outside the vessel (1.01
105 Pa). The stopcock is then closed and the vessel is allowed to stand until the
temperature returns to its original value. In this equilibrium state, the pressure is
1.06 105 Pa. Explain why there was a temperature change as a result of the
adiabatic expansion?

Solution :
When the gas expands adiabatically, it does positive work.
Thus
Q U W and Q 0
U W
The internal energy of the gas is reduced to provide the
necessary energy to do work. Since the internal energy is
proportional to the absolute temperature hence the
temperature decreases and resulting a temperature change.

Example 6 :
a. Write an expression representing
i. the 1st law of thermodynamics and state the meaning of all the
symbols.
ii. the work done by an ideal gas at variable pressure. [3 marks]
b. Sketch a graph of pressure P versus volume V of 1 mole of ideal
gas. Label and show clearly the four thermodynamics process.
[5 marks]
c. A monatomic ideal gas at pressure P and volume V is compressed
isothermally until its new pressure is 3P. The gas is then allowed
to expand adiabatically until its new volume is 9V. If P, V and for
the gas is 1.2 105 Pa,1.0 10 2 m3 and 5/3 respectively, calculate
i. the final pressure of the gas.
ii. the work done on the gas during isothermal compression.
(Examination Question Intake 2003/2004) [7 marks]

Example 6 :
a. Write an expression representing
i. the 1st law of thermodynamics and state the meaning of all the
symbols.
ii. the work done by an ideal gas at variable pressure. [3 marks]
Solution :
a. i. 1st law of thermodynamics:
Q U W
where U : change in internal energy
Q : quantity of heat trans ferred
W : work done
ii. Work done at variable pressure:

V2 V2
W PdV OR W nRT ln
V1 V1

Example 6 :
b. Sketch a graph of pressure P versus volume V of 1 mole of ideal gas. Label and
show clearly the four thermodynamics process.
[5 marks]

Solution :
b. PV diagram below represents four thermodynamic processes:
P
Isobaric process
A E Isothermal process
PA
Isochoric process T4
D B T3
adiabatic process C T2
T1
0 VA V

Example 6 :
c. A monatomic ideal gas at pressure P and volume V is compressed isothermally
until its new pressure is 3P. The gas is then allowed to expand adiabatically until its
new volume is 9V. If P, V and for the gas is 1.2 105 Pa,1.0 10 2 m3 and 5/3
respectively, calculate
i. the work done on the gas during isothermal compression. [7 marks]

Solution :
i. The work done during the isothermal compression is
V1
W nRT ln and nRT PV
V
V
1
W PV ln 3 W 5
1.2 10 1.0 10 2
ln
V 3
W 1.32 103 J


THE END…
Good luck
For Second Semester Examination

PDT STUDENTS,
DO YOUR BEST, BEAT THE REST
PHYSICS ‘A’, INSYAALLAH ...

Chapter 15=Thermodynamics

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