Chapter 15=Thermodynamics

Physics SF 016 Chapter 15Chapter 16

1

CHAPTER 15: Thermodynamics

Physics SF 016 Chapter 15

15.1 Learning Outcome Remarks :

Keypoint :

Distinguish between thermodynamic work done on the system and work done by the system.

State and use first law of thermodynamics,

WUQ

PHYSICS DF 025CHAPTER

15 :THERMODYNAMICSSubtopic: 15.1


0Q3

15.1.1 Signs for heat, Q and work, W

Sign convention for heat, Q :

Q = positive value

Q = negative value

Heat flow into the system

Heat flow out of the system

3

Surroundings(environment)

System

0W

(a)


System

0W

(b)




0Q

4

Sign convention for work, W:

W = positive value

W = negative value

Work done by the system

Work done on the system


System

0W


System

0W




0Q

5


System

0W


System

0W

Q = positive value

Q = negative value

W = positive value

W = negative value




6

Air

Compression

Air

Expansion

Air

Initially

Motion of piston

Motion of piston

Work done by gas (Expansion)

When the air is expanded, the molecule loses kinetic energy and does positive work on piston.

Work done on gas(Compression)

When the air is compressed, the molecule gains kinetic energy and does negative work on piston.




15.1.2 Work done in the thermodynamics system

7

Consider the infinitesimal work done by the gas (system) during the small expansion, dx in a cylinder with a movable piston as shown in Figure 15.3.

Suppose that the cylinder has a cross sectional area, A and the pressure exerted by the gas (system) at the piston face is P.

GasA

A

dx

Initial

Final

Figure 15.3

F




15.1.3 First law of thermodynamicsIt 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.”

WUQ

suppliedheat ofquantity :Q

energy internal initial :1U

where

energy internal final :2U

done work :W

and 12 UUU

energy internal in the change :U

(15.2)

For infinitesimal change in the energy,

dWdUdQ




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

• The change in the internal energy (U) of a system during any thermodynamic process is independent of path. For example a thermodynamics system goes from state 1 to state 2 as shown in Figure 16.5.

2P

1V

3

4

1P

2VFigure 15.4

2

1P

V0

23124121 UUU




10

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. Calculatea. 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.

Example 1 :

0.2

)kPa(P

)m 10( 32V0.4

300

0

B

CD

A150

Figure 15.6




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.

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

BCABABC WWW

but 0BC W

ABAABC VVPW

223ABC 100.2100.410150 W

J 3000ABC W



W = P.dV


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

ABCABCABC WUQ

ABCBCABABC WQQU

J 1007.1 4ABC U

3000102.3105.10 33ABC U








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

DCADADC WWW

but 0AD W

DCDADC VVPW

223ADC 100.2100.410300 W

J 6000ADC W








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

ADCADCADC WUQ

ADCABCADC WUQ

J 1067.1 4ADC Q

60001007.1 4ADC Q

and ABCADC UU








15

Thermodynamics processes (1 hours)

Remarks :

Keypoint :

Define the following thermodynamics processes: i) Isothermal, ΔU= 0ii) Isovolumetric, W = 0 iii) Isobaric, ΔP = 0iv) Adiabatic, Q = 0

Sketch PV graph to distinguish between isothermal process and adiabatic process.

15.2 Learning Outcome




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

Isothermal process

Isovolumetric @ Isochoric process

Isobaric process Adiabatic process




15.2.1 Isothermal processis defined as a process that occurs at constant temperature.

0U

WQ WUQ

Thus,

Isothermal changesWhen a gas expands or compresses isothermally (constant temperature) thus

constantPV (16.3)

Equation (16.3) can be expressed as

If the gas expand isothermally, thus V2>V1

If the gas compress isothermally, thus V2<V1

2211 VPVP

W = positive

W = negative




is defined as a process that occurs without heat transfer into or out of a system i.e.

For example, the compression stroke in an internal combustion engine is an approximately adiabatic process.

WUUU 12

0Q

WUQ thus

Notes :

For Adiabatic expansion (V2>V1), W = positive value but U =negative value hence the internal energy of the system decreases.

For Adiabatic compression (V2<V1), W = negative value but U =positive value hence the internal energy of the system increases.

15.2.2 Adiabatic Process




15.2.3 Isovolumetric @Isochoric

is defined as a process that occurs at constant volume i.e.

In an isochoric process, all the energy added as heat remains in the system as an increase in the internal energy thus the temperature of the system increases.

For example, heating a gas in a closed constant volume container is an isochoric process.

WUQ

0W thus

12 UUUQ

15.2.4 Isobaricis defined as a process that occurs at constant pressure i.e.

For example, boiling water at constant pressure is an isobaric process.

WUQ

VPW

thus

VPUQ

0P and




15.2.4 Pressure-Volume diagram (graph) for thermodynamic processes

Figure 15.5 shows a PV diagram for each thermodynamic process for a constant amount of an ideal gas.

1V

2P

3P

3V

P

V0

1P

4T3T

1T2T

BD

2V

E

1234 TTTT

A

Figure 16.8

C

Path ABIsothermal process (TB=TA)

Path AC

Path AD

Path AE

Adiabatic process (TC<TA)

Isochoric process (TD<TA)

Isobaric process (TE>TA)



Physics SF 016 Chapter 15From the Figure 15.5,

For comparison between the isothermal (AB) and adiabatic expansions (AC):

The temperature fall (TC<TB) which accompanies the adiabatic expansion results in a lower final pressure than that produced by the isothermal expansion (PC<PB).

The area under the isothermal is greater than that under the adiabatic, i.e. more work is done by the isothermal expansion than by the adiabatic expansion.

The adiabatic through any point is steeper than the isothermal through that point.

2P

3P

P

V0

1P

4T3T

1T2T

BD

EA

C

Figure 15.5 shows a PV diagram for each thermodynamic process for a constant amount of an ideal gas.




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 :






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

The work done by the air is:

Pa 100.2

;m 027.0 ;m 015.05

1

32

31

P

VV

121 VVPW 015.0027.0100.2 5 W

J 2400W




Example 3 :b. The final pressure for the isothermal process is

2211 VPVP

027.0015.0100.2 25 P

Pa 1011.1 52 P





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


25

Remarks :

Keypoint :

• Derive expression for work, W =• 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

isobaric process, use

isovolumetric process, use

Thermodynamics work (4 hour)

2

1

1

2 lnlnP

PnRT

V

VnRTW

12 VVPPdVW

0PdVW

pdV



15.3 Learning Outcome



26

GasA

A

dx

Initial

Final

Figure 15.6

F

The work, dW done by the gas is given by

In a finite change of volume from V1 to V2,

PAF 0cosFdxdW

where and

PAdxdW and dVAdx PdVdW

2

1

V

VPdVW

2

1

V

VPdVdW

(15.3)

done work :W

where

pressure gas :Pgas theof volumeinitial :1V

gas theof volumefinal :2V





27

1PP

V0

1 2

0121 VVPW

P

V0

2P

1P

1V

1

2

0W

For a change in volume at constant pressure, P

12 VVPW

VPW Work done at constant pressure

For any process in the system which the volume is constant (no change in volume), the work done is

0W Work done at constant volume




1V 2V

1P

2P

P

V0

1

20W

Area under graph = work done by gas

2V 1V

2P

1P

P

V0

2

10W

Compression

Area under graph = work done on gas

Expansion

When a gas is expanded from V1 to V2

Work done by gas, 2

1

V

VW pdV

2

1

lnV

nRTV

2

1

1V

VW nRT dV

V

When a gas is compressed from V1=> V2Work done on gas, '

2

'1

V

VW pdV

'2

'1

1V

VW nRT dV

V

'2'

1

lnV

nRTV

Since V2< V1 the value of work done is (-)




From the equation of state for an ideal gas,

Therefore the work done in the isothermal process which change of volume from V1 to V2, is given

nRTPV V

nRTP then

2

1

V

VPdVW

2

1

V

VdV

V

nRTW

2

1

1V

VdV

VnRTW

1

2lnV

VnRTW

12 lnln VVnRTW

(15.9)

2

1ln V

VVnRTW



15.3.3 Work done in Isothermal Process


For isothermal process, the temperature of the system remains unchanged, thus

2211 VPVP 2

1

1

2

P

P

V

V

2

1lnP

PnRTW (15.10)

The equation (16.9) can be expressed as

By applying the 1st law of Thermodynamics,thus

WUQ 0Uand

WQ

2

1

1

2 lnlnP

PnRT

V

VnRTQ



15.3.3 Work done in Isothermal Process


15.3.3 Work done in isobaric process

The work done during the isobaric process which change of volume from V1 to V2 is given by

2

1

V

VPdVW

and constantP

2

1

V

VdVPW

12 VVPW

OR

VPW (15.10)

15.3.3 Work done in isovolumetric process

Since the volume of the system in isovolumetric process remains unchanged, thus

Therefore the work done in the isovolumetric process is

0dV

0PdVW (15.11)




A quantity of ideal gas whose ratio of molar heat capacities is 5/3 has a temperature of 300 K, volume of 64 103 m3 and pressure of 243 kPa. It is made to undergo the following three changes in order:1 : adiabatic compression to a volume 27 103 m3,2 : isothermal expansion to 64 103 m3 ,3 : a return to its original state.

Example 4 :

a. Describe the process 3.b. Sketch and label a graph of pressure against volume for the changes described.

a. Process 3 is a process at constant volume known as isovolumetric (isochoric).

b. The graph of gas pressure (P) against gas volume (V) for the changes described is shown in Figure 15.7.

3P

27

Pa) 10( 4P

)m 10( 33V0

102

K 533

K 3003.24

64

1

2

3Process 2

Process 3Process 1

Figure 15.7




A vessel of volume 8.00 103 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?

Example 5 :

Solution :

Adiabatic expansion

Isochoric process

1V1T

1PInitial

23 VV 13 TT

3PFinal

2V2T

2P




Solution : When the gas expands adiabatically, it does positive work. Thus

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.

WUQ WU

0Qand



A vessel of volume 8.00 103 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?

Example 5 :


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,1.0 102 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]

Example 6 :

Solution :

a. i. 1st law of thermodynamics:

ii. Work done at variable pressure:

WUQ

ferredheat trans ofquantity :Qenergy internalin change :Uwhere

done work :W

1

2lnV

VnRTW 2

1

V

VPdVW OR




Example 6 :

Solution :

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]

b. PV diagram below represents four thermodynamic processes:

3T

1T

P

V

AP

0 AV

4T

2TB

E

DC

AIsobaric process

Isochoric process

Isothermal process

adiabatic process




Example 6 :

Solution :

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,1.0 102 m3 and 5/3 respectively, calculate i. the work done on the gas during isothermal compression. [7 marks]

V

VnRTW 1ln

J 1032.1 3W

V

V

PVW 3ln

PVnRT and

3

1ln100.1102.1 25W

i. The work done during the isothermal compression is




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Chapter 15=Thermodynamics