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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,

W U Q




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)

Surroundings

(environment)

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

Surroundings

(environment)

System

0W

Surroundings

(environment)

System

0W




0Q

5

Surroundings

(environment)

System

0W

Surroundings

(environment)

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 , themolecule gains kinetic energy and

does negative work on piston.




15.1.3 First law of thermodynamics

It states that : “The heat (Q) supplied to a system is equal to the increase in theinternal energy (U ) of the system plus the work done (W ) by the system on its

surroundings.”

W U Q

suppliedheatof quantity:Q

energyinternalinitial:1U

where

energyinternalfinal:2U

donework :W

and12 U U U

energyinternalin thechange:U

(15.2)

For infinitesimal change in the energy,

dW dU dQ




•The first law of thermodynamics is

a generalization of the principle of conservation of energy to include

energy transfer through heat as wellas mechanical work.

•The change in the internal energy

(U ) of a system during anythermodynamic process is

independent of path. For example athermodynamics system goes fromstate 1 to state 2 as shown in Figure

16.5.

2P

1V

3

4

1P

2V

Figure 15.4

2

1

P

V 0

23124121 U U U




Calculate:

a. the work done in the process ABC,

b. the change in the internal energyof the gas in the process ABC,

c. the work done in the process ADC,

d. the total amount of heattransferred in the process ADC.

a. The work done in the process ABC is

given by :

BCABABC W W W

but 0BC W

ABAABC V V PW

223

ABC 100.2100.410150 W

J3000ABC W

W = P.dV




c. The work done in the process ADC isgiven by

DCADADC W W W

but 0AD W

DCDADC V V PW

223

ADC100.2100.410300

W

J6000ADC W

Calculate:








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

ADCADCADC W U Q

ADCABCADC W U Q

J1067.1 4

ADC Q

60001007.1 4

ADCQ

andABCADC U U

Calculate:








15

Thermodynamics processes (1 hours)

Remarks :

Keypoint :

Define the following thermodynamicsprocesses:

i) Isothermal, ΔU = 0

ii) Isovolumetric, W = 0iii) Isobaric, Δ P = 0 iv) Adiabatic, Q = 0

Sketch P V graph to distinguishbetween isothermal process andadiabatic process.

15.2 Learning Outcome




There are four specific kinds of thermodynamicprocesses. It is :

Isothermal process

Isovolumetric @ Isochoric process

Isobaric process

Adiabatic process




15.2.1 Isothermal process

is defined as a process that occurs atconstant temperature.

0

U

W Q W U Q

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 V PV P

W = positive

W = negative




is defined as a process that occurswithout 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.

W U U U 12

0Q

W U Q thus

Notes :

For Adiabatic expansion

(V 2>V 1), W = positive value

but U =negative value hence

the internal energy of thesystem decreases.

For Adiabatic compression

(V 2<V 1), W = negative value but U =positive value hencethe internal energy of the

system increases.

15.2.2 Adiabatic Process




15.2.3 Isovolumetric @

Isochoricis defined as a process that occurs atconstant volume i.e.

In an isochoric process, all the energyadded as heat remains in the systemas an increase in the internal energy thusthe temperature of the system increases.

For example, heating a gas in a closedconstant volume container is an isochoric

process.

W U Q

0W thus

12 U U U Q

15.2.4 Isobaric

is defined as a process that occurs atconstant pressure i.e.

For example, boiling water at constantpressure is an isobaric process.

W U Q

V PW

thus

V PU Q

0P and




15.2.4 Pressure-Volume diagram (graph) for thermodynamic processes

Figure 15.5 shows a P V diagram for eachthermodynamic process for a constant

amount of an ideal gas.

1V

2P

3P

3

V

P

V 0

1P

4T 3T

1T 2T

BD

2

V

E

1234 T T T T

A

Figure 16.8

C

Path AB

Isothermal process (T B=T A)

Path AC

Path AD

Path AE

Adiabatic process (T C<T A)

Isochoric process (T D<T A)

Isobaric process (T E>T A)




From the Figure 15.5,

For comparison between the

isothermal (AB) and adiabaticexpansions (AC):

The temperature fall (T C<T B) which

accompanies the adiabatic

expansion results in a lower final

pressure than that produced by the

isothermal expansion (P C<P B).

The area under the isothermal is

greater than that under the

adiabatic, i.e. more work is done by

the isothermal expansion than bythe adiabatic expansion.

The adiabatic through any point is

steeper than the isothermal through

that point.

2P

3P

P

V 0

1P

4T

3T

1T 2T

B

D

EA

C

Figure 15.5 shows a P V diagram for eachthermodynamic 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 thesame initial conditions as in (a) and expanding by the same

amount, the change occurs isothermally.

Example 3 :




Air is contained in a cylinder by africtionless 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 initialconditions 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:

Pa100.2

;m027.0;m015.0

5

1

3

2

3

1

P

V V

121 V V PW

015.0027.0100.2 5 W

J2400W




Example 3 :b. The final pressure for the isothermal

process is

2211 V PV P

027.0015.0100.2 2

5

PPa1011.1 5

2 P

Air is contained in a cylinder by africtionless 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 initialconditions 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

2lnln

P

PnRT

V

V nRT W

12V V PPdV W

0 PdV W

pdV

15.3 Learning Outcome




15.3.1 Work done in the thermodynamics system

26

Gas

A

A

dx

Initial

Final

Figure 15.6

F

The work, dW done by the gas is

given by

In a finite change of volume fromV 1 to V 2,

PAF 0

cosFdxdW where and

PAdxdW and dV Adx PdV dW

2

1

V

V PdV W

2

1

V

V PdV dW

(15.3)

donework :W

where

pressuregas:P

gastheof volumeinitial:1

V gastheof volumefinal:2V




15.3.2 Work done in the thermodynamics system

27

1P

P

V 0

1 2

0121

V V PW

P

V 0

2P

1P

1V

1

2

0W

For a change in volume at constantpressure, P

12

V V PW

V PW Work done atconstant

pressure

For any process in the system which the

volume is constant (no change in volume),

the work done is

0W Work done at constantvolume




1V 2V

1P

2P

P

V 0

1

2

0W

Area under graph

= work done by gas

2V 1V

2P

1P

P

V 0

2

1

0W

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

V W pdV

2

1ln

V

nRT V

2

1

1V

V W nRT dV

V

When a gas is compressed from V1=> V2

Work done on gas, '2

'1

V

V W pdV

'2

'1

1V

V W nRT dV

V

'

2

'

1

lnV

nRT V

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 V 1 to V 2, is given

nRT PV V

nRT P then

2

1

V

V PdV W

2

1

V

V dV

V

nRT W

2

1

1V

V dV

V nRT W

1

2lnV

V nRT W

12 lnln V V nRT W

(15.9)

2

1ln

V

V V nRT W

15.3.3 Work done in Isothermal Process




For isothermal process, the temperature

of the system remains unchanged, thus

2211 V PV P 2

1

1

2

P

P

V

V

2

1ln P

PnRT W (15.10)

The equation (16.9) can be expressed as

By applying the 1st law of

Thermodynamics,thus

W U Q 0U and

W Q

2

1

1

2 lnlnP

PnRT

V

V nRT Q

15.3.3 Work done in Isothermal Process




15.3.3 Work done in isobaric process

The work done during the isobaricprocess which change of volume from V 1

to V 2 is given by

2

1

V

V PdV W

and constantP

2

1

V

V dV PW

12 V V PW

OR

V PW (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

0 PdV W (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

)m10( 33V 0

102

K533

K3003.24

64

1

2

3

Process 2

Process 3Process 1

Figure 15.7




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.

W U Q W U

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, expellingsome 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]

Example 6 :

Solution :

a. i. 1st law of thermodynamics:

ii. Work done at variable pressure:

W U Q

ferredheat transof quantity:Qenergyinternalinchange:U where

donework :W

1

2lnV

V nRT W

2

1

V

V

PdV W 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

2T B

E

DC

A

Isobaric 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

V nRT W 1ln

J1032.1 3W

V

V

PV W 3ln

PV nRT and

3

1ln100.1102.1

25W

i. The work done during the isothermal compression is




THE END…

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