Chapter 15 Students Physics

8/12/2019 Chapter 15 Students Physics

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The study of relationships

involving heat,

mechanical work, and

other aspects of energyand energy transferfor

the system.

Thermodynamic systemisany collection of objects that is

convenient to regard as a unit,

and that may have the

potential energy to exchange

energy with its surroundings.

CHAPTER 15:Thermodynamics

(4 Hours)


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15.1 First Law of Thermodynamics

15.2 Thermodynamics Processes

15.3 Thermodynamics Work

SUBTOPIC

2


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Learning Outcome:

At the end of this chapter, students should be able to:

Distinguishbetween work done on the system and work

done by the system.

State and usefirst law of thermodynamics,

3

15.1 First law of thermodynamics (1 hour)

WUQ


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Thermodynamicsis the study of energy relationships that

involve heat, mechanical work, and other aspects of energyand energy transfer.

15.1 First Law of Thermodynamics

3 quantitiesinvolved in a thermodynamic system :

The first law of thermodynamicsis the extension of the

principle of conservation of energyto include both heat

and mechanical energy.

1. Heat , Q

2. Internal energy , U

3. Work , W

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F

Gas A

A

dx

Initial

Final

When a gas expands, its pushes out on its boundarysurfaces as they move outward; an expanding gasalways

does positive work.

Work done by the system (+)

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Figure above shows a gas in a cylinder with a moveable

piston.

Suppose that the cylinder has a cross sectional area,Aand the pressure exerted by the gas (system) at the

piston face isP.

The forceFexerted on the piston by the system isF=PA.

When the piston moves out a small distance dx, the workdWdone by the force is

dW=Fdx=PAdx

but Adx= dV

dVis the small change of volume of the system (gas)

The work done by the systemisdW=PdV

P, pressure is constant 6


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Work done on the system (-)

dx

Suppose that the cylinder has

a cross sectional area,Aand

the pressure of the gas isP.

The external forceF

exerted on the system is

F

F=PA

The magnitude of external

force F exerted on the

system equal to PA

because the piston isalways in equilibrium

between the external force

and the force from the gas.

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When the piston moves in a small distance dx, the work

dWdone by the force is

dW= -Fdx= -PAdx

negative signbecause

initial value is greater than

final value

but - Adx= - dV

The work done on the systemis

dW= -PdV

P, pressure is constant8


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In both cases if the volume of the gas changes from V1to

V2, the work done is given by

donework:W

2

1

V

V

PdVdW

pressuregas:Pvolumeinitial:

1V

volumefinal:2

V

12 VVPW

VPW VPW -=

Work done by the system

Gas expandsVolume increases

+ W

Work done on the system

Gas is compressedVolume decreases

- W

9

dW P dV


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The First Law of Thermodynamics (flot)

The change in internal energy of a closed

system, U

, will be equal to the heat added tothe system minus the work done by the system

WQU

systemtheondoneiswork:

systemthebydoneiswork:

systemthefromremovedisheat:

systemthetoaddedisheat

energyinternalindecrease

energyinternalinincrease

W-

W

- Q

Q

U

U

+

:+

:-

:+

system

+Q -Q

-W +W

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WQU

Rearrange WUQ +=

When Qis added to a system (gas) , the temperature of the

gas increases, thus causing the internal energy to increase

by an amount ofUjoule.

At the same time, when its temperature increases, its

volume increases too.

When the volume of the gas increases, work is done by the

gas (W).

Translation :

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Example 15.1

A 2500 J heat is added to a system and 1800 J work is

done on the system. Calculate the change in internal

energy of the system.

Solution

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Example 15.2

The work done to compress one mole of a monoatomic

ideal gas is 6200 J. The temperature of the gas changes

from 350 to 550 K.

a) How much heat flows between the gas and its

surroundings ?

b) Determine whether the heat flows into or out of the gas.

Solution

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Solution 15.2

K550K350

J6200

21 ,

,3,1

TT

Wfn

a)

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Example 15.3

A gas in a cylinder expands from a volume of 0.400 m3to

0.700 m3. Heat is added just rapidly enough to keep the

pressure constant at 2.00 x 105Pa during the expansion.The total heat added is 1.40 x 105 J.

Calculate the work done by the gas and the change in

internal energy of the gas.

Solution

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Exercise

1. A system absorbs 200 J of heat as the internal energy

increases by 150 J. What work is done by the gas ?

50 J

2. In a chemical laboratory, a technician applies 340 J of

energy to a gas while the system surrounding the gas

does 140 J of work on the gas. What is the change ininternal energy ?

480 J

17

3. 8000 J of heat is removed from a refrigerator by a

compressor which has done 5000 J of work. What is the

change in internal energy of the gas in the system?

-3000 J


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Learning Outcome:


Define the followingthermodynamics processes:

Isothermal, U= 0

Isovolumetric,W= 0

Isobaric, P= 0

Adiabatic,Q= 0

SketchPVgraph for all the thermodynamic processes.

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15.2 Thermodynamics processes (1 hour)


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15.2 Thermodynamics processes

1) Isothermalprocess

2) Isochoric (isovolumetric)process3) Isobaricprocess

4) Adiabaticprocess

There are 4 common processes of thermodynamics:

(iso = same)T V,P

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1) Isothermalprocess

Isothermalprocess is defined as a process that occurs

at constant temperature.

0U

0=22

=

-=

12

12

nRTf

nRTf

U

UUU

-

T1= T2

WQU From flot,

then0If =U WQ =

PV = constant

2211 VPVP

From the Boyles law :

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Isochoric (isovolumetric)process is defined as a

process that occurs at constant volume.

0=W 0- == 12 VVPW V1= V2

WQU From flot,

then0If =W

QU=

2) Isochoric (isovolumetric)process

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Adiabaticprocess is defined as a process that occurs

without the transfer of heat (into or out of the system).

0=Q WQU From flot,

then0If =Q

WU -=

Isobaricprocess is defined as a process that occursat constant pressure.

12 VVPQU

WQU From flot,

VPW

3) Isobaric process

4) Adiabatic process

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Pressure-Volume Diagram (graph) for

Thermodynamic Processes

3T

1T

P

V

AP

0AV

4T

2TB

E

DC

A

1234>>> TTTT

Path AB Isothermal process (TB=TA)

Path AC

Path AD

Path AE

Adiabatic process (TC


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24

Exercise

1. A gas system which undergoes an adiabatic process

does 5.0kJ of work against an external force. What is

the change in its internal energy?5000 J

2. A gas is compressed under constant pressure,

i) Sketch the pressurevolume graph.

ii) How is the work done in compressing the gascalculated?

iii) Explain what will happen to the final temperature of

the gas.

3. A gas undergoes the following thermodynamicsprocesses: isobaric expansion, heated at constant

volume, compressed isothermally, and finally expands

adiabatically back its initial pressure and volume. Sketch

all the processes given on the same P-V graph.


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Learning Outcome:


Deriveexpression for work ,

Determinework from the area under the p-V graph Derivethe equation of work done in isothermal,

isovolumetric, and isobaric processes.

Calculatework done in

isothermal process and use

isobaric process, use

isovolumetric process, use 25

15.3 Thermodynamics work (2 hours)

W PdV

2

1

1

2 lnlnppnRT

VVnRTW

)( 12 VVpPdVW

0 PdVW


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GasA

A

dx

Initial

Final

Consider the infinitesimal work done by the gas (system) during

the small expansion, dxin a cylinder with a movable piston asshown in Figure 15.3.

Suppose that the cylinder has a cross sectional area,Aand the

pressure exerted by the gas (system) at the piston face isP.

Work done in the thermodynamics system

Figure 15.3

F

15.3 Thermodynamics work


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The work, dWdone by the gas is given by

In a finite change of volume from V1to V

2,

PAF0cosFdxdW where andPAdxdW and dVAdx PdVdW

2

1

V

V

PdVW

donework:Wwhere

2

1

V

V PdVdW

pressuregas:Pgastheofvolumeinitial:1V

gastheofvolumefinal:2V

(16.1)


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1V

P

V2V

1P

2P

0

1

20W

PVdiagram Work done = area under the P-V graph

2V

P

V1V

2P

1P

0

2

1

0W

1V

P

V2V

1P

0

1 2

0VVPW 121

1V

P

V

2P

1P

0

1

2

0W

isothermal

expansion

isothermal

compression

isobaric expansionisochoric

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1) Isothermal

WQ =

0U

2

1

2

1

=== V

V

V

V

dVV

nRTPdVWQ

1

2ln=

V

VnRTW

2

1ln=

PPnRTW

From Boyles law :

2

1

1

2

P

P

V

V

2211 VPVP

0== WQU -

Equation of work done in thermodynamic processes

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2) Isochoric (isovolumetric)

3) Isobaric

)(==12

VVPVPW -

Since the volume of the system in isovolumetric process

remains unchanged, thus

Therefore the work done in the isovolumetric process is

0 PdVW

0dV

The work done during the isobaric process which change of

volume from V1to V

2is given by

2

1

V

V PdVW and constantP

2

1

V

V

dVPW

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Work done at constant

volume

Work done at

constant

pressure


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Example 15.4

How much work is done by an ideal gas in expanding

isothermally from an initial volume of 3.00 liters at 20.0

atm to a final volume of 24.0 liters?

SolutionV1= 3.00 liters, V2= 24.0 liters ,

P= 20.0 atm

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Two liters of an ideal gas have a temperature of 300 K

and a pressure of 20.0 atm. The gas undergoes an

isobaric expansion while its temperature is increased to

500 K. What work is done by the gas ?

Example 15.5

Solution T1= 300 K, T2= 500 K ,P= 20.0 atm, V1 =2 liters

32

E l 15 6


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Example 15.6

(a) Write an expression representing

i. the 1stlaw 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 pressurePversus volume Vof 1 mole of

ideal gas. Label and show clearly the four thermodynamics

process. [5 marks]

(Exam.Ques.intake 2003/2004)

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Solution 15.6

a) i. 1st law of thermodynamics:

ii. Work done at variable pressure:

where

or

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b)PVdiagram below represents four thermodynamic

processes:

35

Example 15 7


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Example 15.7

In a thermodynamic system, the changing of state for that

system shows by thePV-diagram below.

0.2

PaPx /104

33 /10 mVx 0.5

0.8

0

B

C

D

A0.3

In process AB, 150 J of heat is added to the system

and in process BD, 600 J of heat is added. Determine

a. the change in internal energy in process AB.

b. the change in internal energy in process ABD.

c. the total heat added in process ACD. 36

Solution 15 7


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QAB= 150 J,QBD= 600 J, VA=VB= 2.0x10-3m3,

VC=VD= 5.0x10-3m3, PA=PC= 3x10

4Pa, PB=PD= 8x104Pa

Solution 15.7

37

PaPx /104Solution 15 7


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b. The work done in process

ABD is

Therefore, the change in internal energy :

The total heat transferred

in process ABD is given

by

0.2

PaPx /10

33

/10 mVx

0.5

0.8

0

B

C

D

A0.3

Solution 15.7

QAB= 150 J,QBD= 600 J,VA=VB= 2.0x10

-3m3,

VC=VD= 5.0x10-3m3,PA=PC= 3x10

4Pa,PB=PD= 8x10

4Pa

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c. The change in internal

energy in process ACD is

0.2

PaPx /104

3 /10 mVx 0.5

0.8

0

B

C

D

A0.3

Solution 15.7

QAB= 150 J,QBD= 600 J,V

A

=VB

= 2.0x10-3m3,

VC=VD= 5.0x10-3m3,

PA=PC= 3x104Pa,

PB=PD= 8x104Pa

The work done inprocess ACD is given by

Therefore, the total

heat transferred :

39

Example 15 8


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Example 15.8

A gas in the cylinder of a diesel engine can undergo cyclic

processes. Figure below shows one cycle ABCDA that is

executed by an ideal gas in the engine mentioned.

401.

Pa10xP 5/

34

m10xV

/

01.

0010.0

B C

D

A

006.

87.

016.

a. If the temperature of the gas in states A and B are 300 K

and 660 K, respectively. Calculate the temperature in

states C and D.

b. Determine the work done by the gas in process BC.

40

Solution 15 8


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Solution 15.8

PB=PC= 16.0x105Pa,PA= 1.0x10

5Pa, PD= 7.8x105Pa,

VA=VD= 10.0x10-4m3, VB= 1.40x10

-4m3, VC= 6.00x10-4m3

a. Given TA= 300 Kand TB= 660 K

401.

Pa10xP 5/

34

m10xV

/

01.

0010.0

B C

D

A

006.

87.

016.

41

Solution 15 8


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Solution 15.8

PB=PC= 16.0x105Pa,PA= 1.0x10

5Pa,

PD= 7.8x105Pa, VA=VD= 10.0x10

-4m3, VB= 1.40x10-4m3,

VC= 6.00x10-4m3

401.

Pa10xP 5/

34 m10xV /

01.

0010.0

B C

D

A

006.

87.

016.

b. Process BC occurs at constant pressure, thus the work

done by the gas is given by

42

Example 15 9


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Example 15.9

(a) Define (i) the adiabatic compression process

(ii) the reversible process [2 marks]

(b) One mole of an ideal monatomic gas is at the initial temperature of

650 K. The initial pressure and volume of the gas isP0and V0,

respectively. At initial stage, the gas undergoes isothermal

expansion and its volume increase to 2V0. Then, this gas through

the isochoric process and return to its initial pressure. Finally, the

gas undergoes isobaric compression so that it return to its initial

temperature, pressure and volume.(i) Sketch the pressure against volume graph for the entire

process. [4 marks]

(ii) By using the 1st law of thermodynamics, proved that the

heat,

where nis the number of moles,Ris molar gas

constant and T is the absolute temperature. Then, calculate

the total heat for the entire process. [8 marks]

(iii) State whether the heat is absorbed or released. [1 mark]

(UseR= 8.31 J K-1 mol-1)(Exam.Ques.intake 2001/2002)

000 VP2nRTQ ln

43

Solution 15 9


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Solution 15.9

44

Solution 15 9


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Solution 15.9

(b) One mole of an ideal monatomic gas is at the initial temperature of

650 K. The initial pressure and volume of the gas isP0and V0,

respectively. At initial stage, the gas undergoes isothermal

expansion and its volume increase to 2V0. Then, this gas through

the isochoric process and return to its initial pressure. Finally, the

gas undergoes isobaric compression so that it return to its initial

temperature, pressure and volume.

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Solution 15 9


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0T

0P

PPressure,

V

Volume,0 0V

1T

B

CA

1P

0V2

(i) Sketch the pressure against volume graph for the entire

process.

Solution 15.9

46

Solution 15 9


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ii. From thePVdiagram,

Solution 15.9

0T

0P

PPressure,

VVolume,0

0V

1T

B

CA

1P

0

V2

In process AB

(Isothermal process):

In process CA (isobaric

process):

Using Charless law, hence

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Solution15 9


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The work done inisobaric process:

Solution15.9

0T

0P

PPressure,

VVolume,0

0V

1T

B

CA

1P

0

V2

In process BC (isochoric process):

48

Solution 15 9


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Solution 15.9

0T

0P

PPressure,

VVolume,0

0V

1T

B

CA

1P

0V2

The total heat, Qfor

entire process is given

by

.. proved

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Example 15 10


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Example 15.10

(a) State i. the isobaric process.

ii. the isothermal process.

iii. the adiabatic process. [3 marks]

(b) State the 1stlaw of thermodynamics. [2 marks]

50

Solution 15 10


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Solution 15.10

b. 1stlaw of thermodynamics :

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Exercise


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52

Exercise

Two moles of ideal gas are at a temperature of 300K and

pressure 2.5 x 105Pa. The gas expands isothermally to

twice its initial volume, and then undergoes isobariccompression to its initial volume.

i) Calculate the initial volume of the gas.

ii) What is the pressure of the gas after the gasexpands isothermally to twice its initial volume?

iii) What is the final temperature of the gas after being

compressed isobarically?

iv) Calculate the work done in the isothermal

expansion.v) Draw the P-V graph for the processes above.

0.02 m3,1.3 x 105Pa, 150 K, 3.5 x 103J


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Chapter 15 Students Physics

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