Physics Thermo

8/12/2019 Physics Thermo

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PHYSICS CHAPTER 16

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The study of relationships

involving heat,heat,

mechanical work, andmechanical work, and

other aspects of ener!other aspects of ener!and ener! transferand ener! transferfor

the system.

Thermod!namic s!stemThermod!namic s!stemisany collection of objects that is

convenient to regard as a unit,

and that ma! ha"e thema! ha"e the

potential ener! to e#chanepotential ener! to e#chane

ener! with its s$rro$ndinsener! with its s$rro$ndins.

CHAPTER 16%CHAPTER 16%Thermod!namicsThermod!namics

&' Ho$rs(&' Ho$rs(


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PHYSICS CHAPTER 16

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At the end of this chapter, st$dents sho$ld )e a)le to%At the end of this chapter, st$dents sho$ld )e a)le to%

*istin$ish*istin$ish)etween thermod!namic work done on the)etween thermod!namic work done on the

s!stem and work done )! the s!stem+s!stem and work done )! the s!stem+

*eri"e*eri"ee#pression for work,e#pression for work,

andand determinedeterminework from the area $nder thework from the area $nder the p-Vp-Vraph+raph+

State and $seState and $sefirst law of thermod!namics,first law of thermod!namics,

earnin -$tcome%

16+1 .irst law of thermod!namics &/ ho$rs(

WUQ +=

= PdVW


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PHYSICS CHAPTER 16

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16+1 .irst law of thermod!namics16+1+1 Sins for heat, Qand work, W Sign convention for heat, Q:

Q= positi"epositi"evalue

Q= neati"eneati"evalue Sign convention for work, W:

W= positi"epositi"evalueW= neati"eneati"evalue

Figures !. show a thermodynamic system may e"changeenergy with its surroundings #environment$.

Heat flow into the s!stemHeat flow into the s!stem

Heat flow o$t of the s!stemHeat flow o$t of the s!stem

0ork done )! the s!stem0ork done )! the s!stem

0ork done on the s!stem0ork done on the s!stem

Surroundings

#environment$

System

0>Q 0=W

#a$

Surroundings

#environment$

System

0


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PHYSICS CHAPTER 16

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Surroundings

#environment$

System

0=Q 0>W

Surroundings

#environment$

System

0

=Q

0Q 0>W

Surroundings

#environment$

System

0


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PHYSICS CHAPTER 16

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%"ample for work done by the system and work done on thesystem are shown in Figure !.&.

'hen the air is e#pandede#panded, the molecule loses kineticloses kinetic

ener!ener!and does positi"e workpositi"e workon piston.

'hen the air is compressedcompressed, the molecule ains kineticains kineticener!ener!and does neati"e workneati"e workon piston.

(ir

)ompression

(ir

%"pansion

(ir

*nitially

.i$re 16+/.i$re 16+/

+otion ofpiston

+otion ofpiston


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PHYSICS CHAPTER 16

7

asA

A

dx

*nitial

Final

)onsider the infinitesimal work done by the gas #system$ during

the small e"pansion, dxin a cylinder with a movable piston asshown in Figure !.-.

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

pressure e"erted by the gas #system$ at the piston face isP.

16+1+/ 0ork done in the thermod!namics s!stem

.i$re 16+.i$re 16+

F


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PHYSICS CHAPTER 16

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The work, dWdone by the gas is given by

*n a finite change of volume from V1to V

2,

PAF

=

0=cosFdxdW= where andPAdxdW= and dVAdx=PdVdW=

=2

1

V

VPdVW

donework:Wwhere

=2

1

V

V PdVdW

pressuregas:P

gastheofvolumeinitial:1Vgastheofvolumefinal:2V

#!.$


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PHYSICS CHAPTER 16

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For a change in volume at constant pressure,P

For any process in the system which the volume is constant #no

change in volume$, the work done is

Figures !. show the pressure,Pagainst volume, Vgraph

#PVdiagram$ where

( )12

VVPW =

VPW =0ork done at constant0ork done at constant

press$repress$re

0=W 0ork done at constant0ork done at constant"ol$me"ol$me

Area $nder theArea $nder the PPVVraph 2 0ork doneraph 2 0ork done


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PHYSICS CHAPTER 16

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.i$re 16+3.i$re 16+3

#a$ #b$

1V 2V

1P

2P

P

V0

1

2

0>W2V 1V

2P

1P

P

V0

2

1

0= VVPW

P

V0

2P

1P

1V

1

2

0=W

Stimulation !.
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PHYSICS CHAPTER 16

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states : /The heat &The heat &QQ( s$pplied to a s!stem is e4$al to the( s$pplied to a s!stem is e4$al to theincrease in the internal ener! & increase in the internal ener! &UU( of the s!stem( of the s!stempl$s the work done & pl$s the work done &WW( )! the s!stem on its( )! the s!stem on itss$rro$ndins s$rro$ndins.0

12

For infinitesimal change in the energy,

16+1+ .irst law of thermod!namics

WUQ +=suppliedheatofquantity:Q

energyinternalinitial:1U

where

energyinternalfinal:2U

donework:W

and 12 UUU =

energyinternalin thechange:U

dWdUdQ +=

#!.&$


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PHYSICS CHAPTER 16

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2P

The first law of thermodynamics is a enerali5ation of theenerali5ation of the

principle of conser"ation of ener!principle of conser"ation of ener!to include energy transfer

through heat as well as mechanical work. The chane in the internal ener!chane in the internal ener!&&UU((of a system during

any thermodynamic process is independent of pathindependent of path. Fore"ample a thermodynamics system goes from state to state &as shown in Figure !.3.

23124121 == UUU

1V

3

4

1P

2V.i$re 16+'.i$re 16+'

2

1

P

V0


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PHYSICS CHAPTER 16

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The internal ener! dependsinternal ener! dependsonly on temperat$retemperat$reof thesystem. *f the initial and final temperat$re &state(initial and final temperat$re &state(of the

system is the samesame, hence

because

thus

The chane in the internal ener!chane in the internal ener!also 5ero in the c!clic5ero in the c!clic

thermod!namic process &repeated process(thermod!namic process &repeated process(because the

initial and final state of the system is the same. The internal ener!internal ener! isnot dependnot dependon the "ol$me"ol$meof the

system.

nRT2

fU

=

012 == UUU

12 UU =


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PHYSICS CHAPTER 16

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( vessel contains an ideal gas at pressure 34 k5a. 'hen the gas

is heated it e"pands at constant pressure until the temperature

increases by 44 6. The amount of heat absorbed by the

gas is

.-! k7. 8owever, if the gas at its initial condition is heated at

constant volume until the temperature increases by 44 6, theamount of heat absorbed is -. k7. 9etermine

a. the value of ,

b. the work done by the gas when it e"pands at constant

pressure,c. the change in volume of the gas when the gas is heated at

constant pressure and the temperature rises by 44 6.

E#ample 1 %


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PHYSICS CHAPTER 16

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Sol$tion %Sol$tion %

a. y applying the e;uation for molar heat capacities, For constant pressure:

For constant volume:

#&$ #$:

PPP TnCQ =

K;100a;101!0 3 ==== VPVP TTPP";1011#3";103$#4 33 == VP QQ

( )100PP nCQ = #$

VVV TnCQ =( )100VV nCQ = #&$

V

P

V

P

C

C

Q

Q=

3

3

1011#3103$#4

== VP

CC

40#1=


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PHYSICS CHAPTER 16

16


b. y using the stlaw of thermodynamics, For constant pressure:

For constant volume:

#&$ %#$:

c. The change in the volume of the constant pressure process is

K;100a;101!0 3 ==== VPVP TTPP";1011#3";103$#4 33 == VP QQ

VV WUQ +=VQU= #&$

and 0=VW

PPV WQQ =

PW=33 103$#41011#3

"102!#1

3

=PWPPP VPW =

( ) PV=33 101!0102!#1

33

m1033#&

= PV

PP WUQ +=PP WQU = #$


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PHYSICS CHAPTER 16

17

( vessel contains an ideal gas in condition (, as shown in Figure

!.!. 'hen the condition of the gas changes from ( to that of ,

the gas system undergoes a heat transfer of 4.3 k7. 'hen the gas

in condition changes to condition ), there is a heat transfer of-.& k7. )alculate

a. the work done in the process (),

b. the change in the internal energy of the gas in the process (),

c. the work done in the process (9),

d. the total amount of heat transferred in the process (9).

E#ample / %

0#2

'ka(P

'm10(32

V0#4

300

0)

*+

,

1!0.i$re 16+6.i$re 16+6


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PHYSICS CHAPTER 16

18


a. The work done in the process () is given by

b. y applying the stlaw of thermodynamics for (), thus

)*,),)* WWW +=

a;10300a;101!0 3+*3

), ==== PPPP32

*)

32

+, m100#4;m100#2 ==== VVVV

"102#3";10!#10 3)*3,) +=+= QQ

but 0)*=W( )

,),,)*

VVPW =( )223,)* 100#2100#4101!0

=W"3000,)*=W

,)*,)*,)* WUQ += ( ) ,)*)*,),)* WQQU +=

"100-#1

4

,)* =U

( ) 3000102#310!#10 33,)* +=U


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PHYSICS CHAPTER 16

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c. The work done in the process (9) is given by

d. y applying the stlaw of thermodynamics for (9), thus

+*,+,+* WWW +=

a;10300a;101!0 3+*3

), ==== PPPP32

*)

32

+, m100#4;m100#2 ==== VVVV

"102#3";10!#10 3)*3,) +=+= QQ

but 0,+=W( )

+*+,+*

VVPW =( )223,+* 100#2100#410300

=W"$000,+*=W

,+*,+*,+* WUQ +=,+*,)*,+* WUQ +=

"10$-#1

4

,+* =Q

$000100-#1 4,+* +=Q

and,)*,+* UU =


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PHYSICS CHAPTER 16

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E#ercise 16+1 %

ivenR= (during process

( is &4.4 7, no energy is transferred asheat during process ), and the nett work

done during the cycle is 3.4 7.

AS+ %AS+ % '+7 8'+7 8.i$re 16+:.i$re 16+:


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PHYSICS CHAPTER 16

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At the end of this chapter, st$dents sho$ld )e a)le to%At the end of this chapter, st$dents sho$ld )e a)le to% State and e#plainState and e#plainthermod!namics processes%thermod!namics processes%

Isothermal,Isothermal, UU= 0= 0

Adia)atic,Adia)atic, QQ= 0= 0 Iso"ol$metric,Iso"ol$metric, WW= 0= 0 Iso)aric,Iso)aric, PP= 0= 0

SketchSketchPPVVraph toraph to distin$ishdistin$ish)etween isothermal)etween isothermalprocess and adia)atic process+process and adia)atic process+

*etermine*eterminethe initial and final state of thermod!namicthe initial and final state of thermod!namicprocesses )! $sin the followin form$la%processes )! $sin the followin form$la%

earnin -$tcome%

16+/ Thermod!namics processes &/ ho$rs(

constant=PV.or isothermal process%.or isothermal process%

.or adia)atic process %.or adia)atic process % constant=

PV andand constant1

=

TV


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PHYSICS CHAPTER 16

22

16+/ Thermod!namics processes There are four specific kinds of thermodynamic processes. *t is

*sothermalprocess (diabatic process

*sochoric#isovolumetric$ process

*sobaric process

16+/+1 Isothermalprocess is defined as a process that occ$rs at constant temperat$rea process that occ$rs at constant temperat$re.

i.e.

ThusIsothermal chanesIsothermal chanes

'hen a gas e"pands or compresses isothermally #constanttemperature$ thus

0=U

WQ=WUQ +=

constant=PV #!.-$


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PHYSICS CHAPTER 16

23

%;uation #!.-$ can be e"pressed as

*f the gas e#pand isothermall!e#pand isothermall!, thus V2.V

1

*f the gas compress isothermall!compress isothermall!, thus V2/V

1

16+/+/ Adia)atic process is defined as a process that occ$rs witho$t heat transfer intoa process that occ$rs witho$t heat transfer into

or o$t of a s!stemor o$t of a s!stemi.e.

For e"ample, the compression stroke in an internal combustionengine is an appro"imately adiabatic process.

2211 VPVP =WW 2 positi"e2 positi"e

WW 2 neati"e2 neati"e

(nimation !.

WUUU

== 12

0=Q WUQ +=thus
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PHYSICS CHAPTER 16

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?ote :

For (diabatic e"pansion #V2.V

1$, W2 positi"e "al$e2 positi"e "al$ebut

U2neati"e "al$e2neati"e "al$ehence the internal ener!internal ener!of thesystem decreasesdecreases.

For (diabatic compression #V2/V

1$, W2 neati"e "al$eneati"e "al$ebut

U2positi"e "al$epositi"e "al$ehence the internal ener!internal ener!of the

system increasesincreases.Adia)atic chanesAdia)atic chanes

)onsider the stlaw of thermodynamics written in differential#infinitesimal$ form:

Since that the internal ener!internal ener!is independent of the "ol$meindependent of the "ol$meand is related to the temperat$re )! the molar heat capacit!related to the temperat$re )! the molar heat capacit!

at constant "ol$meat constant "ol$me, therefore

dWdUdQ +=

dTnCdQdU VV==

#!.$


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PHYSICS CHAPTER 16

25

Then e;uation #!.$ can be e"pressed as

E4$ation of adia)atic chanes in temperat$re and "ol$meE4$ation of adia)atic chanes in temperat$re and "ol$me

From the e;uation of state for an ideal gas,

Therefore

dWdTnCdQV

+= PdVdW=#!.3$

where and 0

=dQ

0=+PdVdTnCV

nRTPV= then VnRT

P= substitute in e;. #!.3$

0=

+ dVV

nRTdTnCV

0=

+

V

dV

C

R

T

dT

V

dVV

RT

dTCV

=VP CCR =and


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PHYSICS CHAPTER 16

26

For finite changes in temperature and volume, integrate e;.#!.!$, hence

0=

+

V

dV

C

CC

T

dT

V

VP

and01 =

+

V

dV

C

C

T

dT

V

P =V

P

C

C

( ) 01 =+ VdV

T

dT #!.!$

( )

=+ 01

V

dV

T

dT

( ) constantlnln 1 =+ VT( ) constantln1ln =+ VT

( ) constantln 1 =TV


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PHYSICS CHAPTER 16

27

?ote :

Adia)atic e#pansion &Adia)atic e#pansion &dVdV > 0> 0((always occurs with adrop in temperat$re &drop in temperat$re &dTdT < 0< 0((.

Adia)atic compression &Adia)atic compression &dVdV < 0< 0((always occurs with a

rise in temperat$re &rise in temperat$re &dTdT > 0> 0((.

constant1 =TV

12

1

22

1

11

= VTVT

#!.$

where

retemperatuasoluteinitial:1Tretemperatuasolutefinal:2T

volumeinitial:1V volumefinal:2V


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PHYSICS CHAPTER 16

28

E4$ation of adia)atic chanes in press$re and "ol$meE4$ation of adia)atic chanes in press$re and "ol$me

2earrange the e;uation of state for an ideal gas,

Therefore

nRTPV= thennRPVT= substitute in e;. #!.$

constant1 =

V

nR

PV

constant=PV

12

2211 VPVP =

where pressureinitial:1P

pressurefinal:2P

#!.


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PHYSICS CHAPTER 16

29

is defined as a process that occ$rs at constant "ol$mea process that occ$rs at constant "ol$mei.e.

*n an isochoric process, all the ener! addedener! addedas heat remainsin the system as an

increaseincrease in the

internal ener!internal ener!thus the

temperat$retemperat$reof the system increasesincreases. For e"ample, heating a gas in a closed constant volume

container is an isochoric process.

16+/+3 Iso)ar)aric is defined as a process that occ$rs at constant press$rea process that occ$rs at constant press$rei.e.

For e"ample, boiling water at constant pressure is an isobaricprocess.

16+/+ Isochoric &Iso"ol$metric"ol$metric(

WUQ +=0=W thus

12 UUUQ ==

WUQ +=VPW = thus

VPUQ +=

(nimation !.&

(nimation !.-

0=P and
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PHYSICS CHAPTER 16

30

1V

2P

3P

3V

Figure !.< shows aPVdiagram for each thermodynamicprocess for a constant amount of an ideal gas.

16+/+3 Press$re;,

.i$re 16+=.i$re 16+=

*


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PHYSICS CHAPTER 16

31

From the Figure !.>$ andadiabatic e"pansions #AACC$: The temperat$re falltemperat$re fall&&TT

CC


32/57

PHYSICS CHAPTER 16

32

(ir is contained in a cylinder by a frictionless gas@tight piston.

a. )alculate the work done by the air as it e"pands from a volume of

4.43 m- to a volume of 4.4& m- at a constant pressure of

&.4 435a.

b. 9etermine the final pressure of the air if it starts from the same

initial conditions as in #a$ and e"panding by the same amount, the

change occurs

i. isothermally,

ii. adiabatically.

#iven for air is .4$

E#ample %


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PHYSICS CHAPTER 16

33


a. iven

The work done by the air is

b. i. The final pressure for the isothermal process is

ii. The final pressure for the adiabatic process is given by

a100#2;m02-#0;m01!#0 !13

2

3

1 === PVV

( )121 VVPW =( )01!#002-#0100#2 ! =W

"2400=W

2211 VPVP =( )( ) ( )02-#001!#0100#2 2

! P=a1011#1 !2 =P

2211 VPVP =( )( ) ( ) 40#12

40#1! 02-#001!#0100#2 P=

a10-&

4

2 =P


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PHYSICS CHAPTER 16

34

( vessel contains an ideal gas of volume &.4 cm-at pressure

44 k5a and temperature &3 ). The gas e"pands adiabatically

until the volume becomes .4 cm-. (fter that, it is compressed

isothermally until the volume becomes -.4 cm-.

a. )alculate

i. the pressure and ii. the temperature of the gas in the final condition.

b. Sketch and label a graph of gas pressure #P$ against gas volume

#V$ to show how the pressure and volume changes when the

condition of the gas changes from the initial condition to finalcondition.

#iven for gas is .!$

E#ample 3 %


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PHYSICS CHAPTER 16

35


a. i. For the adiabatic e"pansion,

For the isothermal compression,

K1!#2&a;10100;m100#2 13

1

3$

1 === TPV

$-#1;m100#3;m100#4 3$33$

2 === VV

2V

2T2P

Adia)aticAdia)atic

e#pansione#pansion

IsothermalIsothermal

compressioncompression

1V

1T1P

InitialInitial

3V

23 TT =3P

.inal.inal

2211 VPVP =

a1014#3 42 =P( )( ) ( ) $-#1$2

$-#1$3 100#4100#210100 = P

3322 VPVP =

a101#4

4

3 =P

( )( ) ( )$3$4 100#3100#41014#3 = P


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PHYSICS CHAPTER 16

36


a. ii. For the adiabatic e"pansion,

For the isothermal compression, the temperature along the

process remains $nchaned$nchanedhence

K1!#2&a;10100;m100#2 13

1

3$

1 === TPV

$-#1;m100#3;m100#4 3$33$

2 === VV

1

22

1

11

= VTVT

K1&-2=T

( ) ( ) ( ) 1$-#1$21$-#1$ 100#4100#21!#2& = T

K1&-23 == TT


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PHYSICS CHAPTER 16

37


b. The graph of gas pressure #P$ against gas volume #V$ is shown

in Figure !.A.

K1!#2&a;10100;m100#2 13

1

3$

1 === TPV

$-#1;m100#3;m100#4 3$33$

2 === VV

0#3

1#4

0#2

a'10( 4P

'm10( 3$V0

00#10

1T

2T

3

1

14#3

0#4

2

.i$re 16+?.i$re 16+?

S CS C 6


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PHYSICS CHAPTER 16

38

At the end of this chapter, st$dents sho$ld )e a)le to%At the end of this chapter, st$dents sho$ld )e a)le to% *eri"e*eri"ethe e4$ation of work done in isothermal,the e4$ation of work done in isothermal,

iso"ol$metric and iso)aric processes+iso"ol$metric and iso)aric processes+

Calc$lateCalc$latework done inwork done in

isothermal process and $seisothermal process and $se

iso)aric process, $seiso)aric process, $se

iso"ol$metric process, $seiso"ol$metric process, $se

earnin -$tcome%16+ Thermod!namics work &1 ho$r(

=

=

2

1

1

2 lnlnP

PnRT

V

VnRTW

( )12 VVPPdVW ==

0==

PdVW

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

39

16+ Thermod!namics work

16++1 0ork done in Isothermalprocess From the e;uation of state for an ideal gas,

Therefore the work done in the isothermal process which

change of volume from V1to V2, is given by

nRTPV=V

nRTP=then

=2

1

V

VPdVW

=2

1

V

V dVV

nRTW

= 21

1V

VdV

VnRTW

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

40

For isothermal process, the temperature of the system remains

unchanged, thus

The e;uation #!.A$ can be e"pressed as

=

1

2lnV

VnRTW

( )12 lnln VVnRTW =

#!.A$

2211 VPVP =2

1

1

2

P

P

V

V=

=

2

1lnP

PnRTW #!.4$

[ ] 21

ln V

VVnRTW=

PHYSICS CHAPTER 16


41/57

PHYSICS CHAPTER 16

41

y applying the stlaw of thermodynamics, thus

16++/ 0ork done in iso)ar)aric process The work done during the isobaric process which change of

volume from V1

to V2

is given by

WUQ

+=0=Uand

WQ=

=

=

2

1

1

2 lnln

P

PnRT

V

VnRTQ

=2

1

V

VPdVW and constant=P

= 2

1

V

V

dVPW

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

42

16++ 0ork done in Iso"ol$metric"ol$metric process Since the volume of the system in isovolumetric process remains

unchanged, thus

Therefore the work done in the isovolumetric process is

( )12 VVPW =

12

VPW =

0== PdVW

0=dV

#!.4$

#!.$

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

43

( ;uantity of ideal gas whose ratio of molar heat capacities is !3has a temperature of -44 6, volume of ! 4-m- and pressure of

&- k5a. *t is made to undergo the following three changes in order:

: adiabatic compression to a volume & 4-m-,

& : isothermal e"pansion to ! 4-m- ,

- : a return to its original state.a. )alculate

i. the pressure on completion of process ,

ii. the temperature at which the process & occurs.

b. 9escribe the process -.

c. Sketch and label a graph of pressure against volume for the

changes described.

E#ample ' %

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

44


a. i. The pressure on completion of the adiabatic compression is

2V

2T2P

;m10$4K;300a;10243 33113

1

=== VTP

3

!;m102- 332 ==

V

2211 VPVP =

Adia)aticAdia)atic

compressioncompression

&Process 1(&Process 1(


e#pansione#pansion

&Process /(&Process /(

1V

1T1P

InitialInitial

13 VV =23 TT =

3P

.inal.inal

&Process (&Process (

( )( ) ( ) 3!

3!

3

2

33 102-10$410243 = P

a1002#1 $2 =P

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

45


a. ii. y applying the e;uation of adiabatic changes in temperature

and volume for process ,

b. 5rocess - is a process at constant "ol$me known asprocess at constant "ol$me known as

iso"ol$metric &isochoric(iso"ol$metric &isochoric(.

;m10$4K;300a;10243 33113

1

=== VTP

3

!;m102- 332 ==

V

1

22

1

11

= VTVT( ) ( ) ( )

13

2

13 3!3! 102-10$4300 = TK!332=T

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

46


c. The graph of gas pressure #P$ against gas volume #V$ for the

changes described is shown in Figure !.4.

;m10$4K;300a;10243 33113

1

=== VTP

3

!;m102- 332 ==

V

3

P

2-

a'10( 4P

'm10( 33V0

102

K!33

K3003#24

$4

.i$re 16+17.i$re 16+17

1

2

3

Process /Process /

Process Process

Process 1Process 1

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

47

( vessel of volume


48/57

PHYSICS CHAPTER 16

48


a. 'hen the gas e"pands adiabatically, it does positi"e workpositi"e work.

Thus

The internal ener!internal ener!of the gas is red$cedred$cedto provide thenecessary energy to do work. Since the internal ener!internal ener!is

proportionalproportionalto the a)sol$te temperat$rea)sol$te temperat$rehence the

temperat$re decreasestemperat$re decreasesand resulting a temperature change.

WUQ +=

Adia)aticAdia)atic

e#pansione#pansionIsochoricIsochoric

processprocess

1V

1T1P

InitialInitial

23 VV =13 TT =

3P.inal.inal

WU =

2V

2T2P

;m1000#&a;1014#1 331!

1

== VPa100$#1a;1001#1 !3

!

2 == PP

0=Qand

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

49


b. From the adiabatic e"pansion:

From the isochoric #constant volume$ process:

y substituting e;. #-$ into e;. #&$,

2211 VPVP =

;m1000#&a;1014#1 331!

1

== VPa100$#1a;1001#1 !3

!

2 == PP

=2

1

1

2

V

V

P

P#$

122

111

= VTVT1

2

1

1

2

=

VV

TT #&$

3

3

2

2

T

P

T

P

= 32

1

2

P

P

T

T

=13 TT =and #-$1

2

1

3

2

=

V

V

P

P#$

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

50


b. #$ #$ :

1

2

1

2

1

3

2

1

2

=

V

V

V

V

P

P

P

P

;m1000#&a;1014#1 331!

1

== VPa100$#1a;1001#1 !3

!

2 == PP

2

1

1

3

V

V

P

P

=

2

3

!

! 1000#&

1014#1

100$#1

V

=

33

2 m10$0#& =V

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

51


c. y substituting the value of V2 into the e;. #$, therefore thevalue is

d. The gas is monatomicmonatomic.

;m1000#&a;1014#1 331!

1

== VPa100$#1a;1001#1 !3

!

2 == PP

=

2

1

1

2

ln

ln

V

V

PP

=

2

1

1

2 lnlnV

V

P

P

=

3

3

!

!

10$0#&

1000#&

ln

1014#11001#1ln

$-#1=

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

52

a. 'rite an e"pression representing

i. the stlaw of thermodynamics and state the meaning of all the

symbols.

ii. the work done by an ideal gas at variable pressure. C- marksD

b. Sketch a graph of pressurePversus volume Vof mole of ideal

gas. Eabel and show clearly the four thermodynamics process. C3

marksD

c. ( monatomic ideal gas at pressurePand volume Vis compressed

isothermally until its new pressure is -P. The gas is then allowedto e"pand adiabatically until its new volume is AV. *fP, Vand for

the gas is .& 435a,.4 4&m-and 3- respectively, calculate

i. the final pressure of the gas.

ii. the work done on the gas during isothermal compression.

#%"amG>uesGintake &44-&44$ C marksD

E#ample : %

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

53


a. i. stlaw of thermodynamics:

ii. 'ork done at variable pressure:

b. 5H diagram below represents four thermodynamic processes:

WUQ +=ferredheat transofquantity:Q

energyinternalinchange:Uwherewhere

donework:W

=

1

2lnVVnRTW=

2

1

V

VPdVW -R-R

3T

1T

P

V

AP

0 AV

4T

2TB

E

DC

A

*sobaric process*sobaric process

*sochoric*sochoric

processprocess

*sothermal process*sothermal process

adiabaticadiabatic

processprocess

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

54


c. iven

From the isothermal compression process:

i. y using the e;uation of adiabatic changes in pressure and

volume, hence


compressioncompressionAdia)aticAdia)atic

e#pansione#pansion

VT

PInitialInitial

VV 2=2T

2P.inal.inal

1VTT=1PP 31=

$-#13

!;m100#1a;102#1 32! ==== VP

11VPPV=( ) 13 VPPV=

31

VV =

2211 VPVP =

( )

=

2

112

V

VPP

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

55


c. i.

ii. The work done during the isothermal compression is

a104-#1 32 =P( )

= V

V

PP332 ( )

$-#1

!2

2-1102#13 =

P

=V

VnRTW 1ln

"1032#1 3=W

=V

V

PVW 3ln

PVnRT=and

( )( )

=

31ln100#1102#1 2!W

PHYSICS CHAPTER 16


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PHYSICS CHAPTER 16

56

E#ercise 16+/ %

ivenR=


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PHYSICS CHAPTER 16

THE END

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