01 thermodynamic process theory21

8/21/2019 01 Thermodynamic Process Theory21

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88 Thermodynamic Processes

genius by Pradeep Kshetrapal

13.1 Introduction.

(1) Thermodynamics : It is a branch of science which deals with exchange of heat energy

between bodies and conversion of the heat energy into mechanical energy and viceversa.

(!) Thermodynamic system : " collection of an extremely large number of atoms or

molecules con#ned with in certain boundaries such that it has a certain value of pressure$

volume and temperature is called a thermodynamic system. "nything outside the

thermodynamic system to which energy or matter is exchanged is called its surroundings.Example % &as enclosed in a cylinder #tted with a piston forms the thermodynamic system

but the atmospheric air around the cylinder$ movable piston$ burner etc. are all the surroundings.

Thermodynamic system may be of three types

(i) 'pen system % It exchange both energy and matter with the surrounding.

(ii) losed system % It exchange only energy (not matter) with the surroundings.

(iii) Isolated system % It exchange neither energy nor matter with the surrounding.

() Thermodynamic variables and equation of state : " thermodynamic system can be

described by specifying its pressure$ volume$ temperature$ internal energy and the number of moles. These parameters are called thermodynamic variables. The relation between the

thermodynamic variables (P$ V $ T ) of the system is called e*uation of state.

+or µ moles of an ideal gas$ e*uation of state is PV , µ RT and for 1 mole of an it ideal gas is

PV , RT

+or µ moles of a real gas$ e*uation of state is RT bV V

aP µ µ

µ =−

+ )(

!

!

and for 1 mole of a real gas it is

RT bV V

aP =−

+ )(

!

(-) Thermodynamic equilibrium : hen the thermodynamic variables attain a steadyvalue i.e. they are independent of time$ the system is said to be in the state of thermodynamic

e*uilibrium. +or a system to be in thermodynamic e*uilibrium$ the following conditions must be

ful#lled.

(i) /echanical e*uilibrium % There is no unbalanced force between the system and its

surroundings.

(ii) Thermal e*uilibrium % There is a uniform temperature in all parts of the system and is same as

that of surrounding.

(iii) hemical e*uilibrium % There is a uniform chemical composition through out the system and

the surrounding.


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(0) Thermodynamic process : The process of change of state of a system involves change

of thermodynamic variables such as pressure P$ volume V and temperature T of the system. The

process is nown as thermodynamic process. 2ome important processes are

(i) Isothermal process (ii) "diabatic process (iii) Isobaric process (iv) Isochoric

(isovolumic process)

(v) yclic and noncyclic process (vi) 3eversible and irreversible process

13.2 Zeroth a! of Thermodynamics.

If systems A and B are each in thermal e*uilibrium with a third system C$ then A and B are in

thermal e*uilibrium with each other.

(1) The 4eroth law leads to the concept of temperature. "ll bodies in thermal e*uilibrium

must have a common property which has the same value for all of them. This property is called

the temperature.

(!) The 4eroth law came to light long after the #rst and seconds laws of thermodynamicshad been discovered and numbered. 5ecause the concept of temperature is fundamental to

those two laws$ the law that establishes temperature as a valid concept should have the lowest

number. 6ence it is called 4eroth law.

13.3 "uantities Involved in #irst a! of Thermodynamics.

(1) $eat % Q& : It is the energy that is transferred between a system and its environment

because of the temperature di7erence between them. 6eat always 8ow from a body at higher

temperature to a body at lower temperature till their temperatures become e*ual.

Important points

(i) 6eat is a form of energy so it is a scalar *uantity with dimension 9: !! −T ML .

(ii) ;nit % Joule (2.I.)$ Calorie (practical unit) and 1 calorie , -.! Joule

(iii) 6eat is a path dependent *uantity e.g. 6eat re*uired to change the temperature of a given

gas at a constant pressure is di7erent from that re*uired to change the temperature of same gas

through same amount at constant volume.

(iv) +or solids and li*uids % ∆Q , mL :+or change in state9 and ∆Q , mc∆T :+or change in

temperature9

+or gases when heat is absorbed and temperature changes %

T CQ V V ∆=∆ µ )( :+or constant volume9 and T CQ PP ∆=∆ µ )( :+or constantpressure9

(!) 'or( % W & : or can be de#ned as the energy that is transferred from one body to the

other owing to a force that acts between them

If P be the pressure of the gas in the cylinder$ then force

exerted by the gas on the piston of the cylinder F , PA

In a small displacement of piston through dx $ wor done by

the gas

dV Pdx PAdx F dW === .

∴ Total amount of wor done )( i V

ViV V PdV PdW W

−===∆ ∫ ∫

dx

F!PA


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Thermodynamic Processes 9)

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Important points

(i) i.e. system expands against some external force.

∆W , negative if i V V < i.e. system contracts because of some external force exerted

by the surrounding.

(iii) In P"V diagram or indicator diagram$ the area under P"V curve represents wor done.

W , area under P"V diagram

It is positive if volume increases (for expansion)

It is negative if volume decreases (for compression)

(iv) In a cyclic process wor done is e*ual to the area under the cycle.

It is positive if the cycle is clocwise.

It is negative if the cycle is anticlocwise.

(v) ∫ =

i

V

V dV PW

+rom this e*uation it seems as if wor done can be calculated only when P"V e*uation is

nown and limits iV and V are nown to us. 5ut it is not so. e can calculate wor done if we

now the limits of temperature.

+or example$ the temperature of # moles of an ideal gas is increased from =T to =!T

through a processT

P α = and we are interested in #nding the wor done by the gas. Then

PV , #RT (ideal gas e*uation) >..(i)

andT

P α = >..(ii)

?ividing (i) by (ii)$ we getα

!#RT V = or dT

#RT dV

α

!=

∴ =

!

!!=

=

#RT dT #RT

T dV PW

T

T

V

V

i

=

== ∫ ∫

α

α

2o we have found the wor done without putting the limits of volume.

(vi) If mass less piston is attached to a spring of force constant $ and a mass m is placed

over the piston. If the external pressure is =P and due to expansion of gas the piston moves up

through a distance x then

Total wor done by the gas !1 W W W W ++=

P

A

V

Positive

wor

@xpansion

B P

A

V

Aegativ

ewor

ompression

B

x M

M

P

P

V

Positivewor

P1

P!

B C

% A

V 1

V !

locwise cyclicprocess

P

V

Aegativewor

P1

P!

B

C%

A

V 1

V !

"nticlocwise cyclicprocess


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where W 1 , or done against external pressure )( =P

W ! , or done against spring force )($x

W , or done against gravitational force (mg)

∴ mg$x V PW ++= !=!

1

(vii) If the gas expands in such a way that other side of the piston is vacuum then wor done by

the gas will be 4ero

"s ==∆= V PW :6ere P , =9

() Internal ener*y %U& : Internal energy of a system is the energy possessed by the

system due to molecular motion and molecular con#guration.

The energy due to molecular motion is called internal inetic energy &$ and that due to

molecular con#guration is called internal potential energy &P.

i.e. Total internal energy P$ &&& +=

(i) +or an ideal gas$ as there is no molecular attraction == p&

i.e. internal energy of an ideal gas is totally inetic and is given by RT && $ µ !

==

and change in internal energy T R& ∆=∆ µ !

(ii) In case of gases whatever be the process

T R

& ∆=∆!

µ T CV ∆= µ 1

)(

)1( −

−=∆

−=

γ

µ

γ µ

i T T RT

R

1−

−=

γ

µ µ i RT RT

1

)(

−

−=

γ

ii V PV P

(iii) hange in internal energy does not depends on the path of the process. 2o it is called a

point function i.e. it depends only on the initial and #nal states of the system$ i.e. i &&& −=∆

(iv) hange in internal energy in a cyclic process is always 4ero as for cyclic process i && =

2o ==−=∆ i &&&

Sample problems based on Q+ U and W

P roblem 1. " thermodynamic system is taen through the cycle PQR'P process. The net wor done bythe system is

,-rissa // 2))20(a) != J

Bacuum&as

R

1== $p QP

'!== $p

1== cc == cc

P

V


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Thermodynamic Processes 92

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(b) C != J

(c) -== J

(d) C D- J

'olutio# % (b) or done by the system , "rea of shaded portion on P"V diagram

J!=1=)1=!==(1=)1====( E =×−×−= −

and direction of process is anticlocwise so wor done will be negative i.e. ∆W , C != J.

P roblem 2. "n ideal gas is taen around ABCA as shown in the above P"V diagram. The wor done duringa cycle is

,/T %/n**.4ed.& 2))10

(a) !PV

(b) PV

(c) 1F!PV (d) Gero

'olutio# % (a) or done , "rea enclosed by triangle ABC PV PPV V BC AC !)()(!

1

!

1=−×−×=×=

P roblem 3. The P"V diagram shows seven curved paths (connected by vertical paths) that can befollowed by a gas. hich two of them should be parts of a closed cycle if the net wor doneby the gas is to be at its maximum value

,546 %/n**.& 2)))0

(a) ac

(b) cg

(c) a

(d) cd

'olutio# % (c) "rea enclosed between a and is maximum. 2o wor done in closed cycles follows a and is

maximum.

P roblem 7. If $ molecalC( FHE.-= $ then increase in internal energy when temperature of ! moles of this gas is increased from -= $ to -! $

,/T 1990

(a) !D.= cal (b) 1H.- cal (c) 1.H= cal (d) H.H! cal

'olutio# % (b) Increase in internal energy calT C& ( -.1H!HE.-!)-=-!(HE.-!.. =××=−××=∆=∆ µ

P roblem ;. "n ideal gas of mass m in a state A goes to another state B via three di7erent processes asshown in #gure. If !1$QQ and Q denote the heat absorbed by the gas along the three

paths$ then ,4 /T 19920

(a) !1 QQQ =

(d) !1 QQQ >>

'olutio# % (a) "rea enclosed by curve 1 J "rea enclosed by curve ! J "rea enclosed by curve

∴ !1 QQQ


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P roblem


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genius

∴ 3ise in temperature C+L J

( t °

−=∆

!

!

(E) If icebloc falls down through some height and melts partially then its potential energy

gets converted into heat of melting.

+rom W , JQ

Lm Jmg, M= :where m , mass of ice bloc$ m1 , mass which

melts9

2o *g JL

mg,m=M

If icebloc completely melts down then mg, , J mL

∴ 6eight re*uired for complete melting meteg

JL,=

Sample problems based on Joule's law

P roblem . ater falls from a height of !1= m. "ssuming whole of energy due to fall is converted intoheat the rise in temperature of water would be ( J , -. JouleFcal)

,b. 4T 2))20

(a) -!NC (b) -HNC (c) =.-HNC (d) -.HNC

'olutio# % (c)


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9; Thermodynamic Processes


P roblem 11. " bullet moving with a uniform velocity ( $ stops suddenly after hitting the target and thewhole mass melts be m$ speci#c heat '$ initial temperature !0NC$ melting point -D0NC andthe latent heat L. Then

(a)

J

m( m+mL

!

)!0-D0(!

+−= (b) J

m( mLm+

!

)!0-D0(!

=+−

(c) J

m( mLm+

!

)!0-D0( =+− (d)

J

m( mLm+

!)!0-D0(

!

=−−

'olutio# % (b) K.@. of bullet , 6eat re*uired to raise the temperature of bullet from !0NC to -D0NC Q heatre*uired to melt the bullet

⇒ 9)!0-D0(:!

1 ! mLm+ Jm( +−= ⇒ J

m( mLm+

!)!0-D0(

!

=+− .

P roblem 12. " lead bullet at !DNC Rust melts when stopped by an obstacle. "ssuming that !0 of heatis absorbed by the obstacle$ then the velocity of the bullet at the time of striing (/.P. of lead , !DNC$ speci#c heat of lead , =.= calFgmNC$ latent heat of fusion of lead , E

calFgm and J , -.! JFcal) ,IIT?// 19810

(a) -1= mF+ec (b) 1!= mF+ec (c) =D.0 mFsec (d) Aone of these

'olutio# % (a) ;sing expression obtained in problem (11) we get 9!D!D(:!

1PD0 ! mLm+ Jm( +−=

2ubstituting C*gcal+ oF1=.= ×= and *gcalL F1=E ×= we get +m( F-1==

P roblem 13. " drilling machine of power P watts is used to drill a hole in Cu bloc of mass m *g. If the

speci#c heat of Cu is11

0

−−°C*g J and -= of power lost due to heating of machine the

rise in temperature of the bloc in T +ec (will be in NC)

(a)m+

PT E.=(b)

m+T

PE.=(c)

m+

PT -.=(d)

m+T

P-.=

'olutio# % (a) "s we now)( Time

)(or)(Power

T

W P = ∴ W , P × T

"s -= energy is lost due to heating of machine so only E= energy will increase the

temperature of the bloc

∴ E= of W , m × + × ∆t ⇒

m+

PT

m+

W t

E.=E.===∆ .

13.; #irst a! of Thermodynamics.

It is a statement of conservation of energy in thermodynamical process.

"ccording to it heat given to a system (∆Q) is e*ual to the sum of increase in its internal

energy (∆&) and the wor done (∆W ) by the system against the surroundings.

W &Q ∆+∆=∆

Important points

(1) It maes no distinction between wor and heat as according to it the internal energy (and

hence temperature) of a system may be increased either by adding heat to it or doing wor on it

or both.


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Thermodynamic Processes 9<

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(!) ∆Q and ∆W are the path functions but ∆& is the point function.

() In the above e*uation all three *uantities ∆Q$ ∆& and ∆W must be expressed either in

Joule or in calorie.

(-) ust as 4eroth law of thermodynamics introduces the concept of temperature$ the #rst

law introduces the concept of internal energy.(0) 2ign conventions

∆QPositive hen heat is supplied to a system

Aegative hen heat is drawn from the system

∆W Positive hen wor done by the gas (expansion)

Aegative hen wor done on the gas (compression)

∆&Positive hen temperature increases$ internal energy increases

Aegative hen temperature decreases$ internal energy decreases

(E) hen a thermos bottle is vigorously shaen %

Ao heat is transferred to the co7ee ∆Q , = :"s thermos 8as is insulated from

the surrounding9

or is done on the co7ee against viscous force ∆W , (C)

Internal energy of the co7ee increases ∆& , (Q)

and temperature of the co7ee also increases ∆T , (Q)

(D)


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P roblem 1. "n electric fan is switched on in a closed room. The air in the room is ,4 /T 199


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genius

'olutio# % (c) &Q ∆=∆ 9!DD:

1

0!

1−

−×=∆

−

=∆= R

T R

T C( γ

µ µ , == R :"s for monoatomic

gas

0=γ 9

13.< Isothermal rocess.

hen a thermodynamic system undergoes a physical change in such a way that its

temperature remains constant$ then the change is nown as isothermal changes.

In this process$ P and V change but T , constant i.e. change in temperature ∆T , =

(1) /ssential condition for isothermal process

(i) The walls of the container must be perfectly conducting to

allow free exchange of heat between the gas and its surrounding.

(ii) The process of compression or expansion should be so slow

so as to provide time for the exchange of heat.

2ince these two conditions are not fully realised in practice$ therefore$ no process is

perfectly isothermal.

(!) /quation of state : +rom ideal gas e*uation PV , µ RT

If temperature remains constant then PV , constant i.e. in all isothermal process 5oyleLs law

is obeyed.

6ence e*uation of state is PV , constant.

() /Cample of isothermal process

(i) /elting process :Ice melts at constant temperature =NC9

(ii) 5oiling process :water boils at constant temperature 1==NC9.

(-) Indicator dia*ram

(i) urves obtained on PV graph are called isotherms and they are hyperbolic in nature.

(ii) 2lope of isothermal curve % 5y di7erentiating PV , constant. e get

==+ dPV dV P ⇒ dPV dV P −= ⇒ V

P

dV

dP−=

∴ V

P

dV

dP−==θ tan

(iii) "rea between the isotherm and volume axis represents the wor done in isothermal

process.

&as

onducting walls

T

T !

T 1

T 1JT

!J T

P

V

P

V

θ

P

V

or


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If volume increases ∆W , Q "rea under curve and if volume decreases ∆W , C "rea under

curve

(0) >peciDc heat : 2peci#c heat of gas during isothermal change is in#nite.

"s ∞=×=∆= =mQ

T m

QC :"s ∆T , =9

(E) Isothermal elasticity : +or isothermal process PV , constant

?i7erentiating both sides ==+VdPPdV ⇒ dPV dV P −= ⇒ θ EV dV

dPP ==

−=

2train

2tress

F

∴ PE =θ i.e. isothermal elasticity is e*ual to pressure

"t A.T.P. isothermal elasticity of gas , "tmospheric pressure , !0 F1==1.1 m.×

(D) 'or( done in isothermal process

∫ ∫ ==

i

i

V

V

V

V dV

V

RT dV PW µ

:"s PV , µ RT 9

=

=

i

i

eV

V RT

V

V RT W 1=log=.!log µ µ

or

=

=

i

ie

P

PRT

P

PRT W 1=log=.!log µ µ

() #TE in isothermal process

W &Q ∆+∆=∆ but T & ∆∝∆

∴ ==∆& :"s ∆T , =9

∴ W Q ∆=∆ i.e. heat supplied in an isothermal change is used to do wor against

external surrounding.

or if the wor is done on the system than e*ual amount of heat energy will be liberated by

the system.

Sample problems based on Isothermal process

P roblem 21. 'ne mole of !/ gas having a volume e*ual to !!.- litre+ at =NC and 1 atmospheric pressurein compressed isothermally so that its volume reduces to 11.! litre+. The wor done in thisprocess is ,4 /T 1993@ AF 2))30

(a) 1ED!.0 J (b) 1D! J (c) C 1D! J (d) C 10D!.0 J'olutio# % (d) or done in an adiabatic process

JV

V RT W e

i

e 10D!)EH.=(!D.-.!!

!.11log!D.1log −≈−××=

×××=

= µ

P roblem 22. "n ideal gas A and a real gas B have their volumes increased from V to !V under isothermalconditions. The increase in internal energy

,A>/ 4T 1993@ I4/ 2))1+ 2))20

(a) ill be same in both A and B (b) ill be 4ero in both the gases

(c) 'f B will be more than that of A (d) 'f A will be more than that of B

'olutio# % (c) In real gases an additional wor is also done in expansion due to intermolecular attraction.


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Thermodynamic Processes 1))

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P roblem 23. hich of the following graphs correctly represents the variation of V dPdV F)F(−=β with Pfor an ideal gas at constant temperature

,IIT?// %>creenin*& 2))20

(a) (b) (c) (d)

'olutio# % (a) +or an isothermal process PV , constant ⇒ ==+ VdPPdV ⇒ PdP

dV

V

11 =

−

2o$P

1=β ∴ graph will be rectangular hyperbola.

P roblem 27. onsider the following statements

5ssertion ( A)% The internal energy of an ideal gas does not change during an isothermal

processeason %R& % The decrease in volume of a gas is compensated by a corresponding increase

in pressure when its temperature is held constant.

'f these statements ,>5 19970

(a) 5oth A and R are true and R is a correct explanation of A

(b) 5oth A and R are true but R is not a correct explanation of A

(c) A is true but R is false

(d) 5oth A and R are false

(e) A is false but R is true

'olutio# % (b) "s ∆& ∝ ∆T so for isothermal process internal energy will not change and alsoV

P 1∝

(5oyleMs law)

P roblem 2;. 6ow much energy is absorbed by 1= *g molecule of an ideal gas if it expands from an initialpressure of atm to - atm at a constant temperature of !DNC

,oor(ee 19920

(a) D1=D!I.1 × J (b) D1=!I.1D × J (c) H1=D!I.1 × J (d) H1=!I.1D × J

'olutio# % (a) or done in an isothermal process

JP

PRT W e

ie

D- 1=D!.1EH.===.1=-

log==.)1=1=(log ×=×××=

××××=

= µ

P roblem 2


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1)1 Thermodynamic Processes


(a)

−+

−−

!1

!1!

1

!logV V

V V #

#V

#V #RT e α

β

β (b)

−+

−−

!1

!1!

1

!1=log

V V

V V #

V

V #RT α

αβ

αβ

(c)

−+

−

−

!1

!1!

1

!logV V

V V #

#V

#V #RT

e

β α

α (d)

−+

−

−

!1

!1!

!

1logV V

V V #

#V

#V #RT

e

α β

β

'olutio# % (a) 5y Bander aalLs e*uation!

!

V

#

#V

#RT P

α

β −

−=

or done$ ∫ ∫ ∫ −−== !

1

!

1

!

1!

!V

V

V

V

V

V V

dV #

#V

dV #RT PdV W α

β

[ ]

−+

−−

=

+−=!1

!1!

1

!! log1

)(log!

1

!

1 V V

V V #

#V

#V #RT

V ##V #RT e

V

V

V

V e α

β

β α β

13. 5diabatic rocess.

hen a thermodynamic system undergoes a change in such a way that no exchange of heattaes place between it and the surroundings$ the process is nown

as adiabatic process.

In this process P$ V and T changes but ∆Q , =.

(1) /ssential conditions for adiabatic process

(i) There should not be any exchange of heat between the

system and its surroundings. "ll walls of the container and the

piston must be perfectly insulating.

(ii) The system should be compressed or allowed to expand suddenly so that there is no time

for the exchange of heat between the system and its surroundings.

2ince$ these two conditions are not fully realised in practice$ so no process is perfectly

adiabatic.

(!) /Cample of some adiabatic process

(i) 2udden compression or expansion of a gas in a container with perfectly nonconducting

walls.

(ii) 2udden bursting of the tube of bicycle tyre.

(iii) Propagation of sound waves in air and other gases.

(iv) @xpansion of steam in the cylinder of steam engine.

() #TE in adiabatic process : W &Q ∆+∆=∆

but for adiabatic process ==∆Q ∴ ==∆+∆ W &

If ∆W , positive then ∆& , negative so temperature decreases i.e. adiabatic expansion

produce cooling.

If ∆W , negative then ∆& , positive so temperature increases i.e. adiabatic compression

produce heating.

(-) /quation of state : "s in case of adiabatic change #rst law of thermodynamics reducesto$

==∆+∆ W & $ i.e.$ ==+ dW d& >.. (i)5ut as for an ideal gas dT Cd& V µ = and PdV dW =

&as

Insulatingwalls


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Thermodynamic Processes 1)2

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@*uation (i) becomes ==+ dV PdT C( µ >.. (ii)

5ut for a gas as PV , µ RT $ dT RdPV dV P µ =+ >.. (iii)

2o eliminating dT between e*uation (ii) and (iii) =)(

=++

dV PR

dPV dV PC(

µ µ

or =)1(

)( =+−+ dV PdPV dV P

γ

−=

)1( as

γ RC(

or ==+ dPV dV Pγ i.e.$ ==+P

dP

V

dV γ

hich on integration gives

CPV ee =+ loglogγ $

i.e.$ CPV =)log( γ

or consta=γ PV >.. (iv)

@*uation (iv) is called e*uation of state for adiabatic change and can also be rewritten as

constan1 =−γ TV :as P , ( µ RT FV )9 >.. (iv)

and consta1 =

−γ

γ

P

T

=

P

RT V

µ as >.. (vi)

(0) Indicator dia*ram

(i) urve obtained on PV graph are called adiabatic curve.

(ii) 2lope of adiabatic curve % +rom consta=γ PV

5y di7erentiating$ we get =1

=+ −

dV V PV dP γ γ

γ

−=−=

−

V

P

V

PV

dV

dPγ γ

γ

γ 1

∴2lope of adiabatic curve

−=V

Pγ φ tan

5ut we also now that slope of isothermal curveV

P−=φ tan

2o$ 1

)F(

)F(

curveisothermalof2lope

curveadiabaticof2lope>==

−

−=

(

p

C

C

V P

V Pγ

γ

(E) >peciDc heat : 2peci#c heat of a gas during adiabatic change is 4ero

"s ==

=∆

=∆

=T mT m

QC :"s Q , =9

(D) 5diabatic elasticity : +or adiabatic process consta=γ PV

?i7erentiating both sides =1 =+ − dV V PPV d γ γ γ

φ γ EV dV

dPP ==

−=

2train

2tress

F

PE γ φ =

i.e. adiabatic elasticity is γ times that of pressure but we now isothermal elasticity PE =θ

φ

P

V


16/33



2o γ γ

θ

φ ===P

P

E

E

elasticityIsothermal

elasticity"diabatic

i.e. the ratio of two elasticities of gases is e*ual to the ratio of two speci#c heats.

() 'or( done in adiabatic process

∫ =

i

V

V dV PW ∫ =

i

V

V dV

V

$ γ

=

γ V

$ P"s

or

−

−=

−− 1191:

1γ γ γ i

V

$

V

$

+−=∫ +−

)1(B"s

1

γ

γ γ V dV

or)1(

9:

γ −

−= ii

V PV P9"s: γ γ γ ii V PV PPV $ ===

or 9:)1( i T T

R

−−= γ µ

:"s RT V P µ = and iii RT V P µ = 9

2o)1(

)(

)1(

9:

−

−=

−

−=

γ

µ

γ

i ii T T RV PV P

(H) #ree eCpansion : +ree expansion is adiabatic process in which no wor is performed on

or by the system. onsider two vessels placed in a system which is enclosed with thermal

insulation (asbestoscovered). 'ne vessel contains a gas and the other is evacuated. The two

vessels are connected by a stopcoc. hen suddenly the stopcoc is opened$ the gas rushes into

the evacuated vessel and expands freely. The process is adiabatic as the vessels are placed in

thermal insulating system (dQ , =) moreover$ the walls of the vessel are rigid and hence no

external wor is performed (dW , =).Aow according to the #rst law of thermodynamics d& , =

If i& and & be the initial and #nal internal energies of the gas

then ==− i && :"s i && = 9

Thus the #nal and initial energies are e*ual in free expansion.

(1=) >pecial cases of adiabatic process

consta=γ PV ∴ γ V P

1∝ $

consta1 =−γ γ

PT ∴ 1−∝ γ

γ

T P and consta1 =−γ TV ∴

1

1−

∝γ V

T

Type of *asγ V

P 1∝ 1−∝ γ

γ

T P 11−

∝γ V

T

/onoatomic γ , 0F

F0

1

V P ∝

!F0T P ∝F!

1

V T ∝

?iatomic γ , DF0

0FD

1

V P ∝

!FDT P ∝0F!

1

V T ∝

Polyatomic γ , -F

F-1V P ∝

-

T P ∝ F11V T ∝


17/33


genius

(11) omparison bet!een isothermal and adiabatic process

(i) ompression : If a gas is compressed isothermally and

adiabatically from volume V i to V then from the slope of the graph

it is clear that graph 1 represents adiabatic process where as graph

! represent isothermal process.

or done W adiabatic S W isothermal

+inal pressure Padiabatic S Pisothermal

+inal temperature T adiabatic S T isothermal

(ii) /Cpansion : If a gas expands isothermally and adiabatically from volume V i to V then

from the slope of the graph it is clear that graph 1 represent

isothermal process$ graph ! represent adiabatic process.

or done W isothermal S W adiabatic

+inal pressure Pisothermal S Padiabatic

+inal temperature T isothermal S T adiabatic

Sample problems based on Adiabatic process

P roblem 28. ?uring an adiabatic process$ the pressure of a gas is found to be proportional to the cube of

its absolute temperature. The ratio ( p CC F for the gas is

,5I/// 2))30

(a)!

(b)

-(c) ! (d)

0

'olutio# % (a) &iven T P ∝ . 5ut for adiabatic process 1−∝ γ γ

T P. 2o$

1=

−γ γ

⇒ !

=γ ⇒

!

=

(

p

C

C

P roblem 29. "n ideal gas at !DNC is compressed adiabatically to!D

I of its original volume. If

0=γ $ then

the rise in temperature is,4T 1987@ A>/ 4T 1999@ /T 2))3@ 6>/5T 2))30

(a) -0= $ (b) D0 $ (c) !!0 $ (d) -=0 $

'olutio# % (b) +or an adiabatic process =−1γ TV constant

∴

1

1!

!1

−

=

γ

V

V

T

T ⇒ $ V

V T T ED0!D==

!D==

F!1

01

!

11! ====

−−γ

⇒

$ T D0==ED0 =−=∆

P roblem 3). If γ , !.0 and volume is e*ual to 1F times to the initial volume then pressure P ′ is e*ual to(initial pressure , P)

,/T 2))30

(a) PP =′ (b) PP !=′ (c) !F10)!(×=′ PP (d) PP D=′

'olutio# % (c) +or an adiabatic process =γ PV constant ⇒ γ

=

!

1

1

!

V

V

P

P ⇒ !F0I=

′P

P ⇒ !F10)!(×=′ PP

P roblem 31. In adiabatic expansion of a gas ,4 /T 2))30

(a) Its pressure increases (b) Its temperature falls

P

V i

1

!

V

V

P

V

1!

V i

V


18/33

1); Thermodynamic Processes


(c) Its density decreases (d) Its thermal energyincreases

'olutio# % (b) W &Q ∆+∆=∆ $ In an adiabatic process ∆Q , = ∴ W & ∆=∆−In expansion ∆W , positive ∴ ∆& , negative. 6ence internal energy i.e. temperature

decreases.P roblem 32. P"V plots for two gases during adiabatic process are shown in the #gure. Plots 1 and ! should

correspond respectively to,IIT?// %>creenin*& 2))10

(a) 0e and !/

(b) !/ and 0e

(c) 0e and Ar

(d) !/ and !.

'olutio# % (b) 2lope of adiabatic curve∝

γ ∝

gasof the"tomicity

1

. 2o γ is inversely proportional to

atomicity of the gas.

:"s EE.1=γ for monoatomic gas$ γ , 1.- for diatomic gas and γ , 1. for triatomic non

linear gas.9

+rom the graph it is clear that slope of the curve 1 is less so this should be adiabatic curve for

diatomic gas (i.e. /!).

2imilarly slope of the curve ! is more so it should be adiabatic curve for monoatomic gas

(i.e. 0e).

P roblem 33. " monoatomic ideal gas$ initially at temperature 1T $ is enclosed in a cylinder #tted with a

frictionless piston. The gas is allowed to expand adiabatically to a temperature !T by

releasing the piston suddenly. If 1L and !L are the lengths of the gas column before and

after expansion respectively$ then !1 FT T is given by

,IIT?// %>creenin*& 2)))0

(a)

F!

!

1

L

L(b)

!

1

L

L(c)

1

!

L

L(d)

F!

1

!

L

L

'olutio# % (d) +or an adiabatic process1

!!1

11−− = γ γ V T V T ⇒

F!

1

!

1F0

1

!

1

1

!

!

1

=

=

=

−−

L

L

AL

AL

V

V

T

T γ

.

P roblem 37. +our curves A$ B$ C and % are drawn in the adRoining #gure for a given amount of gas. Thecurves which represent adiabatic and isothermal changes are

,4T 198/5T 19990

(a) C and % respectively

(b) % and C respectively

(c) A and B respectively

(d) B and A respectively

'olutio# % (c) "s we now that slope of isothermal and adiabatic curves are always negative and slope of

adiabatic curve is always greater than that of isothermal curve so is the given graph curve A

and curve B represents adiabatic and isothermal changes respectively.

P

V

1

!

V

P

B

A

C

%


19/33

Thermodynamic Processes 1)<

genius

P roblem 3;. " thermally insulated container is divided into two parts by a screen. In one part thepressure and temperature are P and T for an ideal gas #lled. In the second part it is vacuum.

If now a small hole is created in the screen$ then the temperature of the gas will ,/T 19990

(a) ?ecrease (b) Increase (c) 3emain same (d) Aone of these

'olutio# % (c) In second part there is a vacuum i.e. P , =. 2o wor done in expansion , P∆V , =

P roblem 3peciDc heat : 2peci#c heat of gas during isobaric process R

CP

+= 1!

P

V

P1

P!

A I B

CII%


20/33

1) Thermodynamic Processes


(-) Aul( modulus of elasticity : =F

=∆−∆

=V V

P$ :"s ∆P , =9

(0) 'or( done in isobaric process : 9: i V

V

V

V V V PdV PdV PW

i

i

−===∆ ∫ ∫ :"s P

, constant9

∴ T RT T RV V PW i i ∆=−=−=∆ µ µ 9:)(

(E) #TE in isobaric process : T C& V ∆=∆ µ T R

∆−

=)1(γ

µ and T RW ∆=∆ µ

+rom +


21/33


genius

e*uivalent of heat , -.1H JFcalorie$ the energy spent in this process in overcomingintermolecular forces is ,4 /T 1999+ 2))1@ -rissa // 2))20

(a) 0-= calorie (b) -= calorie (c) 0== calorie (d) Gero

'olutio# % (c) @nergy spent in overcoming inter molecular forces W Q& ∆−∆=∆

caloriV V PQ 0==!.-

1=)11ED1(1==1.1

0-=)(

E0

1! ≈×−×

−=−−∆=

−

P roblem 39. " gas expands !0.= m at constant pressure ! F1= m. $ the wor done is

,4T 199@ 6>/5T 1999@ I4/ 2))1+ 2))20

(a) !.0 erg+ (b) !0= J (c) !0= W (d) !0=

'olutio# % (b) "s we now$ wor done JV P !0=!0.=1= =×=∆=

P roblem 7). 0 mole of hydrogen gas is heated from =NC to E=NC at constant pressure. 6eat given tothe gas is (given R , ! calFmole degree)

,4 /T 2))20

(a) D0= calorie (b) E= calorie (c) 1=0= calorie (d) 1-D= calorie

'olutio# % (c) T RT CQ p p ∆

−=∆=∆

1)(

γ γ µ µ

∴ caloriQ p 1=0==!

0

0

D!0=!

10

D0

D

0)( =××××=××

−×=∆ :"s µ , 0 mole and γ ,

0

D

for 0!9

P roblem 71. The latent heat of vaporisation of water is !!-= JFgm. If the wor done in the process of expansion of 1g is 1E J$ then increase in internal energy is

,4T 2)))0

(a) !-= J (b) !!-= J (c) !=D! J (d) 1H=- J

'olutio# % (c) W &Q ∆+∆=∆ ⇒ 1EI!!-= +∆= & ⇒ J& !=D!=∆

P roblem 72. hen an ideal gas (γ , 0F) is heated under constant pressure$ then what percentage of given heat energy will be utilised in doing external wor

,/T 19990

(a) -= (b) = (c) E= (d) !=

'olutio# % (a) 0

!

01

11

11 =

−−=−=

−=

∆∆−∆

=∆∆

γ p

( p

C

CC

Q

&Q

Q

W

i.e. percentage energy utilised in doing external wor , 1==0! × , -=.

P roblem 73. "t 1==NC the volume of 1*g of water is 1= m− and volume of 1 *g of steam at normal

pressure is ED1.1 m . The latent heat of steam is *g J F1=.! E× and the normal pressure

is !0 F1= m. . If 0 *g of water at 1==NC is converted into steam$ the increase in the internal

energy of water in this process will be

(a) 01=0.I × J (b) E1=EE.1= × J (c) E1=0.11 × J (d) Gero

'olutio# % (b) 6eat re*uired to convert 0 *g of water into steam JmLQ EE 1=0.111=.!0 ×=××==∆

or done in expanding volume$ JV PW E0 1=I0.=91=ED1.1:1=0 ×=−×=∆=∆ −


22/33



Aow by #rst law of thermodynamics W Q& ∆−∆=∆ ⇒

J& EEE 1=EE.1=1=I0.=1=0.11 ×=×−×=∆

13.9 Isochoric or Isometric rocess.hen a thermodynamic process undergoes a physical change in such a way that its volume

remains constant$ then the change is nown as isochoric process.

In this process P and T changes but V , constant. 6ence &aylussacLs law is obeyed in this

process.

(1) /quation of state

+rom ideal gas e*uation PV , µ RT

If volume remains constant P∝

T or == !!

1

1

T

P

T

P

constant

(!) Indicator dia*ram : &raph I and II represent isometric

decrease in pressure at volume 1V and isometric increase in

pressure at volume !V respectively and slope of indicator diagram

∞=dV

dP

() >peciDc heat : 2peci#c heat of gas during isochoric process R

CV !

=

(-) Aul( modulus of elasticity % ∞=∆=∆−∆=

=F

P

V V

P$

(0) 'or( done in isobaric process : 9: i V V PV PW −=∆=∆ :"s V , =9

∴ ==∆W

(E) #TE in isochoric process % W &Q ∆+∆=∆ &∆= :"s ∆W , =9

T CQ V ∆=∆ µ T R

∆−

=1γ

µ 1−

−=

γ

ii V PV P

ote % Isometric expansion of the pressure of a gas is given by)1(

= t PP

Pt γ

+=

where =°

= C per P!D

1γ coeOcient of pressure expansion.

Sample problems based on Isochoric process

P roblem 77. In pressurevolume diagram given below$ the isochoric$ isothermal$ and isobaric partsrespectively$ are

,4anipal 4// 199;0

(a) BA$ A%$ %C

(b) %C$ CB$ BA

(c) AB$ BC$ C%

(d) C%$ %A$ AB

P

V V 1 V !

A

1

BC

11

%

C

B A

%V

P


23/33


genius

'olutio# % (d) Process C% is isochoric as volume is constant) Process %A is isothermal as temperatureconstant and Process AB is isobaric as pressure is constant.

P roblem 7;. /olar speci#c heat of oxygen at constant pressure CmolcalC p °= FF!.D and R , . JFmolF$ . "t constant volume$ 0 mol of oxygen is heated from 1=NC to !=NC$ the *uantity of heat re*uired is approximately

,4 4T 1980(a) !0 cal (b) 0= cal (c) !0= cal (d) 0== cal

'olutio# % (c) 5y /ayerLs formula CmolcalRCC p( °≈−=−= F0!!.D

"t constant volume calT cQ ( !0=1=00 =××=∆=∆ µ

13.1) eversible and Irreversible rocess.

(1) eversible process : " reversible process is one which can be reversed in such a way

that all changes occurring in the direct process are exactly repeated in the opposite order and

inverse sense and no change is left in any of the bodies taing part in the process or in the

surroundings. +or example if heat is absorbed in the direct process$ the same amount of heat

should be given out in the reverse process$ if wor is done on the woring substance in the directprocess then the same amount of wor should be done by the woring substance in the reverse

process. The conditions for reversibility are

(i) There must be complete absence of dissipative forces such as friction$ viscosity$ electricresistance etc.

(ii) The direct and reverse processes must tae place in#nitely slowly.

(iii) The temperature of the system must not di7er appreciably from its surroundings.

2ome examples of reversible process are

(a) "ll isothermal and adiabatic changes are reversible if they are performed very slowly.

(b) hen a certain amount of heat is absorbed by ice$ it melts. If the same amount of heat is

removed from it$ the water formed in the direct process will be converted into ice.

(c) "n extremely slow extension or contraction of a spring without setting up oscillations.

(d) hen a perfectly elastic ball falls from some height on a perfectly elastic hori4ontal

plane$ the ball rises to the initial height.

(e) If the resistance of a thermocouple is negligible there will be no heat produced due to

ouleLs heating e7ect. In such a case heating or cooling is reversible. "t a Runction where a

cooling e7ect is produced due to Peltier e7ect when current 8ows in one direction and e*ual

heating e7ect is produced when the current is reversed.

(f) Bery slow evaporation or condensation.

It should be remembered that the conditions mentioned for a reversible process can never

be realised in practice. 6ence$ a reversible process is only an ideal concept. In actual process$

there is always loss of heat due to friction$ conduction$ radiation etc.

(!) Irreversible process : "ny process which is not reversible exactly is an irreversible

process. "ll natural processes such as conduction$ radiation$ radioactive decay etc. are

irreversible. "ll practical processes such as free expansion$ ouleThomson expansion$ electrical

heating of a wire are also irreversible. 2ome examples of irreversible processes are given below

(i) hen a steel ball is allowed to fall on an inelastic lead sheet$ its inetic energy changes

into heat energy by friction. The heat energy raises the temperature of lead sheet. Ao reverse

transformation of heat energy occurs.

(ii) The sudden and fast stretching of a spring may produce vibrations in it. Aow a part of the

energy is dissipated. This is the case of irreversible process.


24/33



(iii) 2udden expansion or contraction and rapid evaporation or condensation are examples of

irreversible processes.

(iv) Produced by the passage of an electric current through a resistance is irreversible.

(v) 6eat transfer between bodies at di7erent temperatures is also irreversible.

(vi) ouleThomson e7ect is irreversible because on reversing the 8ow of gas a similar

cooling or heating e7ect is not observed.

13.11 yclic and Bon?cyclic rocess.

" cyclic process consists of a series of changes which return the system bac to its initial

state.

In noncyclic process the series of changes involved do not return the system bac to its

initial state.

(1) In case of cyclic process as i && =

∴ ==−=∆ i &&& i.e. change in internal energy for cyclic process is 4ero and also T & ∆∝∆∴ ∆T , = i.e. temperature of system remains constant.

(!) +rom #rst law of thermodynamics W &Q ∆+∆=∆

W Q ∆=∆ i.e. heat supplied is e*ual to the wor done by the system. :"s ∆& , =9

() +or cyclic process P"V graph is a closed curve and area enclosed by the closed path

represents the wor done.

If the cycle is clocwise wor done is positive and if the cycle is anticlocwise wor done is

negative.

(-) or done in non cyclic process depends upon the path chosen or the series of changes

involved and can be calculated by the area covered between the curve and volume axis on PV

diagram.

Sample problems based on Cyclic and noncyclic process

P roblem 7


25/33


genius

(a) - J

(b) D= J

(c) - J

(d) 1- J

'olutio# % (d) 6eat given ∆Q Jcal I-!.-!=!= =×== . or done ∆W , C 0= J :"s process

is anticlocwise95y #rst law of thermodynamics ⇒ JW Q& 1-)0=(I- =−−=∆−∆=∆

P roblem 7. "n ideal gas is taen through the cycle A → B → C → A$ as shown in the #gure. If the netheat supplied to the gas in the cycle is 0 J$ the wor done by the gas in the process C → A is,IIT?// %>creenin*& 2))20

(a) C 0 J

(b) C 1= J

(c) C 10 J

(d) C != J 'olutio# % (a) +or a cyclic process. Total wor done CABC AB W W W ++=

⇒ CAW ++−×=×× =)1!(1=1==.1!

1:W BC , = since there is no change in volume

along BC9

⇒ CAW J J += 1=0 ⇒ JW CA 0−=

P roblem 78. In the following indicator diagram$ the net amount of wor done will be

(a) Positive

(b) Aegative

(c) Gero

(d) In#nity

'olutio# % (b) or done during process 1 is positive while during process ! it is negative. 5ecause process1 is clocwise while process ! is anticlocwise. 5ut area enclosed by P"V graph (i.e. wordone) in process 1 is smaller so$ net wor done will be negative.

P roblem 79. " cyclic process for 1 mole of an ideal gas is shown in #gure in the V T $ diagram. The wordone in AB$ BC and CA respectively

(a) )($ln$= !1!

1! T T R

V

V RT −

(b)!

11!1 ln$=)$(

V V RT T T R −

(c) )($ln$= !11

!! T T R

V

V RT −

(d) )($ln$= 1!1

!! T T R

V

V RT −

'olutio# % (c) Process AB is isochoric$ ∴ ==∆= V PW AB

Process BC is isothermal ∴

=

1

!! ln.

V

V RT W BC

Process CA is isobaric ∴ V PW CA ∆= T R∆= )( 1! T T R −=

1 !

P

V

V !

/

V 1

A

C

B

T 1

T !

T

V

C

A

B

1=

!

1

V (m)

P(-m!

)


26/33



P roblem ;). " cyclic process ABC% is shown in the #gure P"V diagram. hich of the following curvesrepresent the same process

(a) (b) (c) (d)

'olutio# % (a) AB is isobaric process$ BC is isothermal process$ C% is isometric process and %A isisothermal process

These process are correctly represented by graph (a). 13.12 Graphical epresentation of Farious rocesses.

13.13 $eat /n*ine.

6eat engine is a device which converts heat into wor continuously through a cyclic process.

The essential parts of a heat engine are

>ource : It is a reservoir of heat at high temperature and in#nite thermal capacity. "nyamount of heat can be extracted from it.

'or(in* substance : 2team$ petrol etc.

>in( : It is a reservoir of heat at low temperature and in#nite thermal capacity. "ny amountof heat can be given to the sin.

The woring substance absorbs heat Q1 from the source$ doesan amount of wor W $ returns the remaining amount of heat to thesin and comes bac to its original state and there occurs no

change in its internal energy.5y repeating the same cycle over and over again$ wor is

continuously obtained.

The performance of heat engine is expressed by means of

eOciencyU η which is de#ned as the ratio of useful wor obtained

from the engine to the heat supplied to it.

1input6eat

doneor

Q

W ==η

5ut by #rst law of thermodynamics for cyclic process ∆& , =. ∴ ∆Q , ∆W so

!1 QQW −=

V

P Isochoric

IsobarIsotherm

"diabatic

/ T

P

Isotherm

"diabatic

Isochoric

Isobar

/ T

V Isotherm

Isobar

Isochoric

"diabatic

/

W ! Q1 C Q!

Q1

Q!

6eat

@ngine

2ource (T 1)

(T !)2in

A B

C

P

T

%

A B

C

P

T

%

A

B

C

P

T

%

A B

C

P

T

%

A B

C

P

V

%


27/33


genius

∴ 1

!

1

!1 1Q

Q

Q

QQ−=

−=η

" perfect heat engine is one which converts all heat into wor i.e. 1QW = so that =! =Q andhence 1=η .

5ut practically eOciency of an engine is always less than 1. 13.17 efri*erator or $eat ump.

" refrigerator or heat pump is basically a heat engine run in reverse direction.

It essentially consists of three parts

>ource : "t higher temperature T 1.

'or(in* substance : It is called refrigerant li*uid ammonia and freon wors as a woring

substance.

>in( : "t lower temperature T !.

The woring substance taes heat Q! from a sin (contents of refrigerator) at lower

temperature$ has a net amount of wor done W on it by an external

agent (usually compressor of refrigerator) and gives out a larger

amount of heat Q1 to a hot body at temperature T 1 (usually

atmosphere). Thus$ it transfers heat from a cold to a hot body at

the expense of mechanical energy supplied to it by an external

agent. The cold body is thus cooled more and more.

The performance of a refrigerator is expressed by means of

coeOcient of performanceU β which is de#ned as the ratio of the

heat extracted from the cold body to the wor needed to transfer it to the hot body.

i.e. !1

!!

donewor

extracted6eat

QQ

Q

W

Q

−===β

∴!1

!

QQ

Q

−=β

" perfect refrigerator is one which transfers heat from cold to hot body without doing wor

i.e. W , = so that !1 QQ = and hence ∞=β

(1) arnot refri*erator

+or arnot refrigerator!

1

!

1

T T

QQ = ∴

!

!1

!

!1

T T T

QQQ −=− or

!1

!

!1

!

T T T

QQQ

−=

−

2o coeOcient of performance!1

!

T T

T

−=β

where T 1 , temperature of surrounding$ T ! , temperature of cold body

It is clear that β , = when T ! , =

i.e. the coeOcient of performance will be 4ero if the cold body is at the temperature e*ual to

absolute 4ero.

(!) elation bet!een coeHcient of performance and eHciency of refri*erator

W ! Q1C Q

!

Q1

Q!

6eat@ngin

e

2ource("tmosphere)(T 1)

(T !)

2in(ontents ofrefrigerator)


28/33

11; Thermodynamic Processes


e now!1

!

QQ

Q

−=β or

1!

1!

F1

F

QQ

QQ

−=β >.. (i)

5ut the eOciency

1

!1

Q

Q−=η or η −= 1

1

!

Q

Q>..(ii)

+rom (i) and (ii) we getη

η β

−= 1

13.1; >econd a! of Thermodynamics.

+irst law of thermodynamics merely explains the e*uivalence of wor and heat. It does notexplain why heat 8ows from bodies at higher temperatures to those at lower temperatures. Itcannot tell us why the converse is possible. It cannot explain why the eOciency of a heat engineis always less than unity. It is also unable to explain why cool water on stirring gets hotterwhereas there is no such e7ect on stirring warm water in a beaer. 2econd law of thermodynamics provides answers to these *uestions. 2tatement of this law is as follows

(1) lausius statement : It is impossible for a self acting machine to transfer heat from acolder body to a hotter one without the aid of an external agency.

+rom lausius statement it is clear that heat cannot 8ow from a body at low temperature to

one at higher temperature unless wor is done by an external agency. This statement is in fair

agreement with our experiences in di7erent branches of physics. +or example$ electrical current

cannot 8ow from a conductor at lower electrostatic potential to that at higher potential unless an

external wor is done. 2imilarly$ a body at a lower gravitational potential level cannot move up to

higher level without wor done by an external agency.

(!) elvins statement : It is impossible for a body or system to perform continuous worby cooling it to a temperature lower than the temperature of the coldest one of its surroundings." arnot engine cannot wor if the source and sin are at the same temperature because wordone by the engine will result into cooling the source and heating the surroundings more andmore.

() elvin?lanc(s statement : It is impossible to design an engine that extracts heatand fully utilises into wor without producing any other e7ect.

+rom this statement it is clear that any amount of heat can never be converted completelyinto wor. It is essential for an engine to return some amount of heat to the sin. "n engineessentially re*uires a source as well as sin. The eOciency of an engine is always less than unitybecause heat cannot be fully converted into wor.

13.1< arnot /n*ine.

arnot designed a theoretical engine which is free from all the defects of a practical engine.

This engine cannot be realised in actual practice$ however$ this can be taen as a standard

against which the performance of an actual engine can be Rudged.

It consists of the following parts

(i) " cylinder with perfectly nonconducting wallsand a perfectly conducting base containing a perfectgas as woring substance and #tted with a nonconducting frictionless piston

(ii) " source of in#nite thermal capacity maintainedat constant higher temperature T 1.

(iii) " sin of in#nite thermal capacity maintained atconstant lower temperature T !.

2ourceT

1 $

Insulating stand

2inT

! $

Idealgas


29/33

Thermodynamic Processes 11<

genius

(iv) " perfectly nonconducting stand for the cylinder.

(1) arnot cycle : "s the engine wors$ the woring substance of the engine undergoes a

cycle nown as arnot cycle. The arnot cycle consists

of the following four stroes

(i) +irst stroe (Isothermal expansion) (curve AB) % The cylinder containing ideal gas as woring

substance allowed to expand slowly at this constant

temperature T 1.

or done , 6eat absorbed by the system

=

=== ∫

1

!111 log

!

1 V

V RT dV PQW e

V

V "rea

AB2E

(ii) 2econd stroe ("diabatic expansion) (curve BC) %

The cylinder is then placed on the non conducting stand and the gas is allowed to expand

adiabatically till the temperature falls from T 1 to T !.

∫ =

!

!

V

V dV PW , =−−

9:)1(

!1 T T R

γ "rea BC02

(iii) Third stroe (Isothermal compression) (curve C%) %

The cylinder is placed on the sin and the gas is compressed at constant temperature T !.

or done , 6eat released by the system

C%F0V

V RT V

V RT dV PQW ee

V

V "realoglog-

!

-!!

-

==−=−== ∫ (iv) +ourth stroe (adiabatic compression) (curve %A) % +inally the cylinder is again placed on

nonconducting stand and the compression is continued so that gas returns to its initial stage.

A%FET T R

T T R

dV PW V

V "rea)(

1)(

1 !11!-

1

-

=−−

=−−

−=−= ∫ γ γ (!) /Hciency of arnot cycle : The eOciency of engine is de#ned as the ratio of wor

done to the heat supplied i.e. 1input6eat

donewor

Q

W ==η

Aet wor done during the complete cycle)()( -!1 W W W W W −+−++= ABC%W W "rea1 =−= :"s -! W W = 9

∴ 1

!

1

1

!1

1

1

1

11Q

Q

W

W

Q

QQ

W

W W

Q

W −=−=

−=

−==η

or)F(log

)F(log1

1!1

-!

V V RT

V V RT

e

e−=η

2ince points B and C lie on same adiabatic curve ∴ 1

!1

!1−− = γ γ V T V T or

1

!

!

1

−

=

γ

V

V

T

T >..(i)

Q1

Q!

W , Q1C

Q!


30/33



"lso point % and A lie on the same adiabatic curve ∴ 1-!1

11−− = γ γ V T V T or

1

1

-

!

1

−

=

γ

V

V

T

T >..(ii)

+rom (i) and (ii) 1

-

!

V

V

V

V =

or 1

!

-

V

V

V

V =

⇒

=

1

!

-

loglogV

V

V

V

ee

2o eOciency of arnot engine1

!1T

T −=η

(i) @Ociency of a heat engine depends only on temperatures of source and sin and isindependent of all other factors.

(ii) "ll reversible heat engines woring between same temperatures are e*ually eOcient andno heat engine can be more eOcient than arnot engine (as it is ideal).

(iii) "s on Kelvin scale$ temperature can never be negative (as = $ is de#ned as the lowestpossible temperature) and T l and T ! are #nite$ eOciency of a heat engine is always lesser thanunity$ i.e.$ whole of heat can never be converted into wor which is in accordance with secondlaw.

ote % The eOciency of an actual engine is much lesser than that of an ideal engine."ctually the practical eOciency of a steam engine is about (10) while that of apetrol engine is -=. The eOciency of a diesel engine is maximum and is about (0=00).

() arnot theorem : The eOciency of arnotLs heat engine depends only on the

temperature of source (T 1) and temperature of sin (T !)$ i.e.$1

!1T

T −=η .

arnot stated that no heat engine woring between two given temperatures of source andsin can be more eOcient than a perfectly reversible engine (arnot engine) woring betweenthe same two temperatures. arnotMs reversible engine woring between two given temperaturesis considered to be the most eOcient engine.

13.1 EiJerence Aet!een etrol /n*ine 5nd Eiesel /n*ine.

etrol en*ine Eiesel en*ine

(i) oring substance is a mixture of petrol vapour

and air.

(i) oring substance in this engine is a mixture

of diesel vapour and air.

(ii) @Ociency is smaller (V-D). (ii) @Ociency is larger (V00).

(iii) It wors with a spar plug. (iii) It wors with an oil plug.

(iv) It is associated with the ris of explosion$

because petrol vapour and air is compressed. 2o$ low

compression ratio is ept.

(iv) Ao ris of explosion$ because only air is

compressed. 6ence compression ratio is ept

large.

(v) Petrol vapour and air is created with spar plug. 2pray of diesel is obtained through the Ret.

Sample problems based on Second law of thermodynamic

P roblem ;1. If the door of a refrigerator is ept open$ then which of the following is true ,A$6 2))1@ I4/ 2))2@ 5I/// 2))2@ 4T 2))30

(a) 3oom is cooled (b) 3oom is heated

(c) 3oom is either cooled or heated (d) 3oom is neither cooled nor heated


31/33


genius

'olutio# % (b) In a refrigerator$ the woring substance taes heat !Q from the sin at lower temperature

!T and gives out a larger amount of heat 1Q to a hot body at higher temperature 1T .

Therefore the room gets heated if the door of a refrigerator is ept open.

P roblem ;2. The coeOcient of performance of a arnot refrigerator woring between =oC and =oC is,6>/5T 2))20

(a) 1= (b) 1 (c) H (d) =

'olutio# % (c) oeOcient of performance of a arnot refrigerator woring between =NC and =NC is

H=

!D

!D=

!D

!1

! ≈°°

=°−°

°=

−=

C

C

CC

C

T T

T β

P roblem ;3. " arnot engine woring between == $ and E== $ has wor output of == J per cycle. hatis amount of heat energy supplied to the engine from source per cycle

,b. 4T 2))20

(a) 1== JFcycle (b) 1=== JFcycle (c) !=== JFcycle (d) 1E== JFcycle

'olutio# % (d) @Ociency of arnot engine1

!1

T

T T −=η

!

1

E==

==E===

−

"gaininput6eat

doneor=η ∴ 6eat input

η

doneor= .1E==

!F1

I== J==

P roblem ;7. arnot cycle (reversible) of a gas represented by a PressureBolume curve is shown in thediagram

onsider the following statements

I. "rea ABC% , or done on the gas

II. "rea ABC% , Aet heat absorbed

III. hange in the internal energy in cycle , =hich of these are correct

,546 %4ed.& 2))10

(a) I only (b) II only (c) II and III (d) I$ II and III

'olutio# % (c) or done by the gas (as cyclic process is clocwise) ∴ ∆W , "rea ABC%

2o from the #rst law of thermodynamics ∆Q (net heat absorbed) , ∆W , "rea ABC%

"s change in internal energy in cycle ∆& , =.

P roblem ;;. " arnot engine taes 1= *cal of heat from a reservoir at E!DNC and exhausts it to a sin at!DNC. The eOciency of the engine will be

(a) !!.! (b) . (c) --.- (d) EE.E

'olutio# % (d) @Ociency of arnot engineH

E

H==

==H==

1

!1 =−=−=T

T T or EE.E.

13.18 /ntropy.

@ntropy is a measure of disorder of molecular motion of a system. &reater is the disorder$

greater is the entropy.

The change in entropy i.e. retemperatu"bsolute

systebyabsorbed6eat=d' or

T

dQd' =

The relation is called the mathematical form of 2econd


32/33



(i) hen heat given to a substance changes its state at constant temperature$ then changein entropy

T

mL

T

dQd' ±==

where positive sign refers to heat absorption and negative sign to heat evolution.

(ii) hen heat given to a substance raises its temperature from T 1 to T !$ then change inentropy

=== ∫ ∫

1

!log!

1 T

T mc

T

dT mc

T

dQd' e

T

T

⇒

=∆

1

!log=.!T

T mc' e .

(!) #or a perfect *as : Perfect gas e*uation for # moles is PV , #RT

∫ ∫ +==∆ T dV PdT #CT dQ' V :"s dQ , d& Q dW W

⇒ ∫ +

=∆T

dV V

#RT dT #C

'V ∫ ∫ +=

!

1

!

1

V

V

T

T V

V

dV #R

T

dT #C :"s PV , #RT 9

∴

+

=∆

1

!

1

! loglogV

V #R

T

T #C' eeV

2imilarly in terms of T and P

−

=∆

1

!

1

! loglogP

P#R

T

T #C' eeP

and in terms of P and V

+

=∆

1

!

1

! loglogV

V #C

P

P#C' ePeV

Sample problems !iscellaneous

P roblem ;


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genius

'olutio# % (c) +rom the given VT diagram$ we can see that

In process AB$ V ∝ T ∴ Pressure is constant ("s *uantity of the gas remains same)

In process BC$ V , constant and in process CA$ T , constant

∴ These processes are correctly represented on PV diagram by graph (c).

01 thermodynamic process theory21

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