01-Thermodynamic Process (Theory)

genius PHYSICS  Thermodynamic  Thermodynamic Processes 1

genius Notes by  : Pradeep Kshetrapal

P

φ 

V 

dx 

F=PA

genius genius PHYSICS

2  Thermodynamic Processes Processes

13.1 Introduction. (1) Thermodynamics : It is a ranch o! science "hich deals "ith e#change o! heat energy et"een odies and con$ersion o! the heat energy into mechanical energy and $ice%$ersa. (&) Therm Thermody odyna nami mic c system system : ' collection o! an e#tremely large numer o! atoms or molecul molecules es conne conned d "ith "ith in certai certain n ounda oundaries ries such that that it has a certai certain n $alue $alue o! press pressur ure e $olu $olume me and and temp temper erat atur ure e is call called ed a ther hermody modyna nami mic c sys ysttem. em. 'n 'ny ythin thing g outs utside ide the the thermodynamic system to "hich energy or matter is e#changed is called i ts surroundings. Example : *as enclosed enclosed in a cylinder tted "ith a piston piston !orms the thermodyn thermodynamic amic system ut the atmospheric air around the cylinder mo$ale piston urner etc. etc. are all the surroundings.  Thermodynamic system may e o! three types (i) +pen system : It e#change oth energy and matter "ith the surrounding. (ii) Closed system : It e#change only energy (not matter) "ith the surroundings. (iii) Isolated system : It e#change neither energy nor matter " ith the surrounding. (,) Thermodynamic Thermodynamic variables and equation of state : ' thermodynamic system can e descried y speci!ying its pressure $olume temperature internal energy and the numer o!  moles moles.. Thes These e para parame mete ters rs are are calle called d ther thermo mody dyna nami mic c $aria $arial les es.. The The rela relati tion on et" et"een een the the thermodynamic $ariales (P (P V  T ) o! the system is called e-uation o! state. or  µ  moles  moles o! an ideal gas e-uation o! state is PV  /  /  µ RT  RT  and  and !or 1 mole o! an it ideal gas is PV  /  / RT 

  or µ  or  µ  moles  moles o! a real gas e-uation o! state is  P +  

a µ & V &

     (V  −  µ  b) = µ  RT  and !or 1 mole o! a real gas it is  

 P + a  (V − b) = RT       V &   (0) Thermody Thermodynam namic ic equilibri equilibrium um : hen the thermodynamic $ariales attain a steady $alue i.e. they i.e. they are independent o! time the system is said to e in the state o! thermodynamic e-uilirium. or a system to e in thermodynamic e-uilirium the !ollo"ing conditions must e !ullled. (i) 2echanic 2echanical al e-uili e-uiliriu rium m : There There is no unala unalance nced d !orce !orce et"ee et"een n the system system and its surroundings. (ii) Thermal e-uilirium : There is a uni!orm temperature in all parts o! the system and is same as that o! surrounding. surrounding. (iii) Chemical e-uilirium : There is i s a uni!orm chemical composition through out the system and the surrounding.

genius genius PHYSICS

2  Thermodynamic Processes Processes

13.1 Introduction. (1) Thermodynamics : It is a ranch o! science "hich deals "ith e#change o! heat energy et"een odies and con$ersion o! the heat energy into mechanical energy and $ice%$ersa. (&) Therm Thermody odyna nami mic c system system : ' collection o! an e#tremely large numer o! atoms or molecul molecules es conne conned d "ith "ith in certai certain n ounda oundaries ries such that that it has a certai certain n $alue $alue o! press pressur ure e $olu $olume me and and temp temper erat atur ure e is call called ed a ther hermody modyna nami mic c sys ysttem. em. 'n 'ny ythin thing g outs utside ide the the thermodynamic system to "hich energy or matter is e#changed is called i ts surroundings. Example : *as enclosed enclosed in a cylinder tted "ith a piston piston !orms the thermodyn thermodynamic amic system ut the atmospheric air around the cylinder mo$ale piston urner etc. etc. are all the surroundings.  Thermodynamic system may e o! three types (i) +pen system : It e#change oth energy and matter "ith the surrounding. (ii) Closed system : It e#change only energy (not matter) "ith the surroundings. (iii) Isolated system : It e#change neither energy nor matter " ith the surrounding. (,) Thermodynamic Thermodynamic variables and equation of state : ' thermodynamic system can e descried y speci!ying its pressure $olume temperature internal energy and the numer o!  moles moles.. Thes These e para parame mete ters rs are are calle called d ther thermo mody dyna nami mic c $aria $arial les es.. The The rela relati tion on et" et"een een the the thermodynamic $ariales (P (P V  T ) o! the system is called e-uation o! state. or  µ  moles  moles o! an ideal gas e-uation o! state is PV  /  /  µ RT  RT  and  and !or 1 mole o! an it ideal gas is PV  /  / RT 

  or µ  or  µ  moles  moles o! a real gas e-uation o! state is  P +  

a µ & V &

     (V  −  µ  b) = µ  RT  and !or 1 mole o! a real gas it is  

 P + a  (V − b) = RT       V &   (0) Thermody Thermodynam namic ic equilibri equilibrium um : hen the thermodynamic $ariales attain a steady $alue i.e. they i.e. they are independent o! time the system is said to e in the state o! thermodynamic e-uilirium. or a system to e in thermodynamic e-uilirium the !ollo"ing conditions must e !ullled. (i) 2echanic 2echanical al e-uili e-uiliriu rium m : There There is no unala unalance nced d !orce !orce et"ee et"een n the system system and its surroundings. (ii) Thermal e-uilirium : There is a uni!orm temperature in all parts o! the system and is same as that o! surrounding. surrounding. (iii) Chemical e-uilirium : There is i s a uni!orm chemical composition through out the system and the surrounding.

genius PHYSICS  Thermodynamic  Thermodynamic Processes 3

(3) Thermodynamic Thermodynamic process :  The process o! change o! state o! a system in$ol$es change o! thermodynamic $ariales such as pressure P $olume V  and   and temperature T  o!  o! the system. The process is 4no"n as thermodynamic process. Some important processes are (i) Isotherma Isothermall process process (iso$olumic process)

(ii) 'diaatic 'diaatic process process

($) Cyclic Cyclic and and non%cyclic non%cyclic proc process ess

(iii) Isoaric Isoaric process process

(i$) Isochoric Isochoric

($i) 5e$ers 5e$ersile ile and irre$ersi irre$ersile le process process

 13.2 Zeroth Law of Thermodynamics . I! systems A systems A and  and B are each in thermal e-uilirium "ith a third system C then A then A and  and B are in thermal e-uilirium "ith each other. other. (1) The 6eroth la" leads to the concept o! temperature. 'll odies in thermal e-uilirium must ha$e a common property "hich has the same $alue !or all o! them. This property is called the temperature. (&) The 6eroth la" came to light long a!ter the rst and seconds la"s o! thermodynamics had een disco$ered and numered. 7ecause the concept o! temperature is !undamental to those t"o la"s the la" that estalishes temperature as a $alid concept should ha$e the lo"est numer. Hence it is called 6eroth la".

 13.3 uantities Involved in !irst Law of Thermodynamics . (1) "eat # Q$ : It is the energy that is trans!erred et"een a system and its en$ironment ecause o! the temperature di8erence et"een them. Heat al"ays 9o" !rom a ody at higher temperature to a ody at lo"er temperature till their temperatures ecome e-ual. Important points (i) Heat is a !orm o! energy so it is a scalar -uantity "ith dimension ;ML&T −& : . (ii) to &T > through a process P

=

α 

T 

 and "e are interested in nding the "or4 done y the gas. Then

PV  / &RT  and

P

=

(ideal gas e-uation)

α 

?..(ii)

T 

@i$iding (i) y (ii) "e get V  = ∴

#  =

V $ 

∫ 

V i

?..(i)

P dV  =

&T >

∫ 

T >

&&RT  &RT & dT   or dV  = α  α 

 α    &&RT  dT  = &&RT       >  T     α   

So "e ha$e !ound the "or4 done "ithout putting the limits o! $olume. ($i) I! mass less piston is attached to a spring o! !orce constant '   and a mass m is placed o$er the piston. I! the e#ternal pressure is P>  and due to e#pansion o! gas the piston mo$es up through a distance x  then  Total "or4 done y the gas #  = # 1 + # & + # ,

M M

P

 x 

genius PHYSICS  Thermodynamic Processes (

"here

# 1 / or4 done against e#ternal pressure (P> ) # & / or4 done against spring !orce ('x ) # , / or4 done against gra$itational !orce (m!) #  = P>V  +

∴

1 'x & &

+ m!

($ii) I! the gas e#pands in such a "ay that other side o! the piston is $acuum then "or4 done y the gas "ill e 6ero 's #  = P∆V  = >

;Here P / >

Dacuum

*as

(,) Internal ener)y #U$ : Internal energy o! a system is the energy possessed y the system due to molecular motion and molecular conguration.  The energy due to molecular motion is called internal 4inetic 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 o! an ideal gas is totally 4inetic and is gi$en y ) and change in internal energy ∆) =

, &

= ) '  =

, µ RT  &

µ R ∆T 

(ii) In case o! gases "hate$er e the process

∆) =  µ  =

$  R∆T  &

= µ CV ∆T  =  µ 

R (γ   − 1)

∆T  =

 µ  R(T $ 

− T i )

γ   − 1

=

−  µ RT i γ   − 1

 µ RT $ 

(P $  V $  − Pi V i ) γ  

−1

(iii) Change in internal energy does not depends on the path o! the process. So it is called a point !unction i.e. it depends only on the initial and nal states o! the system i.e. ∆) = ) $  − )i (i$) Change in internal energy in a cyclic process is al"ays 6ero as !or cyclic process ) $  = )i So

∆) = ) $  − )i = > Sample problems based on

Q*

U and  W 

Problem 1. ' thermodynamic system is ta4en through the cycle P"R*P process. The net "or4 done y the system is +,rissa - 2//20

P

(a) &> J

&>> 'p

()  &> J

1>> 'p

*

P 1>> cc

R

" ,>> cc

V 

genius PHYSICS

  Thermodynamic Processes (c) 0>> J (d)  ,E0 J *olutio& : ()

or4 done y the system / 'rea o! shaded portion on P%V  diagram

= (,>>− 1>>)1>−F × (&>>− 1>) × 1>, =

&>  J

and direction o! process is anticloc4"ise so "or4 done "ill e negati$e i.e. ∆# /  &> J.

Problem 2. 'n ideal gas is ta4en around ABCA as sho"n in the ao$e P%V   diagram. The "or4 done during a cycle is P

B

(a) &PV 

(,P, ,V ) C (P, ,V )

 A (P,V )

() PV 

+T #n)).45ed.$ 2//10

E

( V 

(c) 1G&PV  (d) ero *olutio& : (a) or4 done / 1 1 =  AC × BC = × (,V  − V ) × (,P − P ) = & PV  & &

'rea

enclosed

y

triangle

 ABC

Problem 3.  The P%V   diagram sho"s se$en cur$ed paths (connected y $ertical paths) that can e !ollo"ed y a gas. hich t"o o! them should e parts o! a closed cycle i! the net "or4 done y the gas is to e at its ma#imum $alue a P +657 #n)).$ 2///0 b

c

d

(a) ac

e

$ 

!

() c!

V 

(c) a$  (d) cd

*olutio& : (c) 'rea enclosed et"een a  and $   is ma#imum. So "or4 done in closed cycles !ollo"s a and $  is ma#imum.

Problem '. I! C+  = 0.FcalG mole'   then increase in internal energy "hen temperature o! & moles o! this gas is increased !rom ,0> '  to ,0& '  +89T 1;0 (a) &E.J> cal

() 1.J0 cal

(c) 1,.> cal

*olutio& : () Increase in ∆) = µ .C + .∆T  = & × 0.F× (,0&− ,0>) = & × 0.F× & = 1.J0cal

(d) .& cal

internal

energy

Problem (. 'n ideal gas o! mass m in a state  A  goes to another state B  $ia three di8erent processes as sho"n in gure. I! "1 "&   and ",  denote the heat asored y the gas along the three paths then +59 9T 120 (a) "1 < "& < ", () "1 < "& = ",

P

 A

1

&

,

B V 

genius PHYSICS  Thermodynamic Processes ;

(c) "1 = "&

> ",

(d) "1 > "& > ", *olutio& : (a) , ∴

'rea enclosed y cur$e 1  'rea enclosed y cur$e &  'rea enclosed y cur$e "1 < "& < ",

;'s ∆) is same !or all cur$es

Problem .  The relation et"een the internal energy ) and adiaatic constant γ   is (a) ) =

PV  γ   − 1

() ) =

PV γ   γ   − 1

(c) ) =

Change in internal energy / ∆) =

*olutio& : (a)

=et initially T 1 ) c+ 

=

R γ   − 1

µ  c+ ∆T  ⇒

PV  γ  

)&

= >  so )1 = >  and nally T & = T   and

=  µ  c+ T  =

µ  T  × c+  =

PV  R

×

R

γ   − 1

=

PV 

γ   − 1

 

(d) )

;'s

=

γ  

PV 

− )1 = µ c+ (T & − T 1) )&

=)

PV  = µ  RT 

∴

 µ T 

=

PV  R

and



 13.' -oule.0OC

(d) 0.OC

=oss in potential energy o! "ater / Increment in thermal energy o! "ater ⇒ m!.=   J

× ms∆t 

⇒

.J × &1> =  0., × 1>>>∆t 

∴

∆t  = >.0I°C

Problem =. ' loc4 o! mass 1>> !m slides on a rough hori6ontal sur!ace. I! the speed o! the loc4 decreases !rom 1> mGs to 3 mGs the thermal energy de$eloped in the process is +79>6T 2//20 (a) ,.E3 J *olutio& : (a)

=

() ,E.3 J

Thermal

energy

(c) >.,E3 J

de$eloped

/

=oss

(d) >.E3 J in

4inetic

energy

1 1 m(+ && − + 1&) = × >.1× (1>& − 3&) = ,.E3 J & &

Problem .  The "eight o! a person is F> -!. I! he gets 1> 3 caloies heat through !ood and the eciency o! his ody is &JQ then upto ho" much height he can clim (appro#imately) +6!5 1;0 (a) 1>> m *olutio& : ()

() &>> m

(c) 0>> m

(d) 1>>> m

Increment in potential energy o! mass / &JQ o! heat gained ⇒ m!.=  J

  &J × 1>3      ⇒  1>>  

  &J × 1>3     ⇒  1>>  

F>× .J × . = 0.& × 

. = &>>m

Problem 1/. Hailstone at >OC !alls !rom a height o! 1 -m on an insulating sur!ace con$erting "hole o! its 4inetic energy into heat. hat part o! it "ill melt (! = 1>mG s& ) +59 95T 1'0

genius PHYSICS  Thermodynamic Processes 

(a) *olutio& : (a)

1

()

,,

1

(c)

J

1

× 1>−0

,,

Anergy re-uired !or melting / =oss in potential energy ⇒ 0.1J× (mN× J>× 1>−, ) = m× 1>× 1>>> ⇒

⇒

mN m

=

(d) 'll o! it "ill melt  J

× m′L = m!

1   ,,

;'s L / J>

×

1>,

caloie/-!

Problem 11. ' ullet mo$ing "ith a uni!orm $elocity +  stops suddenly a!ter hitting the target and the "hole mass melts e m specic heat * initial temperature &3OC melting point 0E3OC and the latent heat L. Then (a) mL = ms(0E3− &3) +

m+ & & J

(c) ms(0E3− &3) + mL =

m+ &  J

ms(0E3− &3) − mL =

*olutio& : ()

() ms(0E3− &3) + mL =

m+ & & J

(d)

m+ & & J

K.A. o! ullet / Heat re-uired to raise the temperature o! ullet !rom &3OC to 0E3OC R heat re-uired to melt the ullet ⇒

1 m+ & &

=  J ;ms(0E3− &3) + mL:

⇒

ms(0E3− &3) + mL =

m+ & . & J

Problem 12. ' lead ullet at &EOC ust melts "hen stopped y an ostacle. 'ssuming that &3Q o! heat is asored y the ostacle then the $elocity o! the ullet at the time o! stri4ing (2.P. o! lead / ,&EO C specic heat o! lead / >.>, calG!mOC latent heat o! !usion o! lead / F calG!m and J / 0.& JGcal) +IIT?- 1=10 (a) 01> mGsec *olutio& : (a)

mGsec e#pression

(c) ,>E.3 mGsec

otained

in

(d) Bone o! these

prolem

(11)

"e

get

 1 m+ &   =  J ;ms(,&E− &E+ mL:    &  

E3Q

Sustituting s =

,

o >., × 1> calG-! C

 and L =

,

F × 1> calG-!  "e

get +  = 01>m G s

Problem 13. ' drilling machine o! po"er P "atts is used to drill a hole in Cu loc4 o! mass m -!. I! the specic heat o! Cu is 3  J -! 1 C 1  and 0>Q o! po"er lost due to heating o! machine the rise in temperature o! the loc4 in T sec ("ill e in OC) −

(a) *olutio& : (a)

>.F PT 

()

ms

°

>.F P

(c)

msT 

's "e 4no" Po"er(P ) =

−

or4(# )  Time(T )

∴

>.0 PT 

ms

#  / P

×

(d)

>.0 P

msT 

T 

's 0>Q energy is lost due to heating o! machine so only F>Q energy "ill increase the temperature o! the loc4 ∴ F>Q

o! #  / m

×

s

× ∆t ⇒

∆t  =

>.F#  ms

=

>.FPT  ms

.

 13.( !irst Law of Thermodynamics. It is a statement o! conser$ation o! energy in thermodynamical process.

genius PHYSICS

1/  Thermodynamic Processes 'ccording to it heat gi$en to a system ( ∆") is e-ual to the sum o! increase in its internal energy (∆)) and the "or4 done ( ∆# ) y the system against the surroundings.

∆" = ∆) + ∆#  Important points (1) It ma4es no distinction et"een "or4 and heat as according to it the internal energy (and hence temperature) o! a system may e increased either y adding heat to it or doing "or4 on it or oth. (&) ∆" and ∆#  are the path !unctions ut

∆) is

(,) In the ao$e e-uation all three -uantities  Joule or in caloie.

the point !unction. ∆" ∆)  and ∆#   must

e e#pressed either in

(0) Lust as 6eroth la" o! thermodynamics introduces the concept o! temperature the rst la" introduces the concept o! internal energy. (3) Sign con$entions ∆"

∆# 

∆)

Positi$e Begati$e Positi$e Begati$e

hen heat is supplied to a system hen heat is dra"n !rom the system hen "or4 done y the gas (e#pansion) hen "or4 done on the gas (compression)

Positi$e

hen temperature increases

increases

internal

energy

Begati$e

hen temperature decreases

decreases

internal

energy

(F) hen a thermos ottle is $igorously sha4en : Bo heat is trans!erred to the co8ee the surrounding

∆"

/>

;'s thermos 9as4 is insulated !rom

or4 is done on the co8ee against $iscous !orce Internal energy o! the co8ee increases

∆) /

∆#  /

()

(R)

and temperature o! the co8ee also increases ∆T  / (R) (E) =imitation : irst la" o! thermodynamics does not indicate the direction o! heat trans!er. It does not tell anything aout the conditions under "hich heat can e trans!ormed into "or4 and also it does not indicate as to "hy the "hole o! heat energy cannot e con$erted into mechanical "or4 continuously.

Sample problems based on First law of thermodynamics Problem 1'. I! 13> J o! heat is added to a system and the "or4 done y the system is 11>  J then change in internal energy "ill e +657 #n)).$ 1@ A"7 2///0 (a) &F> J *olutio& : (d)

∆" = ∆) + ∆# 

() 13> J ⇒

13> = ∆) + 11>

(c) 11> J ⇒ ∆) = 0>  J

(d) 0> J

genius PHYSICS  Thermodynamic Processes 11

Problem 1(. 11> J o! heat is added to a gaseous system "hose internal energy change is 0>  J then the amount o! e#ternal "or4 done is +A> 95T 13@ 6!5 1@ -I958 2///0 (a) 13> J

() E> J

∆" = ∆) + ∆# 

*olutio& : ()

⇒

(c) 11> J

11>= 0> + ∆# 

(d) 0> J

⇒ ∆#  = E>  J

Problem 1. hen an ideal diatomic gas is heated at constant pressure the !raction o! the heat energy supplied "hich increases the internal energy o! the gas is +IIT?- 1/@ 89T 2///0 (a)

& 3

∆) = ∆"

*olutio& : (d)

Problem 1;.

()

C + ∆T  C P ∆T 

=

,

(c)

3

R G(γ   − 1) γ  R G(γ   − 1)

=

1 1 = γ   E G 3

=

,

(d)

E

3 E

3 E

'n electric !an is s"itched on in a closed room. The air in the room is +59 9T 10

(a) Cooled () Heated (c) 2aintains its temperature (d) Heated or cooled depending on the atmospheric pressure *olutio& : ()

hen an electric !an is s"itched on in a closed room con$entional current o! air 9o"s. Hence due to $iscous !orce mechanical energy is con$erted into heat and some heat is also produced due to thermal e8ect o! electric current in motor o! !an.

Problem 1=.

' gas is compressed at a constant pressure o!

to a $olume o!

0 m,

&

3> N G m

 !rom a $olume o!

1>m,

. Anergy o! 1>> J  is then added to the gas y heating. Its

internal energy is

+5B8 1'0

(a) Increased y 0>> J () Increased y &>> J (c) Increased y 1>> J (d) @ecreased y &>> J

∆" = ∆) + ∆# 

*olutio& : (a)

⇒

∆" = ∆) + ∆dV ⇒ 1>> = ∆) + 3>(0 − 1>)

⇒

1>>= ∆) − ,>>

∴

∆) = 0>>  J

Problem 1.

' thermodynamic process is sho"n in the gure. The pressures and $olumes

corresponding to some points in the gure are : P A V  A

= , × 1>0 Pa PB = J × 1>0 Pa

and

= & × 1>−, m,  V ( = 3 × 1>−, m,

In process AB F>> J o! heat is added to the system and in process BC &>> J o! heat is added to the system. The change in internal energy o! the system in process  AC "ould e +A> 95T 120 P

B

C

 A

(

(a) 3F> J () J>> J (c) F>> J (d) F0> J

0

V 

genius PHYSICS

12  Thermodynamic Processes 7y adoining graph #  AB

*olutio& : (a) ∴

#  AC

Bo"

= >  and

# BC

= J × 1>0 ;3 − &: × 1>−, = &0> J

= #  AB + # BC = > + &0> = &0> J

∆" AC = ∆" AB + ∆"BC = F>>+ &>>= J>> J

rom rst la" o! thermodynamics ∆" AC = ∆) AC + ∆#  AC

⇒

J>>= ∆) AC

+ &0>

⇒

∆) AC = 3F> J . Problem 2/. I! R / uni$ersal gas constant the amount o! heat needed to raise the temperature o! & mole o! an ideal monoatomic gas !rom &E, '   to ,E, '   "hen no "or4 is done +59 9T 1/0 (a) 1>> R

() 13> R

  R     =  µ  C+ ∆T  =  µ   ∆T  = & × 1 γ   −    

∆" = ∆)

*olutio& : (c)

monoatomic gas γ   =

3 ,

(c) ,>> R R 3 −1 ,

(d) 3>> R

;,E,− &E,:

/ ,>> R

;'s !or



 13. Isothermal 9rocess. hen a thermodynamic system undergoes a physical change in such a "ay that its temperature remains constant then the change is 4no"n as isothermal changes. In this process P and V  change ut T  / constant i.e. change in temperature ∆T  / > (1) ssential condition for isothermal process (i) The "alls o! the container must e per!ectly conducting to allo" !ree e#change o! heat et"een the gas and its surrounding.

Conductin g "alls *as

(ii) The process o! compression or e#pansion should e so slo" so as to pro$ide time !or the e#change o! heat.

Since these t"o conditions are not !ully realised in practice there!ore no process is per!ectly isothermal. (&) quation of state :  rom ideal gas e-uation PV  /  µ RT  I! temperature remains constant then PV / constant i.e. in all isothermal process 7oyleMs la" is oeyed. Hence e-uation o! state is PV  / constant. (,) Cample of isothermal process (i) 2elting process ;Ice melts at constant temperature >O C (ii) 7oiling process ;"ater oils at constant temperature 1>>O C. (0) Indicator dia)ram P

P

P

T 1T & T , T , T & T 1

θ  or4 V 

V 

V 

genius PHYSICS  Thermodynamic Processes 13

(i) Cur$es otained on PV  graph are called isotherms and they are hyperolic in nature. (ii) Slope o! isothermal cur$e : 7y di8erentiating PV  / constant. e get ⇒ P dV  = − V  dP

P dV + V dP = >

∴

tanθ  =

dP = dV 

−

⇒

dP dV 

=−

P V 

P V 

(iii) 'rea et"een the isotherm and $olume a#is represents the "or4 done in isothermal process. I! $olume increases ∆#  / R 'rea under cur$e and i! $olume decreases cur$e

∆#   /

 'rea under

(3) >peciDc heat : Specic heat o! gas during isothermal change is innite. 's

C

=

" " = m∆T  m× >

=∞

;'s ∆T  / >

(F) Isothermal elasticity :  or isothermal process PV  / constant @i8erentiating oth sides PdV + VdP = > ∴

Eθ 

⇒ P dV  = − V  dP ⇒

P

=

dP Stress = =E − dV G V  Strain θ 

= P i.e. isothermal elasticity is e-ual to pressure

't B.T.P. isothermal elasticity o! gas / 'tmospheric pressure / 1.>1× 1>3 N G m& (E) %or& done in isothermal process #  =

V $ 

∫ 

V i

P dV =

V $ 

∫ 

V i

µ RT  dV  V 

;'s PV  /  µ RT 

 V $     V $         µ  & . ,>, RT  log = 1>     V    i     V i     P     P   #  =  µ RT loge  i   = &.,>,µ RT  log1> i    P$     P$            # =  µ RT loge 

or

(J) !LTE in isothermal process

∆" = ∆) + ∆#   ut

∆) ∝ ∆T 

∴

∆) = >

∴

∆" = ∆#  i.e.  heat supplied in an isothermal change is used to do "or4 against

;'s ∆T  / >

e#ternal surrounding. or i! the "or4 is done on the system than e-ual amount o! heat energy "ill e lierated y the system.

Sample problems based on Isothermal process Problem 21. +ne mole o! 0& gas ha$ing a $olume e-ual to &&.0 lites  at >OC  and 1 atmospheric pressure in compressed isothermally so that its $olume reduces to 11.& lites. The "or4 done in this process is +59 9T 13@ AF9 2//30 (a) 1FE&.3 J *olutio& : (d)

() 1E&J J

or4 done in an adiaatic process

(c)  1E&J J

(d)  13E&.3 J

genius PHYSICS

1'  Thermodynamic Processes

 V $    .&    11   1 J . , &E, log   = × × ×    = J., × &E,× (− >.F) ≈ −13E& J e   V  && . 0       i  

#  =  µ  RT loge 

Problem 22. 'n ideal gas A and a real gas B ha$e their $olumes increased !rom V  to &V  under isothermal conditions. The increase in internal energy +A> 95T 13@ -I958 2//1* 2//20 (a) ill e same in oth A and B

() ill e 6ero in oth the gases

(c) +! B "ill e more than that o! A

(d) +! A "ill e more than that o! B

*olutio& : (c) attraction.

In real gases an additional "or4 is also done in e#pansion due to intermolecular

Problem 23.

hich

o!

the !ollo"ing graphs correctly represents the $ariation β  = −(dV G dP) G V   "ith P !or an ideal gas at constant temperature +IIT?- #>creenin)$ 2//20

(a)

()

(c)

β 

(d)

o! 

β 

β 

*olutio& : (a)

P

P

β 

P

or an isothermal process PV  / constant P

So β  =

Problem 2'.

1 P

∴ graph

⇒

PdV +  VdP = >

⇒

−

1  dV   1    = V   dP   P

"ill e rectangular hyperola.

Consider the !ollo"ing statements

6ssertion ( A): The internal energy o! an ideal gas does not change during an isothermal process 8eason #R$ : The decrease in $olume o! a gas is compensated y a corresponding increase in pressure "hen its temperature is held constant. +! these statements

+>86 1'0

(a) 7oth A and R are true and R is a correct e#planation o! A () 7oth A and R are true ut R is not a correct e#planation o! A (c) A is true ut R is !alse (d) 7oth A and R are !alse (e) A is !alse ut R is true *olutio& : ()'s ∆)

∝ ∆T  so

!or isothermal process internal energy "ill not change and also P ∝

1 V 

(7oyleNs la")

Problem 2(. Ho" much energy is asored y 1> -! molecule o! an ideal gas i! it e#pands !rom an initial pressure o! J atm to 0 atm at a constant temperature o! &EOC +8oor&ee 120 (a) 1.E&J× 1>E J *olutio& : (a)

() 1E.&J× 1>E J

or4 done in an isothermal process

(c) 1.E&J× 1>I J

(d) 1E.&J× 1>I J

genius PHYSICS  Thermodynamic Processes 1(

  Pi   J     = (1>× 1>, ) × J., × ,>>× loge      = 1>0 × J., × ,>>× >.F,= 1.E&J× 1>E  J  P$     0      

#  = µ  RT loge 

Problem 2. 3 moles o! an ideal gas undergoes an isothermal process at 3>> '   in "hich its $olume is douled. The "or4 done y the gas system is (a) ,3>> J

() 100>> J

(c) 1EJ>> J

(d) 3&>> J

*olutio& : ()

 V    &V  ∆#  = µ  RT loge  $     = 3 × J., × 3>>× loge      = 3 × J., × 3>>× >.FI ≈ 100>>J .   V      V i  

Problem 2;.

or4 done y a system under isothermal change !rom a $olume V 1  to V &  !or a

 

gas "hich oeys Dander aalNs e-uation (V  − β &) P +

 

 V & − &β    & V 1 − V &     + α &     V  & β  V  V  −   1     1 &    V  − &α          + β && V 1 − V &    &RT loge &  V 1 − &α      V 1V &  

 V & − αβ    & V 1 − V &     + α &     − V  αβ  V  V    1     1 &    V  − &β    &  V 1V &     + α &     (d) &RT loge 1 V  & β  V  V  − −   &     1 &   () &RT log1>

(a) &RT loge (c)

*olutio& : (a)

7y Dander aalMs e-uation P or4 done #  =

V &

∫ 

V 1

= &RT [ loge (V  − &β )]

V &

∫ 

PdV = &RT 

=

V 1

V & V 1

α &&     = &RT  V     

&RT  V  − &β  dV 

V  − &β 

−

α &&

V &

− α &&

V &

dV 

V 1

V &

∫ 

V &

1 + α &   = &RT loge  V  V  &

1

 V  − V    − &β  + α &&  1 &     V 1 − &β    V 1V &  

V &

 13.; 6diabatic 9rocess. hen a thermodynamic system undergoes a change in such a "ay that no e#change o! heat ta4es place et"een it and the surroundings the process is 4no"n as adiaatic process. In this process P V  and T  changes ut

∆" /

Insulating "alls

>.

(1) ssential conditions for adiabatic process

*as

(i) There should not e any e#change o! heat et"een the system and its surroundings. 'll "alls o! the container and the piston must e per!ectly insulating. (ii) The system should e compressed or allo"ed to e#pand suddenly so that there is no time !or the e#change o! heat et"een the system and its surroundings. Since these t"o conditions are not !ully realised in practice so no process is per!ectly adiaatic. (&) Cample of some adiabatic process (i) Sudden compression or e#pansion o! a gas in a container "ith per!ectly non%conducting "alls. (ii) Sudden ursting o! the tue o! icycle tyre. (iii) Propagation o! sound "a$es in air and other gases. (i$) A#pansion o! steam in the cylinder o! steam engine.

genius PHYSICS

1  Thermodynamic Processes (,) !LTE in adiabatic process : ∆" = ∆) + ∆#  ut !or adiaatic process ∆" = > I!

∆# 

/ positi$e then

∆)

∴

∆) + ∆# = >

/ negati$e so temperature decreases i.e.  adiaatic e#pansion

produce cooling. I! ∆#  / negati$e then produce heating.

∆) /

positi$e so temperature increases i.e.  adiaatic compression

(0) quation of state : 's in case o! adiaatic change rst la" o! thermodynamics reduces to

∆) + ∆# = >  7ut as !or an ideal gas d) =  µ C V dT  and A-uation (i) ecomes  µ C+ dT + P dV  = >

i.e. d) + d# = >

?.. (i)

d# = PdV  ?.. (ii)

7ut !or a gas as PV  /  µ RT  P dV + V dP = µ R dT 

?.. (iii) (P dV + V dP)

So eliminating dT  et"een e-uation (ii) and (iii)  µ C+  or or

(P dV + V dP) (γ   − 1) γ  P

i.e.

+ P dV = >

 R  asC+  = (γ   − 1)  

+ P dV = >

dV + V dP = >

 µ R

γ  

dV  dP + V  P

=>

hich on integration gi$es γ  loge V  + loge P

=C

log(PV γ   ) = C

i.e.

PV γ  

or

?.. (i$)

= consta

A-uation (i$) is called e-uation o! state !or adiaatic change and can also e re%"ritten as TV γ  −1

and

T γ   P γ  −1

= constan

;as P / ( µ RT GV )

asV  = 

= consta

?.. (i$)

 

µ RT 

P

?.. ($i)

(3) Indicator dia)ram (i) Cur$e otained on PV graph are called adiaatic cur$e. (ii) Slope o! adiaatic cur$e : rom PV γ   7y di8erentiating "e get

dP V γ  

P

= consta

+ Pγ  V 

−1

γ  

dV  = > φ 

dP = dV  ∴ Slope

−γ  

PV 

−1

γ  

V γ  

P   = −γ        V  

V 

 P      V  

o! adiaatic cur$e tanφ  = −γ  

7ut "e also 4no" that slope o! isothermal cur$e tanφ  = So

Slopeo! adiaatic cur$e = Slopeo! isothermal cur$e

− γ  (P G V ) = γ   = − (P G V )

C p C+ 

>1

−P V 

genius PHYSICS  Thermodynamic Processes 1;

(F) >peciDc heat : Specic heat o! a gas during adiaatic change is 6ero 's

C

" > = => m∆T  m∆T 

=

;'s " / >

(E) 6diabatic elasticity : or adiaatic process PV γ  

= consta

@i8erentiating oth sides d PV γ   + Pγ  V γ  −1dV  = > γ  P

=

dP Stress = =E − dV G V  Strain φ 

= γ  P

Eφ 

i.e. adiaatic elasticity is γ   times that o! pressure ut "e 4no" isothermal elasticity Eθ  So

Eφ  Eθ 

=

'diaaticelasticity γ  P = Isothermal elasticity P

=P

= γ  

i.e. the ratio o! t"o elasticities o! gases is e-ual to the ratio o! t"o specic heats. (J) %or& done in adiabatic process #  =

V $ 

∫ 

V i

P dV  =

' 

V $ 

∫ 

V i

γ  

V 

   '    's = P     γ    V      

dV 

or

 '  '   =  γ  −1 − γ  −1  ;1 − γ  :  V $  V i  

or

=

or

=

So

=

  's 

1

− Pi V i : (1 − γ  )

;P $ V $   µ R

(1 − γ  )

;T $ 

∫ D

dV  =

%γ  

;'s '  = PV γ  

− T i :

;'s P $ V $  =

V −γ  +1   (−γ   + 1)

= P$ V $ γ   =

Pi V iγ   :

µ RT $   and

Pi V i

= µ RT i 

− P$ V $  :  µ R(T i − T $  ) = (γ   − 1) (γ   − 1)

;Pi V i

() !ree eCpansion : ree e#pansion is adiaatic process in "hich no "or4 is per!ormed on or y the system. Consider t"o $essels placed in a system "hich is enclosed "ith thermal insulation (asestos%co$ered). +ne $essel contains a gas and the other is e$acuated. The t"o $essels are connected y a stopcoc4. hen suddenly the stopcoc4 is opened the gas rushes into the e$acuated $essel and e#pands !reely. The process is adiaatic as the $essels are placed in thermal insulating system (d" / >) moreo$er the "alls o! the $essel are rigid and hence no e#ternal "or4 is per!ormed (d#  / >). Bo" according to the rst la" o! thermodynamics d) / > I! ) i  and ) $   e the initial and nal internal energies o! the gas then

) $ 

− )i = >

;'s ) $  = )i 

 Thus the nal and initial energies are e-ual in !ree e#pansion. (1>) >pecial cases of adiabatic process

genius PHYSICS

1=  Thermodynamic Processes PV γ  

T  ∝

= consta

∴

P

∝

1 γ  

V 

γ  



1−γ  

PT 

= consta

∴

P

γ   γ  −1  

∝ T 

TV γ  −1

and

= consta

∴

1 V γ  −1

P

Type of )as 2onoatomic γ   / 3G,

∝

P  ∝

@iatomic γ   / EG3

P  ∝

Polyatomic γ   / 0G,

P  ∝

1 γ  

V 

1

P

γ   γ  −1

T  ∝

∝ T 

P  ∝ T 3 G &

T  ∝

V 3 G , 1

P ∝ T E G &

T  ∝

V E G 3 1

P ∝ T 0

T  ∝

V 0 G ,

1 V γ  −1

1

V & G , 1 V & G 3 1

V 1G ,

(11) omparison between isothermal and adiabatic process (i) ompression : I! a gas is compressed isothermally and adiaatically !rom $olume V i to V $   then !rom the slope o! the graph it is clear that graph 1 represents adiaatic process "here as graph & represent isothermal process.

P

1 &

or4 done

# adiaatic  # isothermal

inal pressure

Padiaatic  Pisothermal

inal temperature

T adiaatic  T isothermal

V $ 

V i

V 

(ii) Cpansion : I! a gas e#pands isothermally and adiaatically !rom $olume V i to V $   then !rom the slope o! the graph it is clear that graph 1 represent P isothermal process graph & represent adiaatic process. or4 done

# isothermal  # adiaatic

inal pressure

Pisothermal  Padiaatic

inal temperature

T isothermal  T adiaatic

1 & V i

V 

V $ 

Sample problems based on Adiabatic process Problem 2=. @uring an adiaatic process the pressure o! a gas is !ound to e proportional to the cue o! its asolute temperature. The ratio C p G C+   !or the gas is +6I 2//30 (a) *olutio& : (a) C p C + 

=

, &

, &

()

0 ,

(c) &

(d) γ  

3 ,

γ  

=, *i$en P ∝ T , . 7ut !or adiaatic process . So γ   − 1 P ∝ T γ  −1

⇒

γ   =

, ⇒ &

genius PHYSICS  Thermodynamic Processes 1

'n ideal gas at &EOC is compressed adiaatically to

Problem 2. γ   =

J  o! its original $olume. I!  &E

3  then the rise in temperature is ,

+95T 1='@ A> 95T 1@ T 2//3@ 79>6T 2//30 (a) 03> ' 

() ,E3 ' 

(c) &&3 ' 

(d) 0>3 ' 

or an adiaatic process TV γ  −1 =  constant

*olutio& : ()

T 1 T &

∴

∆T  = FE3− ,>>=

 V   =  &  V 1 

γ  −1

⇒

T &

 V   = T 1  1   V & 

γ  −1

3

&E , = ,>>  J

−1

&E = ,>>  J

&G ,

⇒

= FE3' 

,E3' 

Problem 3/. I! γ   / &.3 and $olume is e-ual to 1GJ times to the initial $olume then pressure P is e-ual to (initial pressure / P)

′

+89T 2//30 (a) P′ = P *olutio& : (c)

() P ′ = &P

or an adiaatic process P′

Problem 31.

=

(c) P′ = P × (&)13G &

PV γ  

= 

constant

⇒

P& P1

 V   =  1  V & 

γ  

⇒

P′ P

= J3 G &

⇒

P × (&)13G &

In adiaatic e#pansion o! a gas +59 9T 2//30

(a) Its pressure increases !alls

() Its

(c) Its density decreases increases *olutio& : ()

(d) P′ = EP

(d)

Its

∆" = ∆) + ∆#  In an adiaatic process ∆" / > ∴ In e#pansion ∆#  / positi$e temperature decreases.

∴

∆)

temperature

thermal

energy

− ∆) = ∆# 

/ negati$e. Hence internal energy i.e.

Problem 32. P%V   plots !or t"o gases during adiaatic process are sho"n in the gure. Plots 1 and & should correspond respecti$ely to P 2//10 +IIT?- #>creenin)$ (a) 1e and 0&

 1

() 0&  and 1e

& V 

(c) 1e and A  (d) 0&  and N& *olutio& : ()Slope o! adiaatic cur$e

∝

γ  

∝

1 . So γ   is in$ersely proportional to 'tomicity o! thegas

atomicity o! the gas. ;'s γ   = 1.FF !or monoatomic gas γ   / 1.0 !or diatomic gas and γ   / 1.,, !or triatomic non%linear gas. rom the graph it is clear that slope o! the cur$e 1 is less so this should e adiaatic cur$e !or diatomic gas (i.e. 0&). Similarly slope o! the cur$e & is more so it should e adiaatic cur$e !or monoatomic gas (i.e. 1e).

genius PHYSICS

2/  Thermodynamic Processes Problem 33.

' monoatomic ideal gas initially at temperature T 1  is enclosed in a cylinder

tted "ith a !rictionless piston. The gas is allo"ed to e#pand adiaatically to a temperature T &  y releasing the piston suddenly. I! L1  and L&  are the lengths o!  the gas column e!ore and a!ter e#pansion respecti$ely then T 1 G T &  is gi$en y

+IIT?- #>creenin)$ 2///0 &G ,

  L   (a)  1      L&   *olutio& : (d)

()

L1

(c)

L& γ   −1

or an adiaatic process T 1V 1

γ   −1

= T &V &

&G ,

L&

 L   (d)  &       L1  

L1 ⇒

T 1 T &

 V   =  &  V 1 

γ  −1

 L  A  = &   L1 A 

3 G ,−1

L  =  &  L1 

&G ,

.

Problem 3'. our cur$es A B C and ( are dra"n in the adoining gure !or a gi$en amount o!  gas. The cur$es "hich represent adiaatic and isothermal changes are +95T 1=@ 79>6T 10

(a) C and ( respecti$ely

P B

() ( and C respecti$ely (c) A and B respecti$ely

C

 A

(d) B and A respecti$ely

( V 

*olutio& : (c) 's "e 4no" that slope o! isothermal and adiaatic cur$es are al"ays negati$e and slope o! adiaatic cur$e is al"ays greater than that o! isothermal cur$e so is the gi$en graph cur$e  A  and cur$e B represents adiaatic and isothermal changes respecti$ely.

Problem 3(. ' thermally insulated container is di$ided into t"o parts y a screen. In one part the pressure and temperature are P and T   !or an ideal gas lled. In the second part it is $acuum. I! no" a small hole is created in the screen then the temperature o!  the gas "ill +89T 10 (a) @ecrease *olutio& : (c) >

() Increase

(c) 5emain same

(d) Bone o! these

In second part there is a $acuum i.e. P / >. So "or4 done in e#pansion / P∆V  /

Problem 3.  T"o samples A and B o! a gas initially at the same pressure and temperature are compressed !rom $olume V  to V G& ( A isothermally and B adiaatically). The nal pressure o! A is +59 9T 1* @ 59 95T 1;* 0

*olutio& : (c)

(a) *reater than the nal pressure o! B

() A-ual to the nal pressure o! B

(c) =ess than the nal pressure o! B

(d) T"ice the nal pressure o! B

or isothermal process P1V  = P&N

V  &

N

⇒ P&

= &P1

?..(i)

γ  

γ  

or adiaatic process P1V 

V    = P&       &  

Since γ   > 1 . There!ore P& > P&N

⇒ P&

= &γ   P1

?..(ii)

genius PHYSICS  Thermodynamic Processes 21

Problem 3;.

' gas has pressure P and $olume V . It is no" compressed adiaatically to

1 ,&

times the original $olume. I! (,&)1.0 = 1&J the nal pressure is (a) ,& P

() 1&J P

(c)

or an adiaatic process P1V 1 = P&V & γ  

*olutio& : ()

∴ inal

γ  

⇒

P

(d)

1&J

P& P1

P ,&

γ  

 V 1     V   1.0 =     =    = (,&)1.0 = 1&J V  G ,&    V &    

pressure / 1&J P.

 13.= Isobaric 9rocess. hen a thermodynamic system undergoes a physical change in such a "ay that its pressure remains constant then the change is 4no"n as isoaric process. In this process V  and T  changes ut P remains constant. Hence CharleMs la" is oeyed in this process. (1) quation of state : rom ideal gas e-uation PV  /  µ RT  I! pressure remains constant V ∝ T 

V 1 T 1

=

V & T &

P

or

 A

I

B

P1

= consta

II

P&

(

(&) Indicator dia)ram : *raph I represent isoaric e#pansion graph II represent isoaric compression. Slope o! indicator diagram

C

V 

dP => dV 

  $  + 1 R    &  

(,) >peciDc heat : Specic heat o! gas during isoaric process C P =  (0) Aul& modulus of elasticity : '  =

∆P => − ∆V G V 

(3) %or& done in isobaric process : ∆#  =

V $ 

∫ 

V i

;'s ∆P / >

P dV  = P

V $ 

∫ 

V i

dV  = P;V $ 

− V i :

;'s

P

/ constant ∴

∆#  = P(V $  − V i ) =  µ R;T $  − T i : = µ R ∆T 

(F) !LTE in isobaric process : rom =T@

∆" = ∆) + ∆# 

∴

∆" =  µ 

R

∆) = µ  C V  ∆ T  =  µ  (γ   − 1) ∆T   

 1  R ∆T  +  µ R ∆T  =  µ R ∆T  + 1 (γ   − 1)  γ   − 1 

=  µ R ∆T 

and

∆#  =

µ R

∆T 

  γ     γ     =  µ   R ∆T  γ   − 1  γ   − 1 

∆" = µ  C P  ∆ T  (E) Camples of isobaric process (i) Con$ersion o! "ater into $apour phase (oiling process) : hen "ater gets con$erted into $apour phase then its $olume increases. Hence some part o! asored heat is used up to increase the $olume against e#ternal pressure and remaining

genius PHYSICS

22  Thermodynamic Processes amount o! heat is used up to increase the internal potential energy o! the molecules (ecause interatomic !orces o! attraction ta4es place et"een the molecules o! system and "hen the distance et"een them increases then its potential energy increases. It must e noted that during change o! state since temperature remains constant hence there "ill e no increase in internal 4inetic energy o! the molecules). rom rst la" o! thermodynamics ∆" = ∆) + ∆# 

= ∆) '  + ∆) P + ∆#  = > + ∆) P + P∆V 

∆) P = ∆" − P;V $  − V i :

∴

∆) P = mL − P;V $  − V i :

or

;'s ∆" / mL

(ii) Con$ersion o! ice into "ater

∆" = ∆) + ∆#  = ∆) P + ∆) '  + ∆#  mL =

; ∆) ' 

∆) P + > + P;V $  − V i :

=>

as

there

is

no

change

in

temperature

∆) P = mL

;"hen ice con$ert into "ater then change in $olume is

negligile Note :

In isoaric compression temperature increases and internal energy 9o"s out



in the !orm o! heat energy "hile in isoaric e#pansion temperature increases and heat 9o"s into the system. 

Isoaric e#pansion o! the $olume o! a gas is gi$en y V t  "here

γ  V 

=

1  pe °C &E,

= V > (1 + γ  V t )

= coecient o! $olume e#pansion.

Sample problems based on Isobaric process , 1 cm  o! "ater at its oiling point asors 30> caloies o! heat to ecome steam Problem 3=. "ith a $olume o! 1FE1cm, . I! the atmospheric pressure is 1.>1,× 1>3 N G m&  and the mechanical e-ui$alent o! heat / 0.1 JGcaloie the energy spent in this process in o$ercoming intermolecular !orces is +59 9T 1* 2//1@ ,rissa - 2//20

(a) 30> caloie

() 0> caloie

(c) 3>> caloie

(d) ero

Anergy spent in o$ercoming inter molecular !orces ∆) = ∆" − ∆# 

*olutio& : (c)

= ∆" − P(V & − V 1) = Problem 3.

' gas e#pands

1.>1,× 1>3 (1FE1− 1) × 1>−F 30>− 0.& >.&3 m,  at

≈ 3>>caloi

constant pressure 1>, N G m&  the "or4 done is

+95T 1;@ 79>6T 1@ -I958 2//1* 2//20 (a) &.3 e!s *olutio& : ()

() &3> J

's "e 4no" "or4 done =

(c) &3> # 

(d) &3> N

P∆V  = 1>, × >.&3= &3>   J

Problem '/. 3 mole  o! hydrogen gas is heated !rom ,>OC  to F>OC  at constant pressure. Heat gi$en to the gas is (gi$en R / & calGmole de!ee) +59 9T 2//20 (a) E3> caloie *olutio& : (c)

(∆") p

() F,> caloie

  γ       =  µ  C p ∆T  =  µ    R ∆T   γ   − 1 

(c) 1>3> caloie

(d) 10E> caloie

genius PHYSICS  Thermodynamic Processes 23

∴

E 3

(∆")p

  E      3   × & × ,>  = 3 × & × E × 3 × ,> = 1>3>caloi = 3×  E − 1  3 &     3  

  ;'s  µ  / 3 mole  and γ  /

 !or 1&

Problem '1.  The latent heat o! $aporisation o! "ater is &&0>  JG!m. I! the "or4 done in the process o! e#pansion o! 1! is 1FJ J then increase in internal energy is +95T 2///0 (a) &0>J J

() &&0> J

∆" = ∆) + ∆# 

*olutio& : (c)

⇒

(c) &>E& J

&&0>= ∆)

+ 1FJ ⇒

(d) 1>0 J

  J ∆) =  &>E&

Problem '2. hen an ideal gas (γ   / 3G,) is heated under constant pressure then "hat percentage o! gi$en heat energy "ill e utilised in doing e#ternal "or4 +89T 10 (a) 0>Q

() ,>Q

∆#  ∆" − ∆) = = ∆" ∆"

*olutio& : (a)

C p

− C+ 

C p

= 1−

(c) F>Q

1 = 1− γ  

1 1−

3 ,

=

(d) &>Q

& 3 & × 1>>/ 0>Q. 3

i.e. percentage energy utilised in doing e#ternal "or4 /

Problem '3.

't 1>>OC the $olume o! 1-! o! "ater is

normal pressure is

1.FE1m, .

, , 1>− m  and

$olume o! 1 -! o! steam at F

&., × 1>   J G -!  and

The latent heat o! steam is

the

normal pressure is 1>3 N G m& . I! 3 -! o! "ater at 1>>OC is con$erted into steam the increase in the internal energy o! "ater in this process "ill e (a) J.,3× 1>3 J *olutio& : () ∆" =

Heat

mL = 3 × &., × 1>F

(c) 11.3× 1>F J

() 1>.FF× 1>F J re-uired

to

con$ert

3

-!

o!

(d) ero "ater

into

steam

= 11.3 × 1>F  J

or4 done in e#panding $olume ∆#  = P∆V  = 3 × 1>3;1.FE1− 1>−, : = >.J,3× 1>F  J Bo"

y

rst

la"

o!

thermodynamics

∆) = ∆" − ∆# 

⇒

∆) = 11.3 × 1>F − >.J,3× 1>F = 1>.FF× 1>F  J

 13. Isochoric or Isometric 9rocess. hen a thermodynamic process undergoes a physical change in such a "ay that its $olume remains constant then the change is 4no"n as isochoric process. In this process P and T  changes ut V   / constant. Hence *ay%lussacMs la" is oeyed in this process. (1) quation of state rom ideal gas e-uation PV  /  µ RT 

genius PHYSICS

2'  Thermodynamic Processes I! $olume remains constant P

∝

T  or

P1 T 1

=

P& T &

= constant

(

P  A

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

22

2

C

decrease in pressure at $olume V 1  and isometric increase in

B

pressure at $olume V &  respecti$ely and slope o! indicator diagram dP dV 

V 1

V &

V 

=∞ (,) >peciDc heat : Specic heat o! gas during isochoric process C V  (0) Aul& modulus of elasticity  : '  =

=

$  R &

∆P ∆P = =∞ − ∆V  G V  >

(3) %or& done in isobaric process : ∆#  = P∆V  = P;V $  − V i :

∆#  = >

∴

(F) !LTE in isochoric process : ∆" = ∆) + ∆# 

= ∆)

;'s ∆#  / >

R

∆" = µ  C V  ∆ T  =  µ  γ   − 1 ∆T  = Note :

;'s V  / >



− Pi V i γ   − 1

P$ V $ 

Isometric e#pansion o! the pressure o! a gas is gi$en y Pt  "here γ  P

= P> (1 + γ  P t )

1   =      pe °C = coecient o! pressure e#pansion.  &E, 

Sample problems based on Isochoric process Problem ''. In pressure%$olume diagram gi$en elo" the isochoric isothermal and isoaric parts respecti$ely are P

 A

B

(a) BA A( (C

+5anipal 5 1(0 C

() (C CB BA (

(c) AB BC C(

V 

(d) C( (A AB *olutio& : (d)Process C( is isochoric as $olume is constant,  Process (A is isothermal as temperature constant and Process AB is isoaric as pressure is constant.

Problem '(. 2olar specic heat o! o#ygen at constant pressure C p = E.& calG molG °C  and R / J., JGmolG' . 't constant $olume 3 mol o! o#ygen is heated !rom 1>OC to &>OC the -uantity o! heat re-uired is appro#imately +59 95T 1=;0 (a) &3 cal *olutio& : (c)

() 3> cal

(c) &3> cal

(d) 3>> cal

7y 2ayerMs !ormula C+  = C p − R = E.& − & ≈ 3 calGmol°C 't constant $olume

∆" = µ c+ ∆T  = 3 × 3 × 1> =

&3>cal

 13.1/ 8eversible and Irreversible 9rocess. (1) 8eversible process : ' re$ersile process is one "hich can e re$ersed in such a "ay that all changes occurring in the direct process are e#actly repeated in the opposite order and

genius PHYSICS  Thermodynamic Processes 2(

in$erse sense and no change is le!t in any o! the odies ta4ing part in the process or in the surroundings. or e#ample i! heat is asored in the direct process the same amount o! heat should e gi$en out in the re$erse process i! "or4 is done on the "or4ing sustance in the direct process then the same amount o! "or4 should e done y the "or4ing sustance in the re$erse process. The conditions !or re$ersiility are (i) There must e complete asence o! dissipati$e !orces such as !riction $iscosity electric resistance etc. (ii) The direct and re$erse processes must ta4e place innitely slo"ly. (iii) The temperature o! the system must not di8er apprecialy !rom its surroundings. Some e#amples o! re$ersile process are (a) 'll isothermal and adiaatic changes are re$ersile i! they are per!ormed $ery slo"ly. () hen a certain amount o! heat is asored y ice it melts. I! the same amount o! heat is remo$ed !rom it the "ater !ormed in the direct process "ill e con$erted into ice. (c) 'n e#tremely slo" e#tension or contraction o! a spring "ithout setting up oscillations. (d) hen a per!ectly elastic all !alls !rom some height on a per!ectly elastic hori6ontal plane the all rises to the initial height. (e) I! the resistance o! a thermocouple is negligile there "ill e no heat produced due to  LouleMs heating e8ect. In such a case heating or cooling is re$ersile. 't a unction "here a cooling e8ect is produced due to Peltier e8ect "hen current 9o"s in one direction and e-ual heating e8ect is produced "hen the current is re$ersed. (!) Dery slo" e$aporation or condensation. It should e rememered that the conditions mentioned !or a re$ersile process can ne$er e realised in practice. Hence a re$ersile process is only an ideal concept. In actual process there is al"ays loss o! heat due to !riction conduction radiation etc. (&) Irreversible process : 'ny process "hich is not re$ersile e#actly is an irre$ersile process. 'll natural processes such as conduction radiation radioacti$e decay etc. are irre$ersile. 'll practical processes such as !ree e#pansion Loule%Thomson e#pansion electrical heating o! a "ire are also irre$ersile. Some e#amples o! irre$ersile processes are gi$en elo" (i) hen a steel all is allo"ed to !all on an inelastic lead sheet its 4inetic energy changes into heat energy y !riction. The heat energy raises the temperature o! lead sheet. Bo re$erse trans!ormation o! heat energy occurs. (ii) The sudden and !ast stretching o! a spring may produce $irations in it. Bo" a part o! the energy is dissipated. This is the case o! irre$ersile process. (iii) Sudden e#pansion or contraction and rapid e$aporation or condensation are e#amples o!  irre$ersile processes. (i$) Produced y the passage o! an electric current through a resistance is irre$ersile. ($) Heat trans!er et"een odies at di8erent temperatures is also irre$ersile. ($i) Loule%Thomson e8ect is irre$ersile ecause on re$ersing the 9o" o! gas a similar cooling or heating e8ect is not oser$ed.

 13.11 yclic and Bon?cyclic 9rocess . ' cyclic process consists o! a series o! changes "hich return the system ac4 to its initial state. In non%cyclic process the series o! changes in$ol$ed do not return the system ac4 to its initial state.

genius PHYSICS

2  Thermodynamic Processes (1) In case o! cyclic process as ) $  = )i ∴

∆) = ) $  − )i = > i.e. change in internal energy !or cyclic process is 6ero and also

∴ ∆T  /

∆) ∝ ∆T 

> i.e. temperature o! system remains constant.

(&) rom rst la" o! thermodynamics ∆" = ∆) + ∆# 

∆" = ∆#  i.e. heat supplied is e-ual to the "or4 done y the system.

;'s ∆) / >

(,) or cyclic process P%V  graph is a closed cur$e and area enclosed y the closed path represents the "or4 done. I! the cycle is cloc4"ise "or4 done is positi$e and i! the cycle is anticloc4"ise "or4 done is negati$e. P P  A

(

 A

B Positi$e "or4

(

Begati$e "or4 B

C

C

V 

V 

(0) or4 done in non cyclic process depends upon the path chosen or the series o! changes in$ol$ed and can e calculated y the area co$ered et"een the cur$e and $olume a#is on PV  diagram. P P

C

B C

B

 A (

 A #  = R Shaded area  ABC

#  =  Shaded area V   ABC(

V 

Sample problems based on Cyclic and noncyclic process Problem '.  The P%V  diagram o! a system undergoing thermodynamic trans!ormation is sho"n in gure. The "or4 done on the system in going !rom A → B → C is 3> J and &> cal heat is gi$en to the system. The change in internal energy et"een  A and C P C is +79>6T 2//20 (a) ,0 J  A

() E> J

B V 

(c) J0 J (d) 1,0 J *olutio& : (d) Heat gi$en ∆" = &>cal = process is anticloc4"ise 7y rst la" o! thermodynamics

&>× 0.& = J0  J

⇒

.

∆) = ∆" − ∆#  =

or4 done ∆# /  3> J

;'s

J0 − (− 3>) = 1,0   J

Problem ';. 'n ideal gas is ta4en through the cycle A → B → C → A as sho"n in the gure. I!  the net heat supplied to the gas in the cycle is 3  J the "or4 done y the gas in the process C → A is +IIT?- #>creenin)$ 2//20 V (m,)

(a)  3 J

& 1

C

B

 A 1>

P(N/m& )

genius PHYSICS  Thermodynamic Processes 2;

()  1> J (c)  13 J (d)  &> J *olutio& : (a)

= #  AB + # BC + # CA

or a cyclic process. Total "or4 done ⇒

1 × 1.> × 1> = 1>× (& − 1) + > + # CA &

⇒

3 J

;# BC / > since there is no change in $olume

along BC

Problem '=.

= 1> J + # CA

⇒ #  CA

= −3  J

In the !ollo"ing indicator diagram the net amount o! "or4 done "ill e P

(a) Positi$e 1

() Begati$e

&

(c) ero

V 

(d) Innity *olutio& : ()

or4 done during process 1 is positi$e "hile during process & it is negati$e. 7ecause process 1 is cloc4"ise "hile process & is anticloc4"ise. 7ut area enclosed y P%V   graph (i.e.  "or4 done) in process 1 is smaller so net "or4 done "ill e negati$e. Problem '. ' cyclic process !or 1 mole o! an ideal gas is sho"n in gure in the V %T  diagram.  The "or4 done in AB BC and CA respecti$ely

  V 1       R (T 1 − T &) V    &  

(a) > RT & ln

() R(T 1 − T &)> RT  1 ln

V  V &

V 1 V &

V 1

 V &       R (T 1 − T &) V    1    V &     (d) > RT & ln   R (T & − T 1) V    1   Process AB is isochoric Process BC is isothermal Process CA is isoaric

∴

 A

0

(c) > RT & ln

*olutio& : (c)

C

∴

#  AB

=

B

T 1

T 

T &

P ∆V  = >

 V    = RT &. ln &      V 1   = P∆V  = R∆T  = R(T & − T 1)

# BC

∴

# CA

Problem (/. ' cyclic process ABC( is sho"n in the gure P%V  diagram. hich o! the !ollo"ing cur$es represent the same process P

B

 A

C ( V 

(a)

P

 A

B

()

P

 A

B

(c)

P

B

(d)

 A C (

 A

B

(

C

C (

T 

P

( T 

C T 

T 

genius PHYSICS

2=  Thermodynamic Processes

*olutio& : (a)

 AB is isoaric process BC is isothermal process C( is isometric process and (A is isothermal process  These process are correctly represented y graph (a).

 13.12 Graphical 8epresentation of Farious 9rocesses .

Isochoric

P

P

'diaati c V 

V 

Isochoric Isoa r

Isoa rIsotherm

 13.13 "eat n)ine. 0

'diaati c

Isotherm Isoa r Isochoric

'diaati c T 

Isotherm 0

T 

0

Heat engine is a de$ice "hich con$erts heat into "or4 continuously through a cyclic process.  The essential parts o! a heat engine are >ource : It is a reser$oir o! heat at high temperature and innite thermal capacity. 'ny amount o! heat can e e#tracted !rom it. %or&in) substance :  Steam petrol etc. >in& : It is a reser$oir o! heat at lo" temperature and innite thermal capacity. 'ny amount o! heat can e gi$en to the sin4.  The "or4ing sustance asors heat "1 !rom the source does an amount o! "or4 #  returns the remaining amount o! heat to the sin4 and comes ac4 to its original state and there occurs no change in its internal energy. 7y repeating the same cycle o$er and o$er again "or4 is continuously otained.  The per!ormance o! heat engine is e#pressed y means o! 

Source

(T 1)

"1 Heat Angin e

# = "1  "& "& Sin4

(T &)

UeciencyV η  "hich is dened as the ratio o! use!ul "or4 otained !rom the engine to the heat supplied to it. η  =

or4done #  = Heatinput "1

7ut y rst la" o! thermodynamics !or cyclic process # = "1 − "& ∴

η  =

"1 − "& "1

= 1−

∆)  /

>.

∴ ∆"

/

∆# 

so

"& "1

' per!ect heat engine is one "hich con$erts all heat into "or4 i.e. #  = "1  so that "&  = >  and hence η  = 1 . 7ut practically eciency o! an engine is al"ays less than 1.  13.1' 8efri)erator or "eat 9ump. ' re!rigerator or heat pump is asically a heat engine run in re$erse direction. It essentially consists o! three parts

>ource : 't higher temperature T 1.

genius PHYSICS  Thermodynamic Processes 2

%or&in) substance : It is called re!rigerant li-uid ammonia and !reon "or4s as a "or4ing sustance. >in& : 't lo"er temperature T &.  The "or4ing sustance ta4es heat "& !rom a sin4 (contents o! re!rigerator) at lo"er temperature has a net amount o! "or4 done #  on it y an e#ternal Source agent (usually compressor o! re!rigerator) and gi$es out a larger ('tmosphere)(T 1) "1 amount o! heat "1 to a hot ody at temperature T 1  (usually Heat atmosphere). Thus it trans!ers heat !rom a cold to a hot ody at # = "1  "& Angin the e#pense o! mechanical energy supplied to it y an e#ternal e "& agent. The cold ody is thus cooled more and more. Sin4 (Contents o! re!rigerator)

 The per!ormance o! a re!rigerator is e#pressed y means o! 

(T &)

Ucoecient o! per!ormanceV β   "hich is dened as the ratio o! the heat e#tracted !rom the cold ody to the "or4 needed to trans!er it to the hot ody. i.e.

β  =

∴

β  =

Heate#tracted "&

=

"or4done

# 

=

"& "1 − "&

"& "1 − "&

' per!ect re!rigerator is one "hich trans!ers heat !rom cold to hot ody "ithout doing "or4 i.e. #  / > so that "1

= "&  and hence

β  =

∞

(1) arnot refri)erator or Carnot re!rigerator

"1 "&

=

T 1 T &

So coecient o! per!ormance β  =

∴

"1 − "& "&

=

T 1 − T & "&  or T & "1 − "&

=

T & T 1 − T &

T & T 1 − T &

"here T 1 / temperature o! surrounding T & / temperature o! cold ody It is clear that β  / > "hen T & / > i.e. the coecient o! per!ormance "ill e 6ero i! the cold ody is at the temperature e-ual to asolute 6ero. (&) 8elation between coeHcient of performance and eHciency of refri)erator e 4no" β  =

"& "1 − "&

 or β  =

7ut the eciency η  = 1 −

"& G "1 1 − "& G "1

"& "&  or "1 "1

= 1 − η 

?.. (i) ?..(ii)

 1− η 

rom (i) and (ii) "e get β  =

η 

 13.1( >econd Law of Thermodynamics . irst la" o! thermodynamics merely e#plains the e-ui$alence o! "or4 and heat. It does not e#plain "hy heat 9o"s !rom odies at higher temperatures to those at lo"er temperatures. It cannot tell us "hy the con$erse is possile. It cannot e#plain "hy the eciency o! a heat engine is al"ays less than unity. It is also unale to e#plain "hy cool "ater on stirring gets hotter

genius PHYSICS

3/  Thermodynamic Processes "hereas there is no such e8ect on stirring "arm "ater in a ea4er. Second la" o!  thermodynamics pro$ides ans"ers to these -uestions. Statement o! this la" is as !ollo"s (1) lausius statement : It is impossile !or a sel! acting machine to trans!er heat !rom a colder ody to a hotter one "ithout the aid o! an e#ternal agency. rom Clausius statement it is clear that heat cannot 9o" !rom a ody at lo" temperature to one at higher temperature unless "or4 is done y an e#ternal agency. This statement is in !air agreement "ith our e#periences in di8erent ranches o! physics. or e#ample electrical current cannot 9o" !rom a conductor at lo"er electrostatic potential to that at higher potential unless an e#ternal "or4 is done. Similarly a ody at a lo"er gra$itational potential le$el cannot mo$e up to higher le$el "ithout "or4 done y an e#ternal agency. (&) elvins statement : It is impossile !or a ody or system to per!orm continuous "or4 y cooling it to a temperature lo"er than the temperature o! the coldest one o! its surroundings. ' Carnot engine cannot "or4 i! the source and sin4 are at the same temperature ecause "or4 done y the engine "ill result into cooling the source and heating the surroundings more and more. (,) elvin?9lanc&s statement : It is impossile to design an engine that e#tracts heat and !ully utilises into "or4 "ithout producing any other e8ect. rom this statement it is clear that any amount o! heat can ne$er e con$erted completely into "or4. It is essential !or an engine to return some amount o! heat to the sin4. 'n engine essentially re-uires a source as "ell as sin4. The eciency o! an engine is al"ays less than unity ecause heat cannot e !ully con$erted into "or4.  13.1 arnot n)ine. Carnot designed a theoretical engine "hich is !ree !rom all the de!ects o! a practical engine.  This engine cannot e realised in actual practice ho"e$er this can e ta4en as a standard against "hich the per!ormance o! an actual engine can e udged. It consists o! the !ollo"ing parts (i) ' cylinder "ith per!ectly non%conducting "alls and a per!ectly conducting ase containing a per!ect gas as "or4ing sustance and tted "ith a non% conducting !rictionless piston (ii) ' source o! innite thermal capacity maintained at constant higher temperature T 1. (iii) ' sin4 o! innite thermal capacity maintained at constant lo"er temperature T &. (i$) ' per!ectly non%conducting stand !or the cylinder.

Ideal gas

Source T 1 ' 

Insulatin g stand

Sin4 T & ' 

(1) arnot cycle : 's the engine "or4s the "or4ing sustance o! the engine undergoes a cycle 4no"n as Carnot cycle. The Carnot cycle consists o! the !ollo"ing !our stro4es "1 (i) irst stro4e (Isothermal e#pansion) (cur$e  AB) :  The cylinder containing ideal gas as "or4ing sustance allo"ed to e#pand slo"ly at this constant temperature T 1.

# / "1  "&

or4 done / Heat asored y the system "&

genius PHYSICS  Thermodynamic Processes 31

# 1

= "1 =

V &

∫ 

V 1

 V &       = 'rea AB3E V    1  

P dV = RT 1 loge 

(ii) Second stro4e ('diaatic e#pansion) (cur$e BC) :  The cylinder is then placed on the non conducting stand and the gas is allo"ed to e#pand adiaatically till the temperature !alls !rom T 1 to T &. # & =

V ,

∫ 

V &

R

P dV /

(γ   − 1)

;T 1

− T & : = 'rea BC13

(iii) Third stro4e (Isothermal compression) (cur$e C() :  The cylinder is placed on the sin4 and the gas is compressed at constant temperature T &. or4 done / Heat released y the system # ,

= "& = −

V 0

∫  P dV = −RT  log &

V ,

e

V 0 V ,

= RT & loge

V , V 0

= 'reaC(F1

(i$) ourth stro4e (adiaatic compression) (cur$e (A) : inally the cylinder is again placed on non%conducting stand and the compression is continued so that gas returns to its initial stage. # 0

=−

R

V 1

∫  P dV = − γ   − 1(T 

&

V 0

− T 1) =

R (T 1 γ   − 1

− T & ) =

'rea A(FE

(&) Hciency of arnot cycle :  The eciency o! engine is dened as the ratio o! "or4 η  =

done to the heat supplied i.e.

"or4done #  = Heatinput "1

Bet "or4 done during the complete cycle

#  = # 1

  0 ) = # 1 − # ,  = 'reaABC( + # & + (−# , ) + (−# 

#  # 1 − # , = "1 # 1

∴

η  =

or

η  = 1 −

=

"1 − "& "1

= 1−

# , # 1

= 1−

"& "1

RT & loge (V , G V 0 ) RT 1 loge (V & G V 1 )

Since points B and C lie on same adiaatic cur$e

V , V &

=

V 0 V 1

 or

V , V 0

=

V & V 1

So eciency o! Carnot engine η  = 1 −

⇒

−1

γ  

T 1V &

∴

'lso point ( and A lie on the same adiaatic cur$e

rom (i) and (ii)

;'s # & = # 0 

∴

= T &V ,

T 1V 1γ  −1

−1

γ  

=

 or

T &V 0γ  −1

 or

T 1 T & T 1 T &

γ  −1

 V    =  ,      V &  

?..(i) γ  −1

 V    =  0     ?..(ii)   V 1  

 V ,         = loge  V &     V 0    V 1  

loge 

T & T 1

(i) Aciency o! a heat engine depends only on temperatures o! source and sin4 and is independent o! all other !actors. (ii) 'll re$ersile heat engines "or4ing et"een same temperatures are e-ually ecient and no heat engine can e more ecient than Carnot engine (as it is ideal). (iii) 's on Kel$in scale temperature can ne$er e negati$e (as > '  is dened as the lo"est possile temperature) and T l and T & are nite eciency o! a heat engine is al"ays lesser than

genius PHYSICS

32  Thermodynamic Processes unity i.e. "hole o! heat can ne$er e con$erted into "or4 "hich is in accordance "ith second la". Note :  The eciency o! an actual engine is much lesser than that o! an ideal engine. 'ctually the practical eciency o! a steam engine is aout (J%13)Q "hile that o! a petrol engine is 0>Q. The eciency o! a diesel engine is ma#imum and is aout (3>% 33)Q. (,) arnot theorem :  The eciency o! CarnotMs heat engine depends only on the T & temperature o! source (T 1) and temperature o! sin4 (T &) i.e. η  = 1 − . T 1 Carnot stated that no heat engine "or4ing et"een t"o gi$en temperatures o! source and sin4 can e more ecient than a per!ectly re$ersile engine (Carnot engine) "or4ing et"een the same t"o temperatures. CarnotNs re$ersile engine "or4ing et"een t"o gi$en temperatures is considered to e the most ecient engine.  13.1; EiJerence Aetween 9etrol n)ine 6nd Eiesel n)ine.

9etrol en)ine

Eiesel en)ine

(i) or4ing sustance is a mi#ture o! petrol $apour and air.

(i) or4ing sustance in this engine is a mi#ture o! diesel $apour and air.

(ii) Aciency is smaller (W0EQ).

(ii) Aciency is larger (W33Q).

(iii) It "or4s "ith a spar4 plug.

(iii) It "or4s "ith an oil plug.

(i$) It is associated "ith the ris4 o! e#plosion ecause petrol $apour and air is compressed. So lo" compression ratio is 4ept.

(i$) Bo ris4 o! e#plosion ecause only air is compressed. Hence compression ratio is 4ept large.

($) Petrol $apour and air is created "ith spar4 plug.

Spray o! diesel is otained through the et.

Sample problems based on Second law of thermodynamic Problem (1.

I! the door o! a re!rigerator is 4ept open then "hich o! the !ollo"ing is true

+A"7 2//1@ -I958 2//2@ 6I 2//2@ 95T 2//30 (a) 5oom is cooled

() 5oom is heated

(c) 5oom is either cooled or heated

(d) 5oom is neither cooled nor heated

In a re!rigerator the "or4ing sustance ta4es heat "&  !rom the sin4 at lo"er

*olutio& : ()

temperature T &  and gi$es out a larger amount o! heat "1  to a hot ody at higher temperature T 1 . There!ore the room gets heated i! the door o! a re!rigerator is 4ept open.

Problem (2.  The coecient o! per!ormance o! a Carnot re!rigerator "or4ing et"een ,>oC and >oC is +79>6T 2//20 (a) 1> *olutio& : (c) >OC is

() 1

(c) 

(d) >

Coecient o! per!ormance o! a Carnot re!rigerator "or4ing et"een ,>OC and

β  =

T & T 1 − T &

=

&E,°C ,>,°C − &E,°C

=

&E,°C ,>°C

≈

genius PHYSICS  Thermodynamic Processes 33

Problem (3. ' Carnot engine "or4ing et"een ,>> '  and F>> '  has "or4 output o! J>> J per cycle. hat is amount o! heat energy supplied to the engine !rom source per cycle +9b. 95T 2//20 (a) 1J>> JGcycle

() 1>>> JGcycle

(c) &>>> JGcycle

Aciency o! Carnot engine η  =

*olutio& : (d) 'gain

η 

=

or4 done Heatinput

∴ Heat

(d) 1F>> JGcycle

T 1 − T & F>>− ,>> 1 = T 1 F>> & or4done η 

input =

=

J>> 1G &

= 1F>>  J .

Problem ('. Carnot cycle (re$ersile) o! a gas represented y a Pressure%Dolume cur$e is sho"n in the diagram P  A

Consider the !ollo"ing statements I.

B

'rea ABC( / or4 done on the gas

(

II. 'rea ABC( / Bet heat asored

C V 

III. Change in the internal energy in cycle / > hich o! these are correct +657 #5ed.$ 2//10 (a) I only *olutio& : (c)

() II only

(c) II and III

(d) I II and III

or4 done y the gas (as cyclic process is cloc4"ise)

∴ ∆#  /

'rea ABC(

So !rom the rst la" o! thermodynamics ∆" (net heat asored) / ∆#  / 'rea ABC( 's change in internal energy in cycle

∆) /

>.

Problem ((. ' Carnot engine ta4es 1>, -cal o! heat !rom a reser$oir at F&EOC and e#hausts it to a sin4 at &EOC. The eciency o! the engine "ill e (a) &&.&Q *olutio& : (d)

() ,,.,Q

Aciency o! Carnot engine

(c) 00.0Q

=

T 1 − T & T 1

=

I>>− ,>> I>>

(d) FF.FQ

=

F I

 or FF.FQ.

 13.1= ntropy. Antropy is a measure o! disorder o! molecular motion o! a system. *reater is the disorder greater is the entropy.  The change in entropy i.e. d* =

Heatasored ysyste 'solute temperatu re

 or d* =

d" T 

 The relation is called the mathematical !orm o! Second =a" o! Thermodynamics. Important points (1) !or solids and liquids (i) hen heat gi$en to a sustance changes its state at constant temperature then change in entropy d* =

d" mL =± T  T 

"here positi$e sign re!ers to heat asorption and negati$e sign to heat e$olution. (ii) hen heat gi$en to a sustance raises its temperature !rom T 1 to T & then change in entropy

genius PHYSICS

3'  Thermodynamic Processes d* =

d"

 T    dT  = mcloge  &   mc   T   T 1  

T &

∫  T  ∫  =

T 1

 T    ∆* = &.,>,mcloge  &   .  T 1  

⇒

(&) !or a perfect )as : Per!ect gas e-uation !or & moles is PV  / &RT 

∆* = ⇒

∆* =

d" = T 

∫ 

&CV dT + P dV 

∫ 

T 

 

&RT  dV  = &CV  V  T 

&CV dT +

∫ 

;'s d" / d) R d# X T &

∫ 

dT  T 

T 1

V &

∫ 

+ &R

V 1

dV 

;'s PV  / &RT 

V 

 T     V    ∆* = &CV  loge  &   + &Rloge  &     T 1    V 1  

∴

 T     P   ∆* = &CP loge  &   − &Rloge  &     T 1    P1    P    V    ∆* = &CV  loge  &   + &CP loge  &     P1    V 1  

Similarly in terms o! T  and P

and in terms o! P and V 

Sample problems !iscellaneous Problem (. 'n ideal gas e#pands in such a manner that its pressure and $olume can e related y e-uation PV & / constant. @uring this process the gas is +79>6T 2//20 (a) Heated

() Cooled

(c) Beither heated nor cooled cooled

(d)

or an adiaatic e#pansion PV γ  

*olutio& : () constant ∴ It

Problem (;.

irst heated and then PV & /

= constan and !or the gi$en process

is also an adiaatic e#pansion and during adiaatic e#pansion the gas is cooled.

' cyclic process ABCA is sho"n in the V%T  diagram. Process on the P%V  diagram is V  C

B  A

(a)

P

C

()

P

 A B

*olutio& : (c)

T 

(c)

B

 A

P  A

C

V 

V 

(d)

B

C

P  A

C V 

B V 

rom the gi$en VT  diagram "e can see that In process  AB same)

V  ∝ T 

∴ Pressure

is constant ('s -uantity o! the gas remains