8. ktg & thermodynamics part-1

Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 154

EXERCISE # 1

1.1 For isothermal, PV = constant, P V1

. As area decreases number of collision per unit area increases

l er ki h; ds fy, PV = fu; r kad , P V1

. {ks=kQy ?kVkus l s i zfr , dkad {ks=kQy i j VDdj ksa dh l a[ ; k c<r h gSA

1.2 Vav = mKT8

, as T = constant fu; r kad Vav = constant fu; r kad

1.3* By energy conservation, energy loss by one molecule is equal to gain by other.Åt kZ l aj {k.k l s , d v.kq } kj k O; f; r mt kZ] nwl j s v.kq } kj k i zkIr mt kZ ds r qY; gksxhA

1.4* avVMP

, As pwafd avV

= 0 (in equilibrium l kE; koLFkk esa) avP

= 0

1.5 avav VMP

, as the average momentum of an ideal gas is zero option D is correct.

avav VMP

pwafd vkn' kZ xSl dk vkSl r l aosx ' kwU; gksr k gS

fodYi D l R; gS

2.1 Real gas behaves as an ideal gas at low pressure and high temperatureokLr fod xSl ] fuEu nkc r Fkk mPp r ki i j vkn' kZ xSl dh Hkkafr O; ogkj dj r h gSA

2.2 Pm

= nRT slope of T1 dh <ky > slope of T2 dh <ky T1 > T2

2.3 one molecule has some single value of speed which is equal most probabla speed and average speed of the gas, d v.kq dh dksbZ , d fuf' pr pky dk eku gksxk t ks bl dh vf/kdr e l EHkkO; pky r Fkk vkSl r pky ds eku ds r qY;gksxkA Vmp = Vav.

2.4 VAV = 0M

RT8 = v

for nitrogen ukbVªkst u ds fy, VAV = 2/MT2R8

0

= v .

2.5 U = 2

kTnfN2

nfRT A

fkT

U2 = nNA = N

2.6 Vav 0M

1

oxygen molecule hits the wall with smaller average speed vkDl ht u v.kq nhokj i j de vkSl r pky l s Vdj kr s gSA

KTG & THERMODYNAMICS


2.7 PV = nRT temperature remains same for all ideal gas

l Hkh vkn' kZ xSl ds fy, r ki l eku j gr k gSA

3.1 Vav = 0M

RT8 , VAVV T

For same temp in vessel A, B and C, Average speed of O2 molecule is same in vessel A and C and is equalto V1.i k=k A, B o C, esa l eku r ki i j O2 v.kq dh vkSl r pky i k=k A o C esa l eku gksxh o V1 ds r qY; gksxhA

4.1 As translation K.E is pawfd LFkkukUr fj ; K.E = 23

nRT E = 23

PV

where E = total translational K.E.t gk¡ E = dqy LFkkukUr fj ; K.E.

4.2 For an ideal gas, the no of molecules of equal moles of gas is same .vkn' kZ xSl ds fy, xSl ds l eku eksy l a[ ; k esa v.kqvksa dh l a[ ; k l eku gksr h gSA

4.3 As U is a state function i.e., it depends initial and final positionpawfd U LFkSfr d Qyu gS vFkkZr ; g i zkj fEHkd o vafr e fLFkfr ; ksa i j fuHkZj dj r k gSAin process A and B initial and final temp are same.i zØe A o B esa i zkj afEHkd o vafr e r ki l eku gSA U, = U2 .

5.1 As pawfd U = nRT For closed path cUn i Fk ds fy,T = 0 U = 0.

5.2 T P or ; k TP

= constant fu; r kad

As pawfd VnR

TP = constant fu; r kad or ; k V = constant fu; r kad W = 0.

5.3* A B constant pressure fu; r nkc B C T = constant fu; r kadC D constant Volume fu; r vk; r u D A T = constant fu; r kad clearly, option A and B are constant Li "Vr % fodYi A o B fu; r kad gSA

5.4 As PV = nRT m = V = constant or V1

and P

pawfd PV = nRT m = V = fu; r kad ; k V1

r Fkk P

A B T = constant, pressure increases or volume decreasesA B T = fu; r kad nkc c<us i j vk; r u ?kVsxkB C Volume is constant, V = constantB C vk; r u fu; r gS P = fu; r kadC D P is decreases or volume increases [T = constant]C D P ?kVsxk vFkok vk; r u c<sxk [T = fu; r kad ]D A Volume is constant V = with constant,D A vk; r u fu; r kad gS V = fu; r gSA

clearly option ‘B’ is constant.Li "V fodYi ‘B’ fu; r gSA


5.5 U = 0 T = constant fu; ror PV = constant or P-V curve is a rectangular hyperbola.; k PV = fu; r kad ; k P-V vkj s[ k vk; fr ; i j oy; kdkj gksxkAclearly, option B is correct. Li "Vr % fodYi B l gh gSA

5.6 PnR

TV

P1

slope <ky or ; k P slope1

P2 < P1

5.7 In isothermal expansion l er ki h; i zl kj esaT = constant fu; r U = 0 W = Q option (D) is correct. fodYi (D) l gh gSA

5.8 W.D. = × Pressure Radius × volume Radius (area of ellipse)W.D. = × nkc f=kT; k × vk; r u f=kT; k (nh?kZoÙk dk {ks=k)

W =

2

P–P 12

2

V–V 12 = 4

(P2 – P1) (V2 –V1)

5.9 L M P = constant fu; r V T. MN T = constant fu; rHere, option B is constant ; gkW fodYi B fu; r gSA

5.10 work done on the gas = negative workxSl i j fd; k x; k dk; Z = _ .kkRed dk; ZW = PdV when t c V decreases ?kVr k gSthen r ks W = – vehence option D is correct. vr % fodYi D l gh gSA

5.11 As volume increases vk; r u c<us i jWD continuously increases dk; Z yxkr kj c<sxk

5.12 As pwafd W = PV V = same is both process nksuks i zØeksa esa l eku gSAs pawfd PB > PA W2 > W1 .

5.13 As initial and final state are same pawfd i zkj afEHkd o vafr e voLFkk l eku gSA

TI = TF As pawfd Vrms, avP

and o avK

depends on temperature r ki i j fuHkZj gS all are equal. l Hkh l eku gSA

6.1 U = same is both process nksuks i zØeksa esa l eku gSA Qacb – Wacb = Qadb – Wadb .200 – 80 = 144 – Wadb. Wadb = 24 J.

6.2 U = Qacb – Wacb = 200 – 80 = 120 JU = Qba – Wba , – 120 = Qba + 52 , Qba = – 172 J.

6.3 Ub – Ua = 120 Ub = 120 + 40 = 160J

6.4 in db. Wdb = 0 Ub – Ud = Qdb.160 – 88 = Qdb Qdb = 72J.


6.5* T = nRPV

initial and final temperature are equal i zkafj EHkd o vafr e r ki l eku gSA

as pawfd U = 2RTnf

Uinitial = Ufinal.

Also the net work done by an ideal gas in the process may be zero.i zØe esa vkn' kZ xSl } kj k fd; k x; k dqy dk; Z ' kwU; gks l dr k gSA

6.6* in cyclic process. pØh; i zØe esU = 0 Q = W as pwafd W = + ve Q = +veor Net heat energy has been supplied to the system. in process CAfudk; dks dqy m"eh; mt kZ i znku dh x; h gSA i zØe CA esaW = 0 U = –ve (As pawfd T = decreases ?kV j gk gS) heat energy is rejected out by system fudk; } kj k m"eh; mt kZ R; kxh xbZ gSATeperature at C is maximum C i j r ki eku vf/kdr e gksxkA

6.7* = 1

211 Q

Q–QQW

P/P/0

= 1 –

12

QQ

.

6.8 In process AB T = constant P = increases P V1

i zØe AB eas T = fu; r P = c<sxk P V1

or ; k V = decreases ?kVsxk Q = W . W = – ve. or ; k Q = – ve heat is rejected out of the system. fudk; } kj k m"ek R; kxh t k, xhA

6.9 Q = W (T = constant fu; r ) if heat is released then ; fn m"ek eqDr gks r ks W = –ve

6.10* W = PdV. then r c W = –veAs pressure and volume both decreases t c nkc o vk; r u nksuksa ?kVr s gSa temperature of system decreases fudk; dk r ki ?kVsxk

6.11* U = Q – W is same in both methods as it is a state functionnksuks i zfØ; k esa l eku gS D; ksfd ; g fLFkfr Qyu gSA

6.12* in cyclic process pfØ; i zØe esa U1 + U2 = 0 UNet = 0Q – W = 0

7.1 dQ = dW + dUdQ = PdV + dUdQ = nRdT + dU

dQ = fdU2

+ dU

2RdTƒndU

dQdU

=

1

f2

1

dQdU

= 75


7.2 As pawfd Cp – Cv = RFor above equation, we can say that both Cp and Cv increase by same amount.mi ; qZDr l ehdj .k esa dg l dr s gS fd Cp o Cv l eku ek=kk esa c<r s gSA

7.3 s = TmQ

For changing state voLFkk i fj or Zu ds fy,T = const fu; r vksj or T = 0 s = (infinite vuUr )

7.4 As f = 5

dU = nCv dT = 2nfRdT

Cv = 2fR

Cv = 2R5

7.5 At constant pressure nCPdT = U+ PdV [by first law of thermodynamics]At constant volume nCvT = U, Cp > Cv

fu; r nkc i j nCPdT = U+ PdV [Å"ek xfr dh ds i zFke fu; e l s]fu; r vk; r u i j nCvT = U, Cp > Cv

7.6 Gas has different specific heat for different processesxSl ksa dh fHkUu fHkUu i zØeksa ds fy, fof' k"V m"ek fHkUu&fHkUu gksr h gSA gas has infinite number of specific heats. xSl dh fof' k"V m"ek vuUr gksr h gSA

7.7 U > 0and vksj W > 0 C > Cv

7.8 As compare to gas solid expand very less. xSl dh r qyuk esa Bksl vR; Yi i zl fj r gksr s gSA Cp is slightly greater then Cv. Cp dk eku Cv l s FkksMk l k vf/kd gksr k gSA

7.9* Specific heat of a substance can be finite, infinite, zero and negative.i nkFkZ dh fof' k"V Å"ek i fj fer vi fj fer ' kwU; ; k _ .kkRed gks l dr h gSA

8.1 As Volume decreases vk; r u ?kVus i j pressure of the gas in the cylinder increases csyu esa xSl nkc c<sxk

8.2 AB isothermal l er ki h;PA VA = PB VB ...(i)BC Adiabatic : } ks"ePB VB

= PC VC ...(ii)

CD Isothermal l er ki h;PCVC = PDVD ...(iii)DA Adiabatic : } ks"ePDVD

= PA VA ...(iv)

From (i), (ii), (iii) and (iv) (i), (ii), (iii) vksj (iv) l s

D

A

C

BVV

VV


8.3* From information, the process may be very nearly adibatic. Hence option (C) is correct.i zkIr t kudkj h l s i zØe : ) ks"e ds l ehi r % gksxkA vr % fodYi (C) l gh gSA

8.4* U = 0 (Adiabatic : } ks"e)U = const fu; rnCvT = const fu; rAs O2 and N2 are diatomic, so there temp are equal but is different from Hepwafd O2 o N2 f} i j ek.kfod gS] vr % buds r ki l eku gksxsa i j Ur q He ds r ki l s fHkUu gksxsaAFor adiabatic : } ks"e ds fy, PV = const fu; r kadFor O2, N2 value of is same pressure of O2, N2 remains same but different from HeO2, N2 ds fy, dk eku l eku gS O2 o N2 dk nkc l eku j gsxk fdar q He ds eku l s fHkUu gksxkA

8.5* Q = U + W 25 = 2TnfR

+ 0

25 = 32225f1

f = 3 (monoatomic , d i j ek.kfod)

8.6 For adiabatic : } ks"e i zØe ds fy,T V–1 = C ( > 1) ....(i)For isothermal l er ki h; ds fy, T = const fu; r kad ....(ii)From (i) and (ii) (i) o (ii) l sT2 < T1

8.7 For isothermal l er ki h; ds fy,

PV = C . or 1

1 V1P ...(i)

For adiabatic : } ks"e ds fy,

PV = C,

2V

1P2 ...(ii)

from (i) and (ii) (i) o (ii) l sP1 > P2

8.8 As W.D. by gas in isothermal is more as compare to adiabatic processl er ki h; i zØe esa xSl } kj k fd; k x; k dk; Z] : ) ks"e i zØe esa fd; s x; s dk; Z dh r qyuk esa vf/kd gksr k gS

W2 < W1

8.9 Isothermal l er ki h; P V1

Adiabatic : } ks"e P V1

Also, slope of adiabatic is more as compare to isothermal: } ks"e oØ dh <ky l er ki h oØ dh r qyuk esa vf/kd gksxhA option (C) is correct. fodYi (C) l gh gSA


8.10 Adiabatic process : } ks"e i zØe esaQ = 0For any process fdl h Hkh i zØe esaU = nCVTHence, option (C) is correct. vr % fodYi (C) l gh gSA

8.11* CP > CV

and o CP – CV = 2 option A and B is correct. fodYi A o B l gh gSA

8.12 B = P (for adiabatic process : } ks"e i zØe ds fy,B = 1.4 × 1 × 105 = 1.4 × 105 N/m2

8.13 B = dVVdP

= – dV)PdV(

(for isothermal process l er kr h i zØe ds fy, )

B = P

8.14* For adiabatic process : } ks"e i zØe ds fy, PV = K.

or ; k P V1

And r Fkk PV = nRT

8.15 Slope <ky = – dVdP

As slope of A > slope of B A dh <ky > B dh <ky of A > of Bor ; k A Helium fgfy; e

B Hydrogen gkbMªgkst u

8.16 B AQ = 00 = – 30 + UBAUBA = 30 J UAB = –UBA = – 30 J

8.17 For free expansion eqDr i zl kj ds fy,U = 0 or T = 0 U or T = const fu; r kad

8.18 For free expansion eqDr izlkj ds fy,Q = 0, W = 0, U = 0

8.19 PdP

= – VdV

(For adiabatic : } ks"e ds fy, )

0.5 = – 1.4 VdV

Volume decrease by 0.36% vk; r u 0.36% ?kVsxk

8.20 XY Adiabatic compresion : } ks"e l ai hMuYZ Isothermal Expansion l er ki h; i zl kjZX Compression at constant pressure fu; r nkc i j l ai hMu


8.21 Self explainatory Lor % Li "V

8.22 As W.D. is isobaric > W.D. in Isothermal > W.D in adiabatic pwafd l enkch; esa WD > l er ki h; esa W.D. : } ks"e esa WDor ; k W2 > W1 > W3

Hence option (A) is correct. vr % fodYi (A) l gh gSA

8.23 Process ...(1) is isobaric izØe ...(1) lenkch; gSAU1 = Q – W = positive /kukRed

process (2) is isothermal i zØe (2) l er ki h; gSAU2 = 0

Process (3) is adiabatic i zØe (3) : } ks"e gSAQ = 0U = – W = negative _ .kkRed

U1 > U2 > U3

8.24 As pwafd Q = U + WU = –W (given fn; k x; k gS)or ; k Q = 0 Process is adiabatic i zØe : } ks"e gSA

8.25 For polytropic process i ksyhVªksfi d i zØe ds fy, PVx = k;

C = CV + x1

R

As pwafd PV2 = K (given fn; k x; k gS) Put x = 2

C = CV + 2–1

R = CV – R. C < CV .

EXERCISE # 2PART - I

1. For larger n, pressure will be smaller, so work done will be smaller for larger n.n ds c<s ekuks ds fy; s nkc de gksxk vr % n ds c<s ekuksa ds fy; s dk; Z de gksxkA

2. For an ideal gas vknZ' k xSl ds fy; sCP – CV = RIf ; fn CP – CV = 1.09 R.or pA > pB TA < TBThen gas will be real . Thus pressure is high and temperature is low for real gas.r c xSl okLr fod xSl gksxhA vr % okLr fod xSl ks ds fy; s nkc vf/kd vkSj r ki eku de gksxkA

3. V = kT2/3

dV = 32

31–

Tk d T

W = PdV = VnRT

dV

= R VT

dV = R 32

31–

KT

TdTKT32

= 32

R (T2 – T1)

= 32

R (30) = 20 (8.31) = 166.2 J


4. VPn = constant. fu; r kaddV Pn + VnPn–1 dP = 0

– dVVdP

= nP

= bulk modulus vk; r u i zR; kLFkr k xq.kkad

5. V = k 33.0

VTnRT

V1.33 = const fu; r kadV = const fu; r kad proceses is isochoric i zØe l evk; r fud gSA

6. Correct graph is shown in option (A)l gh xzkQ fp=kkuql kj (A) esa fn; k x; k gSAProcess 1–2 adiabatic process, Process 2–3 Isochoric process, process 3–1 Isothermal process.i zØe 1–2 : } ks"e gS i zØe 2–3 l evk; r fud gS i zØe 3–1 l er ki h; gSA

7. Work done by gas = Area under P-V diagramxSl } kj k fd; k x; k dk; Z = P-V vkj s[ k vUr xZr {ks=kQy

= 2)2–4()3–4(

+ 2)5.2–3()1–2(

= 25.2

= 45

atm L

W = –

4

5 atm L (Work done by gas is negative as cycle is anticlockwise)

(pØ.k okekor Z gS bl fy, xSl } kj k fd; k dk; Z _ .kkRed gksxkA)

8. For adiabatic process : ) ks"e i zfØ; k ds fy; sP V1

= PA V2

PA = P

2

1VV

......(1)

For isothermal process l er ki h i zfØ; k ds fy; sP V1 = PB V2

PB = P 2

1VV

.....(2)

From (1) and (2) l ehdj .k (1) vkSj (2) l sPA < PB [For expansion i zl kj ds fy, V2 > V1 ]and by vkSj PV = nRT TA < TB

9. (P = constant) fu; r kad

WQ

= TnRTCn P

= RCP = 2

5

10. Process AB is isobaric [V T] i zØe AB l enkch; gS [V T] ATB > TA UB > UA

WBC < WAB (Area under P-V curve) (P-V vkj s[ k ds vUr xZr {ks=kQy )


11. dW = dQ – dUdW = nCdT – nCVdT

W = CdT – dTCV

= dTTa

– CV T

= a ln

0

0TT –

1–R)T–T( 12

W = a ln 1–RT)1–(– 0

12. T = T0 + aV3

nRPV

= T0 + aV3

P = nR

20 aVVT

For minimum P, dVdP

= O U; wur e P ds fy; s dVdP

= O

20

VT–

+ a 2V = 0 V = 31

0a2

T

13. P = R

20 aVVT

and vkSj V = 31

0a2

T

P = 23

31

32

031

2TRa

14. Process12 and Process34 are isochoric process.i zfØ; k 12 vkSj i zfØ; k 34 l evk; r uh i zfØ; k gSW12 = 0W34 = 0W23 = n R (T3 – T2) = 3 R (2400 – 800) = 4800 RW41 = nR (T1 – T4) = 3 R (400 – 1200) = – 2400 RW = (4800 – 2400) R = 2400 R = 20 kJ

15.smr

soundV

V

osxewy ek/;oxZ

/ofu

VV

=

P3

P

= 95

= 35

[Monoatomic gas] (, dy i j ek.kfo; xSl )

PT = const fu; r kadP2V = const fu; r kad PV1/2 = const fu; r kad

x = 21

C = CV + x–1R

= 23

R + 2R = 2R7


16. CP = 3.5 R (At STP i j )As temperature increases, vibrational degree of freedom becomes 2 at higher temperature.t c r ki eku c<r k gS] r c mPp r ki eku i j dEi Uu dh Lokr U=k; dksfV 2 gks t kr h gSA

CP = 29

R = 4.5 R

17.

For equilibrium of piston fi LVu dh l kE; koLFkk ds fy; sPS = Kx0

P = SKx 0

For piston fi LVu ds fy; sWall = KE2 – KE1

Wgas – 21

kx2 = 21

mv2

Wgas = 21

kx2 + 21

mv2 = positive /kukRed

Q = 0Q = U + WU = – W = negative _ .kkRedAs internal energy of gas decreases temperature of gas decreases.pwafd xSl dh vkar fj d Åt kZ ?kVr h gSA xSl dk r ki ?kVsxkA

18. For any process U = n CV T, For Isothermal T = 0fdl h Hkh i fzØ; k ds fy, U = n CV T, l er ki h; ds fy, T = 0or ; k U = constant fu; r kadQ = 0 (For adiabatic process) : ) ks"e i zfØ; k ds fy; s U + W = 0

U = – W

EXERCISE # 11.1 As Q = nCdT

and dT = nCQ

Therefore molar heat constant C is the determining factor for rate of change of temperature of a gas, as

heat is supplied to it. It is minimum for isochoric process of a monoatomic gas

R23Cv , resulting in

greatest slope

QdT

i.e. curve 1.

For isobaric process of monoatomic gas and isochoaic process of diatomic gas, their heat capacities are

same

R

25

, therefore both are represented by curve 2.

For isobaric process of diatomic gas Cp =27

R

that is represented by curve 3.


Q axis represent isothermal process and T axis represent adiabatic process.pwafd Q = nCdT

vkSj dT = nCQ

vr % xSl ds r ki i fj or Zu dh nj ds fy , eksy j Å"eh; fLFkj kad C fu.kkZ; d xq.kd gS] t c xSl dks Å"ek nh t kr h gSA ; g

l evk; r fud çfØ; k esa , d i j ek.kqd xSl ds fy , U; wur e

R23Cv gS] i fj .kkeLo: i <ky

QdT

l okZf/kd gS vFkkZr

oØ 1., d i j ek.kqd xSl dh l enkch çfØ; k r Fkk f} i j ek.kqd xSl dh l evk; r fud çfØ; k ds fy , Å"eh; /kkfj r k l eku , oa

R

25

, gSA vr % nksuksa oØ 2 } kj k fu: fi r gSA

f} i j ek.kqd xSl dh l enkch çfØ; k esa Cp =27

R gS] vr % oØ (3) fu: fi r gksr k gSA

Q–v{k l er ki h çfØ; k n' kkZr h gS r Fkk T v{k : ) ks"e çØe n' kkZr h gSA

1.2 (A) If P = 2V2 , from an ideal gas equation PV = nRT we get2V3 = nRT

with increase in volume(i) Temperature increases implies dU = +ve(ii) dW = +veHence dQ = dU + dW = +ve(B) If PV2 = constant, from an ideal gas equation PV = nRT we get VT = K (constant)Hence with increase in volume, temperature decreases

Now dQ = dU + PdV = nCvdT – T

PV dT [ dV = –

TV

dT]

= nCvdT – T

PVdT = n(Cv – R) dT

with increase in volume dT = –veand since Cv > R for monoatomic gas. Hence dQ = –vewith increases in temperature dV = –ve, W = –ve(C) dQ = nC dT = nCv dT + PdV n (Cv + 2R) dT = nCvdT + PdV

2nRdT = PdV dTdV

= +ve

Hence with increase in temperature volume increases and vice versa. dQ = dU + dW = +ve

(D) dQ = nC dT = nCv dT + PdVor n (Cv – 2R)dT = nCvdT + PdV

or – 2nRdT = PdV dTdV

= –ve

with increase in volume temperature decreases.Also dQ = n(Cv – 2R)dTFor expantion dT = –ve but Cv < 2R for monoatomic gas. Therefore dQ = +vewith increase in temperature dV = –ve, W = –ve(A) ; fn P = 2V2, vkn' kZ xSl l ehdj .k PV = nRT l s

2V3 = nRT vk; r u c<+kus i j


(i) r ki eku c<+r k gS vFkkZr ~ dU = +ve(ii) dW = +vevr % dQ = dU + dW = +ve(B) ; fn PV2 = fu; r ] vkn' kZ xSl l ehdj .k PV = nRT l s VT = K (fu; r )vr % vk; r u c<+kus i j r ki ?kVr k gSA

vc dQ = dU + PdV = nCvdT – T

PVdT [ dV = –

TV–

dT]

= nCvdT – T

PVdT = n(Cv – R) dT

vk; r u c<+us i j dT = –ver Fkk pwafd Cv > R , d i j ek.kqd xSl ds fy , , vr % dQ = –ver ki eku c<+us i j dV = – ve W = – ve(C) dQ = nC dT = nCv dT + PdV n (Cv + 2R) dT = nCvdT + PdV

2nRdT = PdV dTdV

= +ve

vr % r ki eku c<+us i j vk; r u c<+r k gS o bl h r j g vk; r u c<+us i j r ki c<+r k gSA dQ = dU + dW = +ve

(D) dQ = nC dT = nCv dT + PdVor n (Cv – 2R)dT = nCvdT + PdV

or – 2nRdT = PdV dTdV

= –ve

vk; r u c<+us i j r ki ?kVr k gSAvr % dQ = n(Cv – 2R)dTi zl kj ds fy , dT = –ve ysfdu Cv < 2R , d i j ek.kqd xSl ds fy , A vr % dQ = +ver ki c<us l s dV = –ve W = – ve

1.3 (A) PV = nRT

P = (nRT) V1

= (constant) V1

, P V1

T = constant i.e. isothermal process

As V1

decreases or V increases W = positive

and Q = U + W = W > 0 (U = 0)(B) Q = 0 and V = increases W = positive(C) PV = nRT V T (P = constant)As volume increases, T also increasesi.e., U > 0and W > 0 So Q > 0(D) For cyclic process U = 0W < 0 (anticlockwise)Q = U + W < 0(A) PV = nRT

P = (nRT) V1

= (fu; r ) V1

, P V1

T = fu; r vr % l er ki h i zØe

V1

?kVr k gS vkSj V c<+r k gS W = /kukRed

and Q = U + U = W > 0


(B) Q = 0 r Fkk V = of) eku W = /kukRed(C) PV = nRT V T (P = fu; r kad )vk; r u c<+us i j T c<+r k gSi.e., U > 0r Fkk W > 0 vr % Q > 0(D) pØh; i zØe ds fy , U = 0W < 0 (okekor Z)Q = U + W < 0

2.5. Heat given Å"ek nh xbZ : TCnQ1V1 For gas A xSl A ds fy , [Aspwafd V = constant fu; r kad dW =

0]

& for Gas B xSl B ds fy , & Q = TCn2V2

( For same heat given, temperature rises by same value for both the gases.)(D; ksafd l eku Å"ek ds fy , r ki dh c<+ksr j h nksuksa ds fy , cj kcj gS)

21 V2V1 CnCn ................(1)Also, (PB)V = n2RT and (PA)V = n1RTvr % (PB)V = n2RT r Fkk (PA)V = n1RT

2

1

nn

= B

A

PP

= 5.15.2

= 35

n1 = 35

n2

Substituting in (1)l ehdj .k (1) esa j [ kus i j

2n35

1VC = 2v2Cn )R()R(

35

CC

2325

1

2

v

v

Hence, Gas B is diatomic and Gas A is monoatomic.vr % B , d f} i j ek.kqd xSl gS r Fkk A , d i j ek.kqd xSl gSA

2.6 Since pwafd n1 = 35

n2 bl fy , Therefore AM

125 =

BM60

35

(From experiment i z; ksx } kj k 1 : WA = 125 gm & wB = 60 gm) 5MB = 4MAThe above relation holds for the pair–Gas A : Ar and Gas B : O2 .Åi j fn; k x; k l EcU/k nksuksa t ksM+ksa ds fy , ykxw gksr k gS xSl A : vkj xu vkSj xSl B : vkWDl ht u

2.7 No. of molecules in 'A' v.kqvksa dk ua-A esa = nNA = 40125

NA = 3.125 NA

(Since n = 40125

for Ar) (pwafd Ar ds fy , n = 40125

)

2.8 Internal energy at any temperature Tfd l h r ki T i j vkar fj d Åt kZ= nCVT

=

2R3

40125

(300) [ CV for mono atomic gas , d i j ek.kqd xSl ds fy , = 2R3

]

Ui = 2812.5 cal.


2.9 In free expansion, temperature of the gas remains constant, thereforeeqDr çl kj esa xSl dk r ki fLFkj j gr k gSA vr %p0 v0 = p. 3v0 where t gkW v0 = initial volume. i zkj fEHkd vk; r u

p = 3p0

2.10 For adiabatic compression, initial conditions are 3p0 and 3v0 . Final volume and pressure are

v0 and 32/3 p0.

: } ks"e l ai hMu ds fy , çkj fEHkd fLFkfr ; k¡ 3p0 r Fkk 3v0 gSaA vfUr e vk; r u r Fkk nkc v0 r Fkk 32/3 p0. gSA

3p0 .(3v0) = 32/3 p0(v0) 3–1 = 32/3

or – 1 = 32

= 35

i.e. gas is monoatomicxSl , d i j ek.kqd gSA

2.11 KEavg TApplying TV – 1 = K for adiabatic process –: ) ks"e çØe ds fy , TV – 1 = K dk ç; ksx d j us i j –

T1 V1 – 1 = T2 V2

– 1

3/213/5

0

01

2

1

1

2 3vv3

VV

TT

3.1 If the rate at which molecules of same mass having same rms velocity striking a wall decreases, then the rateat which momentum is imparted to the wall decreases. This results in lowering of pressure. Hence statement-2 is correct.In statement-1 the rms velocity of gas remains same on increasing the volume of container by piston, since thegiven process is isothermal. Now the piston is at a greater distance from opposite wall and hence time taken bygas molecules from near the opposite wall to reach the piston will be more. Thus rate of molecules striking thepiston decreases. Hence statement-1 is correct and statement-2 is correct explanation.; fn og nj ft l i j l eku nzO; eku r Fkk l eku oxZek/; ewy osx ds v.kq nhokj i j Vdj kr s gS] ?kVr h gS r ks og nj ft l i j l aosx

nhokj i j fn; k t kr k gS] ?kV t kr h gSA i fj .kkeLo: i nkc ?kV t kr k gSA bl fy; s oDr O; –2 l R; gSAoDr O; –1 esa xSl dk oxZek/; ewy osx fi LVu } kj k cr Zu dk vk; r u c<+kus i j l eku j gr k gS pwafd nh xbZ çfØ; k l er ki h gSAvc fi LVu foi j hr nhokj l s vf/kd nwj h i j gksxk r Fkk bl fy; s xSl v.kqvksa } kj k foi j hr nhokj ds i kl l s fi LVu r d i gq¡pus esafy; k x; k l e; vf/kd gksxkA bl fy; s fi LVu i j v.kqvksa ds Vdj kus dh nj ?kV t kr h gSA bl fy; s oDr O; –1 l gh gS r Fkk oDr O; –2 l gh Li "Vhdj .k gSA

3.2 Vrms = MRT3

For different gases at same temp. have different molar mass so different Vrms. and KEavg = nRT2f

.

So reason is true.

l eku r ki i j fHkUu xSl ksa ds fy , fHkUu fHkUu eksy j nzO; eku gksr k gsS bl fy , Vrms. vkSj KEavg = nRT2f

Hkh fHkUu gksr k

gSA vr % dkj .k l R; gSA.


3.3 By derivation we obtain such results because by squaring higher velocities quantity become much greater.(Vrms > Vav > Vmp)O; qRi fr l s ge bl r j g ds i fj .kke i zkIr d j r s gS D; ksafd mPp osxksa dk oxZ dj us i j ek=kk cgqr vf/kd c<+ t kr h gSA (Vrms> Vav > Vmp)

3.4 Vrms = MRT3

, M = Molecular mass v.kqHkkj .

So at same temp molecules of different gases have different speed.vr % l eku r ki i j fHkUu xSl ksa ds v.kq fHkUu pky j [ kr s gSAStatement-2 is correct.oDr O; -2 l gh gSA

3.5 Because speed of particles increase with temp. Vrms = MRT3

.

So momentum of particles also increase with temperature hence pressure increases. Statement-2 is a correct explanation of Statement-1

D; ksfd r ki eku ds l kFk d.kksa dh pky cM+r h gS Vrms = MRT3

.

vr % d.kksa dk l aosx Hkh r ki eku ; k nkc ds l kFk c<+r k gSAoDr O; -2, oDr O; -1 dh l gh O; k[ ; k gSA

3.6 An isothermal process is an example to show the possibility "for both the pressure and volume of a monoatomicideal gas to change simultaneously without causing the internal energy of the gas to change". Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.l er ki h; i zØe , d mnkgj .k gS t ks ; g cr kr k gS fd nkc r Fkk vk; r u , d l kFk vkn' kZ , d i j ekf.od xSl ds fy, ] cnyr k gS ft l esvkUr fj d Åt kZ i fj ofr Zr ugh gksr h gSA vr % oDr O; -1 r Fkk oDr O; -2 l R; gSA o oDr O; -2 oDr O; -1 dks Li "V dj r k gSA

3.7 In adiabatic process temperature is changed without giving or taking heat.: ) ks"e i zØe esa fcuk Å"ek fn; sa ; k fy ; s r ki eku cnyr k gSABy first law of thermodynamics dQ = dU + dWÅ"ekxfr dh ds i zFke fu; e l s dQ = dU + dWWhen work is done by or on the system then internal energy is not conserved.t c fudk; } kj k vFkok fudk; i j dk; Z fd; k t kr k gS, r c vkar fj d Åt kZ l aj f{kr ugh j gr hA Statement-1 and 2 both are correct oDr O; -1, r Fkk oDr O; -2 nksuksa l gh gSA

3.8 Function of fan is circulating the air. In a closed room by heat loss of motor of fan heat is given to air of room soit increases the temperature.i a[ ksa dk dk; Z gok dks ?kqekuk gS , d can dej s esa i a[ ks dh eksVj l s fudyh Å"ek dej s dk r ki eku cM+k nsr h gSA

3.9 As Q = U + W for constant volume W = 0 Q = Upwafd Q = U + W fu; r vk; r u ds fy, W = 0 Q = U Statement-2 is a correct explanation of statement-1 and both are correct

oDr O; -2, oDr O; -1 dh l gh O; k[ ; k gSA o nksuks dFku l gh gSA

3.10 0 = C1(T – T1) + C2(T – T2)Final temperaturevafr e r ki

T = 21

2211CC

TCTC

[Final temperature depends on specific heat of two different conducting bodies, if the

value of specific heat of two conducting bodies are equal then the temperature is equal to mean temperature]


T = 21

2211CC

TCTC

[vafr e r ki nksuks fHkUu pkyd oLr qvksa dh fof' k"V m"ekvksa i j fuHkZj dj r h gSA ; fn nks pkyd oLr qvksa dh fof' k"V

m"ek, a l eku gS r ks r ki vkSl r r ki ds r qY; gksxk]C1 & C2 heat capacities of two bodies. nks oLr qvksa dh m"ek /kkfj r k, a gSA

3.11 Work in isothermal is more than work in adiabatic because area under P-V of anisothermal expansion is more than as compare to an adiabatic expansion.l er ki h i zØe esa : ) ks"e i zØe l s vf/kd dk; Z gksr k gS, D; ksfd l er ki h; i zl kj esa P-V vkj s[ k ds vUr xZr {kS=kQy , : ) ks"e i zl kjds l axr {kS=kQy l s vf/kd gksr k gSA

3.12 As pwafd T = VnC

Qand o CV (O2) > CV (He)

Temperature rise incase of He is more than in case of O2 He ds i zdj .k esa r ki of} O2 ds i zdj .k l s vf/kd gksxkABoth statements are correct but statement-2 is not a correct explanation of statement-1 nksuks oDr O; l gh gS fdUr q oDr O; -2, oDr O; -1 l gh O; k[ ; k ugh gSA

3.13v

p

CC

=1.67 for monoatomic gas and v

p

CC

=1.4 for diatomic gas. Hence statement1 is false and statement

2 is true.

, d i j ek.kqd xSl ds fy, v

p

CC

=1.67 f} i j ek.kqd xSl ds fy, v

p

CC

=1.4 vr % oDr O; -1 vl R; gS] oDr O; -2 l R;

gSFor monoatomic (f = 3) and diatomic (f = 5) Statement-2 is correct , d i j ek.kqd xSl ds fy, (f = 3) o f} i j ek.kqd xSl ds fy, (f = 5) oDr O; -2 l gh gSA

4.1 Isobaric process must pass through origin. l enkch; i zØe ewy fcUnq l s xqt j uk gh pkfg, A

4.2 Isochoric process must pass through origin. l e vk; r uhd i zØe ewy fcUnq l s xqt j uk gh pkfg, A4.3 Straight line on PV does not represent isothermal process.

PV i j l h/kh j s[ kk l er ki h; i zØe dks i znf' kZr ugha dj r h gSA

4.4 Statement is true. dFku l R; gSANo. If they intersect, then at two different temperature (of an isothermal curve), volume and pressure of agas will be same, which is not possible.ugh, ; fn ; s i zfr PNsfnr gks r ks fHkUu r ki ks (, d l er ki h; ozd), xSl ds vk; r u o nkc l eku gksxsa t ks fd l aHko ugh gSA

5.1 Cv = 21

v2v1

nnCnCn

21

= 11

R25)1(R

23)1(

= 2R

5.2 VP2 = constant fu; r


Putting P = V

nRTj [ kus i j , we have ge i zkIr dj r s gSA

VT2

= constant fu; r

or T V

So, if V is doubled, T becomes 2 times.

vr %] ; fn V dks nqxquk fd; k t k; s r ks T, 2 xquk gks t kr k gSA

5.3 Under isothermal conditionsPV = constant

(Pi) (2V) = PV Pi = P/2Under adiabatic condition

PV = constant where = 1.67 (Pa) (2V)1.67 = PV1.67

Pa = P/(2)1.67

Therefore, the ratio Pa/Pi = (2/21.67) = 67.021

l er ki h; voLFkk esaPV = fu; r

(Pi) (2V) = PV Pi = P/2: nks"e voLFkk esa

PV = fu; r ] t gk¡ = 1.67 (Pa) (2V)1.67 = PV1.67

Pa = P/(2)1.67

vr % vuqi kr Pa/Pi = (2/21.67) = 67.021

5.4 Since, the system is insulated, dQ = 0. Other part is vaccum, therefore, work done by the gas dW is alsozero. Hence, from first law of thermodynamics, dU = 0 i.e. temperature remains constant.

D; ksfd fudk; dqpkyd gS bl fy , dQ = 0 nwl j k Hkkx fuokZr gS] bl fy , xSl } kj k fd; k x; k dk; Z Hkh ' kwU; gksxkkA vr %Å{ek xfr dh ds i zFke fu; e l s dU = 0 gksxk r Fkk r ki eku fu; r j gsxkA

EXERCISE # 41. From Conservation of energy;

Åt kZ l aj {k.k l sIncrease in internal energy = Decrease in Kinetic energyxfr t Åt kZ esa deh = xSl dh vkUr fj d Åt kZ esa of)

n CVT = 21

M v02

Mm

R

23

T = 21

M v02 T =

R3Mv 2

0

2. Slope of adiabatic process at a given state (P,V,T) is more than the slope of isothermal process. The correspondingP–V graph for the two processis as shown in figure.

B

V

C

A

PP3

P1

V1 V2

In the graph, AB is isothermal and BC is adiabatic.WAB

= positive (as volume is increasing)and WBC = negative (as volume is decreasing)


|WBC| > |WAB|, as area under P–V graph gives the work done.Hence, WAB + WBC = W < 0From the graph itself, it is clear that P3 > P1.Hence, the correct option is (A).Note : At point B, slope of adiabatic (process BC) is greater than the slope of isothermal (process AB).nh xbZ voLFkk (P,V,T) i j : ) ks"e i zØe dk <ky l enkch i zØe ds <ky l s vf/kd gSA

B

V

C

A

PP3

P1

V1 V2

nksuksa i zØeksa ds l axr P–V xzkQ fp=k esa i znf' kZr gSAxzkQ esa AB l eknkch gS] o BC : ) ks"e gSA

WAB = /kukRed (D; ksfd vk; r u c<+ j gk gSA)

o WBC = _ .kkRed (D; kssafd vk; r u c<+ j gk gSA) vkSj|WBC| > |WAB|, D; ksafd P–V ds uhps dk {ks=kQy fd; k x; k dk; Z nsr k gSAvr % WAB + WBC = W < 0xzkQ l s ; g Li "V gS fd P3 > P1.bl fy, l gh fodYi (A) gSANote : fcUnq B, i j ] : ) ks"e i zØe (BC) dk <ky l enkch i zØe (AB) ds <ky l s vf/kd gSA

3. At constant pressure, V T

or1

2

VV

= 1

2

TT

or1

2

AhAh

= 1

2

TT

h2 =

1

21 T

Th = (1.0) m 300400

= 34

m

The process is adiabatic when gas is compressed without exchange of heat

2

1

TT

=1

1

2

VV

T1 = 4004.0

34

K.

fu; r nkc i j V T

; k1

2

VV

= 1

2

TT

; k1

2

AhAh

= 1

2

TT

h2 =

1

21 T

Th = (1.0) m 300400

= 34

m

i zØe : ) ks"e gS t c xSl dks fcuk Å"ek fofues; ds l Ei hfM+r fd; k t kr k gSA

2

1

TT

=1

1

2

VV

T1 = 4004.0

34

K

4. U = Q = I2RT = 12 × 100 × 5 × 60 J = 30 KJ

5. Above statement is a definition of one calorie.mi ; qZDr dFku , d dSyksj h dh i fj Hkk"kk gSA


6. A) for J Kat const. V , dW= 0and P T dQ = dU (At constant volume)Now dU is negative , because temperature is decreasing dQ is negativehence Q < 0

(B) for K Lat const P , V TAs volume increases ; Temperature will increase dW = PdVhence W > 0 , Q > 0

(C) for L Mat constant V, dW = 0 P T , Temperature will increaseNow dQ = dUbut dU is positive Q > 0

(D) For M JV is decreasing therefore W < 0(PV)J < (PV)MTJ < TMU < 0hence Q < 0

(A) J K ds fy,fu; r V i j dW = 0rFkk P T dQ = dU (fu; r vk; r u i j )vc dV _ .kkRed gS, D; ksafd r ki eku ?kV j gk gSA dQ _ .kkRed gSA vr % Q < 0

(B) K L ds fy,fu; r P i j , V Tvk; r u c<+r k gS ; r ki c<+sxk dW = PdVvr % W > 0 , Q > 0

(C) L M ds fy,fu; r V i j , dW = 0 P T , r ki eku c<+sxkvc dQ = dUysfdu dV /kukRed gSA Q > 0

(D) M J ds fy,V ?kV j gk gS bl fy, W < 0(PV)J < (PV)MTJ < TMU < 0vr % Q < 0

7. Translation K.E. LFkkukUr fj ; K.E = 23

nRT..

8. PT2 = C


using PV = nRT in PT2 = C

CTV

nRT 2

T3 V

Differentiating we get

3T

dT =

VdV

Coefficient of volume expansion () = dTdV

V1

= T3

Ans. (C)PT2 = C

PT2 = C esa PV = nRT dk mi ; ksx dj us i j

CTV

nRT 2

T3 V

mi j ksä l ehdj .k dk vodyu dj us i j

3T

dT =

VdV

vk; r u i zl kj xq.kkad () = dTdV

V1

= T3

Ans. (C)

9. Column- : Expansion of ideal gasColumn- : Thermodynamic change.(A) Q = 0 (as boundary is non conducting)

in the case of free expansion W = 0Q = U + W0 = U + 0 U = 0

U = const.(A) (q) (As temp remains constant).

(B) P 2V1

PV2 = Cn RTV = CTV = C'Since volume increases the temperature decreases.

Q = n C T , for polytropic process, PVx = constant, C = Cv + x1

R

C = Cv + 12

R

= CV – R 23

R – R

C = 2R

Q = n2RT

T is negative so Q is negativemeans heat is lostSo for (B) (p, r)

(C) PV4/3 = CTV1/3 = C'So when volume increases temperature decreases


Now C = CV + 1

34R

=

23

R – 3R C = –23

R

Q = nCT Q = n )T(R23

as T is negative Q will be positive.Hence (C) p,s

(D) T = nRPV

as product of P and V increases, so temperature increasesQ = U + WU = +ve (T = +ve)W = +ve (As volume increases)So Q = +veHence gas gains heat

(D) (q, s)

Lr EHk- : vkn' kZ xSl ds foLr kj l sLr EHk- : Å"ekxfr d i fj or Zu(A) Q = 0 (pwafd l hek vpkyd gS)

eqä i zl kj dh fLFkfr esa W = 0Q = U + W0 = U + 0 U = 0

U = fu; r(A) (q) (pwafd r ki eku fu; r j gr k gSA).

(B) P 2V1

PV2 = Cn RTV = CTV = C'pwafd vk; r u c<+r i k gS] i j r ki eku ?kVr k gSA

Q = n C T , i ksyhVªksfi d i zØe ds fy, , PVx = fu; r , C = Cv + x1

R

C = Cv + 12

R

= CV – R 23

R – R

C = 2R

Q = n2RT

T _ .kkRed gS vr % Q _ .kkRed gS vFkkZr Å"ek dh gkfu gSvr % (B) (p, r)

(C) PV4/3 = CTV1/3 = C'pwafd t c vkr ; u c<+r k gS r ki ?kVr k gSA

vc C = CV + 1

34R

=

23

R – 3R C = –23

R

Q = nCT Q = n )T(R23

pwafd T _ .kkRed] Q /kukRedvr % (C) p,s


(D) T = nRPV

pwafd P r Fkk V dk xq.kuQy c<+r k gS vr % r ki c<+r k gSQ = U + WU = +ve (T = +ve)W = +ve (pwafd vk; r u c<+r k gSA)So Q = +vevr % Å"ek i zkIr dj sxkA

(D) (q, s)

10. For monoatomic gas, Cp = 25

R, Cv = 23

R. Cp – Cv = R

For diatomic gas Cp = 27

R, Cv = 25

R. Cp – Cv = R

Cp – Cv is same for bothCp + Cv = 6R (for diatomic)Cp + Cv = 4R (for mono)so (Cp + Cv)dia > (Cp + Cv)mono

57

CC

v

p = 1.4 (for diatomic)

35

CC

v

p = 1.66 (for monoatomic)

(Cp) (Cv) = 435

R2 (for diatomic)

(Cp) (Cv) = 415

R2 (for monoatomic)

so (Cp . Cv)diatomic > (Cp . Cv)monoatomic

, d i j ek.kqd xSl ds fy, , Cp = 25

R, Cv = 23

R. Cp – Cv = R

f} i j ek.kqd xSl ds fy, Cp = 27

R, Cv = 25

R. Cp – Cv = R

Cp – Cv nksuksa ds fy, l eku gSCp + Cv = 6R (f} i j ek.kqd ds fy, )Cp + Cv = 4R (, di j ek.kqd ds fy, )so (Cp + Cv)f} i j ek.kqd

> (Cp + Cv), di j ek.kqd

57

CC

v

p = 1.4 (f} i j ek.kqd ds fy, )

35

CC

v

p = 1.66 (, di j ek.kqd ds fy, )

(Cp) (Cv) = 435

R2 (f} i j ek.kqd ds fy, )

(Cp) (Cv) = 415

R2 (, di j ek.kqd ds fy, )

vr % (Cp . Cv)f} i j ek.kqd > (Cp . Cv), di j ek.kqd


11. (A) process is not isothermal(B) volume decreases and temperature decreases

U = negativeW = negative

so Q = negative(C) Work done in process A B C is positive(D) Cycle is clockwise, so work done by the gas is positive.(A) çØe l er ki h; ugha gSA

(B) vk; r u ?kV j gk gS o r ki eku Hkh ?kV j gk gS

U = _ .kkRed

W = _ .kkRed

vr % Q = _ .kkRed

(C) çØe A B C eas fd; k x; k dk; Z /kukRed gSA

(D) pØh; çØe nf{k.kkoÙk gS, bl fy, xSl } kj k fd; k x; k dk; Z /kukRed gSA

12. At low pressure and high temperature inter molecular forces become ineffective. So a real gas behaves like anideal gas.

fuEu nkc vkSj mPp r ki i j vUr j vk.kfod cy] vi zHkkoh gks t kr s gSA bl fy; s , d okLr fod xSl ] vkn' kZ xSl dh r j g O; ogkj

dj sxhA

13.

U = 2f

nRT,, where f,n,R are constants. Also temperature T is same at A & B.

UA = UB

Also, WAB = nRT0

i

f

VV n = nRT0 0

0

VV4

n = nRT0 n4 = P0V0 n4

So, answers are (A) & (B).

U = 2f

nRT,, t gk¡ f,n,R fu; e gSA A vkSj B i j r ki eku T gSA

UA = UB

rFkk WAB = nRT0

i

f

VV n = nRT0 0

0

VV4

n = nRT0 n4 = P0V0 n4

vr % (A) vkSj (B) l gh gSA


14. For adiabatic process, : nks"e çØe ds fy,

TV–1 = constant fu; r

T2 = 1

2

11 V

VT

T2 = 157

1 32T T2 = 4T1 a = 4 Ans.

15. W = 1 2nR(T – T )( –1) T1V1

– 1 = T2 V2 – 1

or

W = 1 1nR(T – 4T )5 – 13

T2 = 4T1

W = – 13nR 3T2

n = 5.6 1

22.4 4

W = – 19nRT2

W = – 19RT8

17. Vrms

= MRT3

)on(argV)helium(V

rms

rms=

helium

onarg

MM

= 104

40

= 3.16

8. Q = nCPT = 2 5 8.31 5

2

Q = 207.75J


PART - II

1. Heat cannot flow itself from a lower temperature to a body of higher temperature. This corresponds to secondlaw of thermodynamics.Å"ek Lor % gh fuEu r ki l s mPp r ki dh vksj i zokfgr ugh gks l dr hA ; g Å"ekxfr dh ds f} r h; fu; e ds l axr vuql kj gSA

2. Given : fn; k x; k gSA P T3 .......(i)In adiabatic process: ) ks"e i zØe esa

TP1-= constant fu; r

T /)1(p1

T(/–1) P .......(ii)Comparing equations (i) and (ii), we getl ehdj .k (i) o (ii) r qyuk l s

31

3– 3 =2 = 3

P

P

CC

= = 23

3. Work does not characterize the thermodynamic state of matter, it is a path function gives only relationshipbetween two quantities.dk; Z i znkFkZ dh Å"ekxfr dh voLFkk dk vfHky{k.k ugh gS, ; g i Fk dk Qyu gS t ks nks j kf' k; ksa esa l EcU/k n' kkZr k gSA

4. T1 = 627 + 273 = 900KQ1 = 3 × 106 calT2 = 27 + 273 = 300K

2

2

1

1

TQ

TQ

Q2 = 1

2

TT

× Q1 = 900300

× 3 × 106

= 1× 106 cal

Work done l Ei Uu dk; Z = Q1 – Q2

= 3 × 106 –1 ×106 = 2 ×106 cal= 2 × 4.2 × 106 J = 8.4 × 106 J

5. Mayer’s formula ises; j l w=k l sCP – CV = R

and r Fkk = V

P

CC

Therefore, using above two relations, we findvr % i z; qDr nksuksa l w=k l s


CV = 1–R

For a mole of monoatomic gas , dy i j ek.kfo; xSl ds , d eksy ds fy, ; = 35

CV = 1–3/53R

= 23

R

When these two moles are mixed, then the heat required to raise the temperature to 1ºC ist c ; s nks eksy fefJr gksr s gS r ks 1ºC r ki c<kus ds fy, vfHk"V Å"ek gksxhA

CV = 23

R + 25

R = 4R

Hence, for one mole, heat required isvr % , d eksy ds fy, vko' ; d Å"ek

= 2R4

= 2R

CV = 2R

1–R = 2R

= 23

Alternative : ossdfYi d fof/k

1–nn 21

= 1–n

1

1

+ 1–n

2

2

Here, ; gk¡ n1 = 1, n2 = 1, 1 = 35

,2 = 57

1–11

= 1–3/51

+ 1–5/71

1–2

= 23

+25

1–2

= 28

1–2

= 4

= 42

+ 1

Hence, vr % = 23

6. In thermodynamic system, entropy and internal energy are state functions, entropy (S) can be zero for adia-batic process. Work done in adiabatic process may be non-zero.Å"ekxfr dh fudk; esa , UVªksi h o vkUr fj d Åt kZ voLFkk Qyu gSA , UVªksi h (S) : ) ks"e i zØe ds fy, ' kwU; gks l dr h gSA : ) ks"ei zØe esa fd; k x; k dk; Z v' kwU; gks l dr k gSA


7. There will be no change in number of moles if the vessels are joined by valve. Therefore, from gas equation; fn i k=k okYo } kj k t qMs gS r ks eksy dh l a[ ; k esa i fj or Zu ugh gksxkA vr % xSl l ehdj .k l sAPV = nRT

1

11

RTVP

+ 2

22

RTVP

=

RTVVP 21

21

122211

TTTVPTVP

=

TVVP 21

T =

122211

2121

TVPTVPTTVVP

Internal energy of the system remains same before and after opening of valve, sookYo dks [ kksyus ds i woZ o i ' pkr fudk; dh vkUr fj d Åt kZ l eku j gxhA vr %

2RT)nnƒ(

2RTnƒ

2RTnƒ 212211

n1T1 + n2 T2 = (n1 + n2)TP1V1 + P2V2 = P(V1+ V2)

Hence, T = 122211

212211

TVPTVPTTVPVP

8.* Statement (a) and (d) are wrong. Concept of entropy is associated with second law of thermodynamics.dFku (a) o (d) vl R; gSA , UVªksi h dh vo/kkj .kk Å"ekxfr dh ds f} r h; fu; e l s l Ecf/kar gSA

9. Cv = 21

v2v1

nnCnCn

21

For helium ghfy; e ds fy, , n1 = 4

16 = 4 and r Fkk1 = 3

5

For oxygen vkWDl ht u ds fy, , n2 = 3216

= 21

and r Fkk 2 = 57

1RC

1v1

= 23

135R

R

25

157

R1

RC2

2v

214

R25,

21R

234

C v

=

29

R45R6

= 18R29

492R29


Now vcCv = 1R

-1 = R

1829

R + 1

1918

CC

v

p + 1

= 291918

= 1.62

10. The Change in internal energy does not depend upon pathfollowedby the process. It only depends on initial and finalstates.Hence, U1= U2

vkUr fj d Åt kZ esa i fj or Zu i zØe } kj k fy, x, s i Fk i j fuHkZj ugh dj r k gS ; g dsoy i zkj fEHkd o vfUr e voLFkk i j fuHkZj dj r kgSAvr % U1= U2

11. According to the figurefp=kkuql kj

Q1 = T0S0 + 21

T0S0 = 23

T0S0

S

T

2T0

S0

T0

2S0

Q2 = T0 (2S0 – S0)= T0S0Q3 = 0

= 1Q

W

= 1

21

QQQ

= 1 – 1

2

QQ

= 1 – 32

= 31

12. Equating internal energy, vkUr fj d Åt kZ dks r qY; dj us i j

1 oRT25 + 1 ffo RT

25RT

231T

37R

23

of T

23T

(3) is correct. l gh gSA

13. For adiabatic : ) ks"e ds fy, , W = 1VPVP 2211

γ =

1TTnR 21

γPutting values, we get = 1.4, hence diatomic.eku j [ kus i j i zkIr gksxk = 1.4, vr % f} i j ek.kfod

14. For Carnot engine using as refrigeratordkuksZ bt au dks j sfÝt j sVj dh Hkkfr i z; ksx dj us ds fy,


1–

TTQW

2

12

It is given fn; k x; k gSA = 101

= 1 – 1

2

TT

109

TT

1

2

So, vr %Q2 = 90 J (as W = 10 J)

15. According to Mayer's relation,es; j l EcU/k l s

Cp – C

v = 28

RmR

16. From first law of thermodynamics,Å"ekxfr dh ds i zFke fu; e l s

Q = U + WFor path iaf,i Fk iaf ds fy,

50 = U + 20 U = Uf – Ui = 30 calFor path ibf,i Fk ibf ds fy,

or ; k Q = U + WW = Q – U = 36 – 30 = 6 cal.

17. As no work is done and system is thermally insulated from surrounding, it means sum of internal energy of gasin two partitions is constant ie, U = U

1 + U

2Assuming both gases have same degree of freedom, thenD; ksfd dk; Z ugh fd; k x; k gS o fudk; i fj os' k l s Å"eh; vo: /k gSA bl fy, nksuks i zdks"B esa xSl dh vkUr fj d Åt kZvksa dk; ksx fu; r j gsxkA U = U

1 + U

2Assuming both gases have same degree of freedom, thennksuks xSl s dh Lokr U=k dksfV; k l eku ekuus i j

U = 2

RT)nnƒ( 21 and U1 =

2RTnƒ 11 , U

2 =

2RTnƒ 22

Solving we get gy dj us i j i zkIr gksxkA T = 122211

212211

TVpTVpTT)VpVp(

18. n (moles) = 2A to B is isobaric processWAB = PV = nRT

= (2) (R) (200)= 400 R

WAB = 400 Rn (eksy ) = 2A l s B r d l enkch çØe gSAWAB = PV = nRT

= (2) (R) (200)= 400 R

WAB = 400 R


19. D to A is isothermal processWork done by the gas in D to A is

WDA = nRT ln 12

VV

= nRT ln 21

PP

= (2) (R) (300) ln 5

5

10210

= (600 R) [– ln 2]= – (600 R ) (0.693)= – 414 R

WDA = – 414 R, it is work done by the gasSo work done on the gas is + 414 RD l s A r d l er ki h; çØe gSAD l s A r d xSl } kj k fd; k x; k dk; Z gksxkA

WDA = nRT ln 12

VV

= nRT ln 21

PP

= (2) (R) (300) ln 5

5

10210

= (600 R) [– ln 2]= – (600 R ) (0.693)= – 414 R

; g xSl } kj k fd; k x; k dk; Z gksxkAvr % xSl i j fd; k x; k dk; Z + 414 R gksxkA

20. WABCDA = WAB + WBC + WCD + WDA

= nR (T)AB + nR (TB) ln CB

PP

+ nR (T)CD + nR (TD) ln AD

PP

= nR (200) + 500 nR ln 2 + nR ( – 200)

+ 300 nR ln 21

= 2 ln 2 [500 R – 300 R]= (400 R ) (ln 2) = (400 R ) (0.693) = 276 R

WABCDA = 276 RWABCDA = WAB + WBC + WCD + WDA

= nR (T)AB + nR (TB) ln CB

PP

+ nR (T)CD + nR (TD) ln AD

PP

= nR (200) + 500 nR ln 2 + nR ( – 200) + 300 R ln 21

= 2 ln 2 [500 R – 300 R]= (400 R ) (ln 2) = (400 R ) (0.693) = 276 R

WABCDA = 276 R


21. 41

dm

= m3

kE = for diatomic f} i j ekf.od ds fy,

kE = 25

PV = 25

× 8 × 104 × 41

= 5 × 104 J

22. TV – 1 = constant fu; r kad

T1 1

57

V

= T21

57

)V32(

1

2

TT

= 5/2)32(1

= 41

= 1 – 1

2

TT

= 1 – 43

41

23.

24.

8. ktg & thermodynamics part-1

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