14-Heat & Thermodynamics
TOPIC
HEAT & THERMODYANMICS
14
SECTION - I : STRAIGHT OBJECTIVE TYPE
[k.M-I: lh/ksoLrqfu"Bizdkj
14.1 A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is: fdlhf}ijek.kqdvkn'kZxSldksfu;rvk;ruijxeZfd;kx;ktcrdmldknkcnqxukughagksx;kvkSjr ~i'pkr~fu;r nkcijxeZfd;kx;ktcrdmldkvk;runqxukughagksx;kAbliwjhizfØ;kdhvkSlreksyjÅ"ek/k fjrkgS%
13R
(A) 6
19R
(B*) 6
23R
(C) 6
17R
(D) 6
Sol. Let initial pressure, volume, temperature be P0, V0, T0 indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0) isobarically.
ekukbldsizkjfEHkdnkc]vk;ruvkSjrkiP0,V0,T0gSAftlsP-Vvkjs[kesavoLFkk'A'lsn'k Z;kx;kgSAblsfQj levk;rfud izfØ;klsvoLFkkB (2P0, V0, 2T0) vkSjfQj voLFkk B ls voLFkkC (2P0, 2V0, 4T0) rd lenkch;izfØ;k
}kjkystk;kx;kA
Total heat absorbed by 1 mole of gas 1eksyxSl}kjkxzg.kdhxbZdqyÅ"ek&
Q = C (2T – T ) + C (4T – 2T )
= 5 R T +
2 0
7 R × 2T =
2 0
19 RT
2 0
Total change in temperature from state A to C is A voLFkklsCvoLFkkrdrkidkdqy ifjorZugS-
T = 3T
Q
19 RT0 19
Molar heat capacity eksyj m"ek /kkfjrk= T
= 2 = R.
3T0
14.2 A gas mixture consists of 2 moles of Oxygen and 4 moles of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is :
,dxSldsfeJ.kesa 2eksyvkWDlhtuvkSj4eksyvkxZurkiT ijgSAlHkhizdkjdsdEiuksadksux.;ekursgq,]fudk; dh dqy vkarfjd mtkZ gS -
(A) 4 R T (B) 5 R T (C) 15 R T (D*) 11 RT
Sol. In an ideal gas internal energy fdlh vkn'kZ xSl dh vkarfjd mtkZ =
f nRT
2
U = 5 × 2 × RT + 4 ×
2
3 RT = 11 RT.
2
14.3 One mole of an ideal gas is taken from state A to state B by three different processes,
(a) ACB (b) ADB (c) AEB as shown in the P V diagram. The heat absorbed by the gas is : 1eksyvkn'kZxSldksAlsBvoLFk rdrhufofHkUuizfØ;kvksads}kjkystk;kx;k]
(a)ACB (b)ADB (c)AEB tSlkfdPVoØesan'kkZ;kx;kgS]xSl}kjkvo'k sf"krÅ"ekgSA
(A) greater in process (b) then in (a) (B) the least in process (b)
(C) the same in (a) and (c) (D*) less in (c) then in (b)
(A) izfØ;k(a)easizfØ;k(b)lsvf/kd (B) izfØ;k(b) esalclsde
(C)(a) vkSj(c) esaleku (D*) (c) esa(b) ls de
Sol. Heat absorbed by gas in three processes is given by
rhuksaizfØ;kvksaesaxSl}kjkxzg.kdhxbZm"ekblizdkjgSA
QACB QADB QAEB
= U + W
= U
= U + W
ACB
AEB
The change in internal energy in all the three cases is same. And W ACB is +ve, WAEB is –ve.
Hence QACB > QADB > QAEB
rhuksaesavkarfjdÅtkZifjorZulekugSvkSjWACB /kukRedgS,WAEB _.k RedgSA vr% QACB > QADB > QAEB
14.4 In an adiabatic expansion the product of pressure and volume : (A*) decreases (B) increases
(C) remains constant (D) first increases, then decreases. =
fdlh :)ks"e izlkj esa nkc vkSj vk;ru dk xq.kuQy - (A*) ?kVrk gS (B) c<+rk gS
(C) fu;r jgrk gS (D) igys c<+rk gS, fQj ?kVrk gS
Sol. In an adiabatic expansion, internal energy decreases and hence temperature decreases.
from equation of state of ideal gas PV = nRT
The product of P and V decreases.
:)ks"eizlkjesa, vkarfjdmtkZ?kVrhgS]blfy,rki?kVrkgS
vkn'kZxSldhvoLFk lehdj.k}kjk- PV = nRT
PvkSjVdkxq.kuQy?kVrkgS.
14.5 One mole of an ideal gas at pressure P0 and temperature T0 is expanded isothermally to twice its volume and then compressed at constant pressure to (V0/2) and the gas is brought back to original state by a process in which P V (Pressure is directly proportional to volume). The correct representation of process is
=
1 eksy vkn'kZ xSl dks vc P0 vkSj rki T0 ls lerkih izlkj }kjk nqxus vk;ru rd ys tk;k x;k rFkk fQj le nkc ij (V0/2) rd laihMu fd;k x;k vkSj fQj mls
,d izfØ;k }kjk izkjfEHkd voLFkk ij ys tk;k x;k ftlesa P V gS (nkc vk;ru
dslekuqikrhgSA) izfØ;kdklghxzkQh;fu:i.kgS
(A)
(B)
(C*)
(D)
Sol.
P0
P0 2
T
T0 T0
4
Process AB is isothermal expansion, BC is isobaric compression In process CA
izfØ;kABlerkihizlkjgSrFk BClenkchlaihMugSA CA izfØ;kesa
P nRT P2 T.
P
14.6 Maxwell’s speed distribution curve is given for two different temperatures. For the given curves.
=
eSDlosydkpkyforj.koØnksfHkUurkiksaijfn;kx;kgSAbuoØksadsfy,
(A) T1 > T2 (B*) T1 < T2 (C) T1 T2 (D) T1 = T2
Sol. Higher is the temperature greater is the most probable speed.
ftrukvf/kdrkigksxkmruhghT;knkvf/kdrelaHk forpkygksxhA
14.7 There are two thin spheres A and B of the same material and same thickness. They emit radiation like black bodies. Radius of A is double that of B. A and B have same temperature T. When A and B are kept in a room of temperature T0 (< T), the ratio of their rates of cooling (rate of fall of temperature) is:
[ assume negligible heat exchange between A and B ] AvkSjBnksdeeksVkbZdsxksysgSaAnksuksalekuinkFkZvkSjlekueksVkbZdsgSaA;svkn'kZÑf".kdkdhrjgmRltZudjrs gSaAAdhf=kT;kBlsnqxuhgSAnksuksadkrkiTgSAtcnksuksadks,ddejsesaT0( T2). Heat current flowing through the rod will be
,dNM+ftldhyEckbZ𝑙vkSjvuqizLFkdkV{ks=kQyAgSdhifjorhZÅ"ekpkydrkk=T,lsnhtkrhgS]tgk¡,d
/kukRedfu;rkadgSvkSjTdsfYouesarkigSAnksuksafdukjsrkiT1vkSjT2(T1>T2)ijgSANM+lsizokfgrÅ"eh;/k jkgksxh&
A (T2 T2 )
(A) 1 2
𝑙
A (T2 T2 )
(B) 1 2
𝑙
A (T2 T2 )
(C) 1 2
𝑙
A (T2 T2 )
(D*) 1 2
𝑙
dT
Sol. Heat current Å"eh; /kkjk: i = – kA dx
idx = – kA dT
𝑙
i dx
0
T2
= – A T dT
T1
(T2 T2 )
A (T2 T2 )
i 𝑙 = – A 2 1
2
i =
1 2
2 𝑙
14.14 P-T diagram is shown below then choose the corresponding V-T diagram P-T xzkQuhpsfn;kgSA bldkV-TxzkQgksxk-
(A) (B)
(C) (D*)
Sol. BC is isochoric. VB > VA , VB = VC , VD > VC
BC ,d le vk;rfud izØe gSVB > VA , VB = VC , VD > VC
14.15 A diatomic ideal gas undergoes a thermodynamic change according to the P–V diagram shown in the figure. The total heat given to the gas is nearly :
,df}&ijek.kqvkn'kZxSluhpsfn;sP–VxzkQdsvuq:iÅ"ekxfrdifjorZudjrhgSAxSldksnhxbZdqyÅ"ek gksxh%
2P0 2P0
P0
V
V0 2V0
P0
V
V0 2V0
(A) 2.5 P0V0 (B) 1.4 P0V0 (C*) 3.9 P0V0 (D) 1.1 P0V0
Sol. QAB = U + WAB
WAB = 0
AB =
2 nR T
2 ( PV) UAB =
5 ( PV)
2
QAB = 2.5 P0 V0
Process izØe BC QBC = UBC + WBC
QBC = 0 + 2P0 V0 log 2
= 1.4 P0 V0
Qnet = QAB + QBC = 3.9 P0 V0
14.16 On an X temperature scale, water freezes at – 125.0° X and boils at 375.0° X. On a Y temperature scale, water freezes at – 70.0°Y and boils at – 30.0°Y. The value of temperature on X-scale equal to the temperature of 50.0°Y on Y-scale is : f d l h X r k i
iSekus ij ikuh – 125.0° X ij terk gS vkSj 375.0° X ij mcyrk gS vkSj fdlh Y rki iSekus ij ikuh – 70.0°Y ij terk gS vkSj – 30.0°Y ij mcyrk gSA X-iSekus ij 50.0°Y rki dk eku gksxk :
(A) 455.0° X (B) – 125.0° X (C*) 1375.0° X (D) 1500.0° X
X (125)
Sol. 500
Y (70)
= 40
For Y = 50 X = 1375.0°X
14.17 A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power ‘P’ and its rate of cooling is R then
,dBkslxksykdkjÑf".kdkftldhf=kT;krvkSjftldkle:inzO;ekuforj.kgS]eqDrvkdk'kesaj[khgqbZgSA;g
‘P’ 'kfDr mRlftZr djrh gS vkSj bldh 'khryu dh nj 'R' gSrks
1
(A) R P r2 (B*) R P r (C) R P 1/r2 (D) R P r
Sol. Rate of radiation per unit area is proportional to (T4)
izfr,dkad{ks=kQylsfofdj.knj(T4)dslekuqikrhgksrhgSA
P AT4
P r2.
dT dT 1
Also lkFk gh ms dt AT4 dt = R
r
(because D;ksafd m = (V) r3 and vkSjA r2)
P R r
14.18 A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is . If at another temperature T the power radiated is P
and wavelength at maximum intensity is 0 then
2
tcrki'T'gS, rc,dÑf".kdk'P'njlsfofdj.kmRltZudjrhgSAblrkiijftlrjaxnS/;Zdhfofdj.kdhrhozrk
vf/kdre gS]og gSAvxjfdlh vkSjrki T´ij'kfDr mRltZdrkP´vkSjvf/kdrefofdj.kdhrjaxnS/;Z 0 gS
0 2
rks-
(A*) P T = 32PT (B) P T = 16PT (C) P T = 8PT (D) P T = 4PT
Sol. For a block body, wavelength for maximum intensity :
fdlhÑf".kdkdsfy,]vf/kdrefofdj.kdsfy,rjaxnS/;ZgS&
1
T
& P T4
P 1
4
P = 16 P. ∴ P T = 32PT
14.19 The emissive power of a black body at T = 300 K is 100 Watt/m2. Consider a body B of area A = 10 m2 coefficient of reflectivity r = 0.3 and coefficient of transmission t = 0.5. Its temperature is 300 K. Then which of the following is incorrect :
(A*) The emissive power of B is 20 W/m2 (B) The emissive power of B is 200 W/m2
(C) The power emitted by B is 200 Watts (D) The emissivity of B is = 0.2
Ñf".kdk dh mRltZu {kerk T = 300 K rki ij 100 Watt/m2 gSA ekuksa ,d oLrq B
gS ftldk {ks=kQy A = 10 m2 ijkorZu xq.kkad r = 0.3 vkSj ikjxeu xq.kkad t =
0.5. gSA mldk rki 300 K gS rks fuEu esa ls dkSulk xyr gS &
(A) B dh mRltZu {kerk 20 W/m2 (B*) B dh mRltZu {kerk 200 W/m2 gS
(C) B }kjk mRltZu dh xbZ 'kfDr gS 200 Watts (D) B dh mRltZdrk gS 0.2
Sol. Since tSls fd , e = a = 0.2
(Since pwafd, a = (1 – r – t) = 0.2 for the body B oLrq ds fy,) E = (100) (0.2) = 20 W/m2
14.20 There are four objects A, B, C and D. It is observed that A and B are in thermal equilibrium and C and D are also in thermal equilibrium. However, A and C are not in thermal equilibrium. We can conclude that :
(A) B and D are in thermal equilibrium
(B) B and D could be in thermal equilibrium (C*) B and D cannot be in thermal equilibrium
(D) The zeroth law of thermodynamics does not apply here because there are more than three objects A,B,CvkSjDpkjoLrq,sagSaAAvkSjBrkih;lkE;koLFk esagSvkSjCvkSjDHkhrkih;lkE;koLFk esagSAAvkSjC rkih;lkE;koLFk esaughagSAge;gfu"d"kZfudkyldrsgSafd&
(A)BvkSjDrkih;lkE;koLFk esagS (B)BvkSjDrkih;lkE;koLFk esagksldrsgSaA (C*)BvkSjDrkih;lkE;koLFk esaughagksldrs
(D)Å"ekxfrdhdk'kwU;kadhfu;e;gk¡ughayxldrkD;ksafdrhulsT;knkoLrq,sagSA
Sol. Thermal equilibrium means same temperature.
rkih;lkE;koLFk dkvFkZlekurkigksrkgSA
14.21 If HC, HK and HF are heat required to raise the temperature of one gram of water by one degree in Celsius, Kelvin and Fahrenheit temperature scales respectively then :
;fn HC, HK vkSj HF og Å"ek gS tks 1xzke ikuh dkrki Øe'k% 1°C, 1K, vkSj 1°F c<+kus ds fy, pkfg, rks %
(A) HK > HC > HF (B) HF > HC > HK (C*) HK = HC > HF (D) HK = HC = HF
C
Sol. 5
= F 32
9
C = F 5 9
C =
5 F
9
C = K
14.22 Find the amount of work done to increase the temperature of one mole of an ideal gas by 30ºC if it is expanding under the condition V T2 / 3 .
,d eksy vkn'kZ xSl dk rki 30ºC c<+kus esa fdruk dk;Z fd;k tk;sxk vxj
mldk izlkj V T2 / 3 ds vuqlkj fd;k tkrk gS -
(A*) 166.2 J (B) 136.2 (C) 126.2 J (D) none of these buesa ls dksbZ ugha
Sol. (A) PV = RT for 1 mole 1 eksy ds fy,
W = PdV RT dV
V = CT2/3
2
dV = 3 CT–1/3 dT or;k
dV 2 dT V = 3 T
T2 RT 2 dT 2
W = T = R(T2 – T1) = 166.2 J
1 3 T 3
14.23 A gas is expanded from volume V0 to 2V0 under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let U1,U2 and U3 be the change in internal energy of the gas in these three processes. Then :
,dxSldksvk;ruV0 ls2V0rdrhufofHkUuizØeksadsvuqlkjizlkfjrfd;ktkrkgSAizØe1lenkch]izØe2lerkih] izØe3:)ks"egSaAekuyhft,burhuizØeksaesavkUrfjdÅtkZesaifjroZuU1,U2vkSjU3gSrks
(A*) U1 > U2 > U3 (B) U1 < U2 < U3
(C) U2 < U1 < U3 (D) U2 < U3 < U1
Sol. (A)
sU nfR T
2
U T;
Since volume is same in all three process therefore temperature will be least having least pressure.
D;ksafdvk;rurhuksaizfØ;kvksaesalekugSblfy,U;wurenkcokyhizfØ;kdkrkiekuU;wuregksxkA
14.24 Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in Figure. By measurring the gradient, it can be shown that the gas may be vkn'kZxSldsfy,nkcovk;rudsy?kqx.kdikB;kadxzkQesafpf=krgSAço.krkdksekidj;gfn[k ;ktkldrk gSfdxSlgksldrhgS&
(A) monoatomic and undergoing an adiabatic change
,dijek.kqdvkSjblesa:)ks"eifjorZugksrsgSA
(B) monoatomic and undergoing an isothermal change
,dijek.kqdvkSjblesalerkihifjorZugksrsgSA
(C*) diatomic and undergoing an adiabatic change
f}&ijek.kqdvkSjblesa:)ks"eifjorZugksrsgSA
(D) triatomic and undergoing an isothermal change.
f=k&ijek.kqdvkSjblesalerkihifjorZugksrsgSA
Sol. log P = m log V where m is slope tgk¡m E > E if temperature in every process decreases
A B C
EA >EB>EC;fn izR;sd izfØ;k esa rki ?kVrk gksA
(C*) E > E > E if temperature in every process increases
A B C
EA> EB> EC;fn izR;sd izfØ;k esa rki c<+rk gksA
(D) E < E < E if temperature in every process increases
B A C
EB < EA< EC;fn izR;sd izfØ;k esa rki c<+rk gksA
Sol. Initial state is same for all three processes (say initial internal energy = E )
In the final state, V = V = V
and P > P > P
P V > P V > P V
E > E > E
if T > T
then E > E for all three processes
and hence (E
– E ) < (E – E ) < (E – E ) E < E < E
If T < T , then E
< E for all three processes
and hence (E
– E ) > (E – E ) > (E – E ) E > E > E
gy lHkhrhuçfØ;kvksadsfy,çkjfEHkdvoLFkklekugSA(ekukçkjfEHkdvkUrfjdÅtkZ=E0) vfUre voLFkk esaVA = VB = VC
vkSj PA > PB > PC
P V
;fn T1 > T2
> P V
> P V
E > E > E
vr% lHkhçfØ;kvksa dsfy, E0 >Ef
vkSj vr% (E0 – EA) < (E0 – EB) < (E0 – EC) EA < EB < EC
;fnT1 (EB – E0) > (EC – E0) EA > EB > EC
14.31 When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%,
tc,drkacsdsflDdsdkrki80oCc<+k;ktkrkgSrksbldkO;kl0.2%c<+tkrkgS& (A*) percentage rise in the area of a face is 0.4% bldhQydh;lrgds{ks=kQyesaçfr'kro`f)0.4%gksxhA
(B) percentage rise in the thickness is 0.4%
eksVkbZeasçfr'kro`f)0.4%gksxhA
(C*) percentage rise in the volume is 0.6%
vk;ru esa çfr'kr o`f) 0.6% gksxhA
(D*) coefficient of linear expansion of copper is 0.25x10-4 / oC.
rkacsdkjs[kh;çlkjxq.k ad0.25x10-4 /oCgSA
Sol.
A
× 100 =
𝑙 × 100
A
% increaseinArea {kS=kQy esa % o`f)= 2× 0.2 = 0.4
V
V × 100 = 3 × 0.2 = 0.6 %
Sincepawfd l = l T
𝑙
× 100 = T × 100 = 0.2
𝑙
= 0.25 × 10–4 / ºC
14.32 A vessel is partly filled with liquid. When the vessel is cooled to a lower temperature, the space in the vessel, unoccupied by the liquid remains constant. Then the volume of the liquid (VL), volume of the vessel (Vv), the coefficients of cubical expansion of the material of the vessel ( ) and of the liquid ( ) are related as
v L
,dik=kvkaf'kd:ilsnzolsHkjkgS]tcik=kfuEurkiijB.Mkfd;ktkrkgS]rksnzo}kjkik=kesa'ks"kLFk ufu;rjgrk gSArcnzodkvk;ru(VL),ik=kdkvk;ru(Vv),ik=kdsinkFkZdkvk;ruçlkjxq.k ad(v)rFkknzodsfy,vk;ruizlkj xq.k ad(L) fuEuçdkjlEcfU/krgSA
(A*) >
(B) <
(C) / =V /V
(D*) / =V /V
L v L v
v L v L
v L L v
Sol. V = V
Y V
= YVVV
YL VV
or YV = VL
butysfdu VV > VL YL > YV
14.33 Two identical objects A and B are at temperatures TA and TB respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature T (TA > T > TB). The objects A and B attain the temperature T eventually. Select the correct statements from the following.
nks,dtSlhoLrq,saAoBØe'k%TAoTB rkiijgSAnksuksaoLrq;saT(TA >T>TB)rkioiw.kZr;kvo'k s"kdnhokjksa okys dejs ds vUnj j[kh gSA oLrq,sa A o B vUr esa T rki izkIr dj ysrh gSA fuEu esa ls lR; dFku pqfu;sA
(A) A only emits radiation, while B only absorbs it until both attain the temperature T.
(B*) A loses more heat by radiation than it absorbs, while B absorbs more radiation than it emits, until they attain the temperature T.
(C) Both A and B only absorb radiation, but do not emit it, until they attain the temperature T. (D*) Each object continues to emit and absorb radiation even after attaining the temperature T.
(A) dsoy A fofdj.k mRlftZr djrk gS] tcfd B dsoy bldks vo'kksf"kr djrk gS tc rd fd nksuksa T rki ugha izkIr dj ysrsA (B*)Aftruhfofdj.k vo'kksf"krdjrk gSmlls vf/kdmRlftZr djrk gS, tcfd B ftruhmRlftZr djrkgS mlls vf/kdvo'kksf"krdjrkgS]tcrdfdosTrkiughaizkIrdjysrsgSA (C)AoBnksuksadsoyfofdj.kvo'k sf"krdjrsgSijUrqblsmRlftZrughadjrs]tcrdfdosTrkiughaizkIrdjrsA (D*)TrkiizkIrdjusdscknHkhizR;sdoLrqfofdj.kmRlftZrovo'kksf"krdjuktkjhj[krhgSA
Sol. Every object emit and absorb the radiations simultaneously, if energy emitted is more than energy
absorbed temperature falls and vice versa. izR;sdoLrqfofdj.k sadkvo'k s"k.kvkSjmRltZulkFk&lkFkdjrhgS];fnmRlftZrÅtkZ, vo'k sf"krÅtkZlsvf/kdgSrks rkifxjrkgSvU;Fk c<+rkgSA
SECTION - III : ASSERTION AND REASON TYPE
[k.M - III : dFku vkSj dkj.k izdkj
14.34. STATEMENT-1 : Two solid cylindrical rods of identical size and different thermal conductivity K1 and K2 are connected in series. Then the equivalent thermal conductivity of two rod system is less than the value of thermal conductivity of either rod.
STATEMENT-2 : For two cylindrical rods of identical size and different thermal conductivity K1 and K2 connected in series, the equivalent thermal conductivity K is given by
2 1 1
K K1 K 2
oDrO;-1: lekuvkdkjrFk fHkUu&fHkUuÅ"ekpkydrkvksaK1rFk K2dhnksBkslcsyukdkjNM+ksdksJs.khØeesatksM+k tkrkgSArksnksNM+fudk;dhrqY;Å"ekpkydrknksuksaNM+ksdh,dSdÅ"ekpkydrklsdegksxhA
oDrO;-2: Js.khØeesatqM+hlekuvkdkjrFk fHkUuÅ"ekpkydrkvksaK1rFk K2dhnkscsyukdkjNM+ksdsfy,]rqY; Å"ek pkydrk nh tkrh gSA
2 1 1
K K1 K 2
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D*) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. Equivalent thermal conductivity of two identical rods in series is given by
nkslekuNM+ksadhJs.khØeesarqY;Å"ekpkydrknhtkrhgS&
2 1 1
K K1 K 2
If ;fn K1 < K2, then rc K1 < K < K2
Hence statement 1 is false.
blfy;s oDrO;-1 vlR; gSA
14.35 STATEMENT-1 : As the temperature of the blackbody increases, the wavelength at which the spectral intensity (E) is maximum decreases.
STATEMENT-2 : The wavelength at which the spectral intensity will be maximum for a black body is
proportional to the fourth power of its absolute temperature.
oDrO;-1: d`f".kdkdkrkic<+usij]ogrjaxnS/;ZftldslaxrLisDVªehrhozrk(E)vf/kdregksrhgSogrjaxnS/
;Z?kVrhgSA
oDrO;-2: ogrjaxnS/;Z, ftldslaxrd`f".kdkdhLisDVªehrhozrkvf/kdregksrhgS, bldsijerkidsprqFkZ?kkr dslekuqikrhgksrhgSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS ; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C*) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. From Wein's law
T = constant i.e., peak emission wavelength 1 . Hence as T increase
decreases.
m m T m
gy% ohufu;e T= fu;rls, f'k[kj mRltZurjaxnS/;Z 1 Avr% Tc<+usaij ?kVsxkA
m m T m
14.36 STATEMENT-1 : An ideal gas is enclosed within a container fitted with a piston. When volume of this enclosed gas is increased at constant temperature, the pressure exerted by the gas on the piston de- creases.
STATEMENT-2 : In the above situation the rate of molecules striking the piston decreases. If the rate at which molecules of a gas having same average speed striking a given area of the wall decreases, the pressure exerted by gas on the wall decreases.
oDrO;-1: ,dvkn'kZ xSlfiLVu ;qDr,d crZu¼ik=k½ dsvUnj HkjhgqbZ gSAtc fu;rrki ij bl xSldk vk;ru c<+k;ktkrkgS]rksxSl}kjkfiLVuijyxk;kx;knkc?kVtkrkgSA
oDrO;-2: blÅijnhxbZfLFkfresafiLVuijv.kqvksadsVdjkusdhnj?kVrhgSA;fnnhokjdsfn;sx;s{ks=kQyij lekuvkSlrpkyokysv.kqvksadsVdjkusdhnj?kVtkrhgS]rksxSl}kjknhokjijyxk;kx;knkc?kVtkrkgSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
(A*) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;-1lR; gS] oDrO;-2vlR; gS ;
(D) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. If the rate at which molecules of same mass having same rms velocity striking a wall decreases, then the rate at 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 the given process is isothermal. Now the piston is at a greater distance from opposite wall and hence time taken by gas molecules from near the opposite wall to reach the piston will be more. Thus rate of molecules striking the piston decreases. Hence statement-1 is correct and statement-2 is correct explanation.
;fnognjftlijlekunzO;ekurFk lekuoxZek/;ewyosxdsv.kqnhokjijVdjkrsgS]?kVrhgSrksognjftlij laosxnhokjijfn;ktkrkgS]?kVtkrhgSAifj.k eLo:inkc?kVtkrkgSAblfy;soDrO;–2lR;gSA
oDrO;–1 esa xSl dk oxZek/;ewy osx fiLVu }kjk crZu dk vk;ru c<+kus ij leku jgrk gSA pwafd nh xbZ çfØ;k lerkih gSA vc fiLVu foijhr nhokj ls vf/kd nwjh ij gksxk, blfy;s xSl v.kqvksa }kjk foijhr nhokj ds ikl ls fiLVu rd igq¡pus esa fy;k x;k le; vf/kd gksxkA blfy;s fiLVu ij v.kqvksa ds Vdjkus dh nj ?kV tkrh gSA blfy;s oDrO;–1 lgh gS rFkk oDrO;– 2 lgh Li"Vhdj.k gSA
SECTION - IV : COMPREHENSION TYPE
[k.M-IV:cks/kuizdkj
Comprehension # 1 vuqPNsn # 1
Two closed identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted. fdlhiqjkusoSKkfuddhiz;ksx'k ykesankscUnpkydh;cDlsfeysAxSldhlEiqf"Vdsfy,nkscDlksaijdqNiz;ksx fd;sx;svkSjifj.k evafdrfd,x,A
Experiment 1. When the two containers are weighed WA = 225 g , WB = 160 g and mass of evacuated container WC = 100 g.
iz;ksx1. tc nksuksa dks rksyk x;k rks WA = 225g, WB = 160gvkSj fuokZfrr cDls dk nzO;eku WC = 100g.
Experiment 2. When the two containers are given same amount of heat same temperature rise is recorded. The pressure change found are
PA = 2.5 atm. PB = 1.5 atm.
iz;ksx2. tcnksuksadkslekuÅ"eknhxbZrksnksuksadsrkiesalekuo`f)ik;hxbZAnkcdkifjorZuik;kx;k
PA = 2.5 atm. PB = 1.5 atm.
Required data for unknown gas :
vKkr xSlds fy,vHkh"B vkadM+s
, d i j ek. kq
¼eksy j nzO; eku½
f} i j ek. kqd
¼eksy j nzO; eku½
14.37 Identify the type of gas filled in container A and B respectively. AvkSjBcDlksaesaxSldsizdkjdksigpkuksa%
(A) Mono, Mono (B) Dia, Dia (C*) Mono, Dia (D) Dia, Mono.
(A) ,d ijek.kqd, ,d ijek.kqd (B) f}ijek.kqd, f}ijek.kqd
(C*) ,d ijek.kqd, f}ijek.kqd (D) f}ijek.kqd, ,d ijek.kqd
Sol. Heat given Å"ek nh xbZ : Q n1CV1T
& for Gas B xSl B ds fy, Q = n2CV2 T
For gas A xSl A ds fy,
( For same heat given, temperature rises by same value for both the gases.) (D;ksafdlekuÅ"ekdsfy,rkidhc<+ksrjhnksuksadsfy,cjkcjgS)
n1CV
n2CV
................(1)
Also, (PB)V = n2RT and (PA)V = n1RT
vr% (PB)V = n2RT rFkk(PA)V = n1RT
n1
n2 =
PA
PB =
2.5
1.5
= 5 n =
3 1
5
3 n2
Substituting in (1)
lehdj.k (1) esa j[kus ij
5
Cv 5
( 5 R)
n C = n C
2 2
3 2 V1 2 v2
C 3 ( 3 R)
Hence, Gas B is diatomic and Gas A is monoatomic.
vr% B ,d f}ijek.kqdxSlgSrFkkA,dijek.kqd xSlgSA
14.38 Identify the gas filled in the container A and B.
AvkSjBesaxSldksigpkuksa
(A) N2, Ne (B) He, H2 (C) O2 , Ar (D*) Ar, O2
5 125 5 60
Sol. Sincepqafd n1 =
n2 Therefore vr%
=
3 MA
3 MB
(From experiment iz;ksx }kjk 1 : WA = 25 gm & WB = 60 gm)
5MB = 4MA
The above relation holds for the pair–Gas A : Ar and Gas B : O2 .
Åijfn;kx;klEcU/knksuksatksM+ksadsfy,ykxwgksrkgSxSlA:vkjxuvkSjxSlB:vkWDlhtu
14.39 If the gases have initial temperature 300 K and they are mixed in an adiabatic container having the same volume as the previous containers. Now the temperature of the mixture is T and pressure is P.
Then
vxjxSlksadkizkjfEHkdrki300KFk vkSjnksuksadks:)ks"ecDlsesafeyk;kx;kftldkvk;ruigysokyscDlksa ds leku gSA vc feJ.k dk rki T vkSj nkc P gS rc&
(A) P > PA , T > 300 K (B*) P > PB , T = 300 K
(C) P < PA , T = 300 K (D) P > PA , T < 300 K
Sol.
nACV
300 nBCV
300 =
nACV T nBCV T
A B A B
(number of molesremainssame) ¼eksyksa dh la[;k leku gSA½
T = 300 K
(It could also be seen directly that temperature finally will be 300 K, since no heat exchange takes place between those gases as their initial temperatures are same)
Since, volume remains same but number of moles increases. Therefore, pressure increases.
;g ns[kk tk ldrk gS fd var esa Hkh rki 300K gksxk D;ksafd Å"ek dk vknku iznku nksuksa xSlksa esa ugha gqvk (mudk izkjfEHkd rki cjkcj gS) pqafd vk;ru leku jgrk gS, ij eksy dh la[;k c<+h gS vr% nkc c<+sxkA
Comprehension # 2 vuqPNsn # 2
A body cools in a surrounding of constant temperature 30 ºC. Its heat capacity is 2J/ºC. Initial temperature of the body is 40ºC . Assume Newton’s law of cooling is valid. The body cools to 38ºC in 10 minutes. 30ºCfu;rrkiekudsokrkoj.kesa,doLrqB.MhgksrhgSAÅ"ek/k fjrk2J/ºCgSAoLrqdkçkjfEHkdrkieku40ºC gSA ekukfd U;wVudk'khryudkfu;e ;gk¡ ekU;gksrkgSAoLrq10feuVesa38ºCrdB.MhgksrhgSA
14.40 In further 10 minutes it will cool from 38ºC to _____ :
vxys10 feuV esa ;g 38ºC ls rd B.Mh gks tk,xh &
(A) 36ºC (B*) 36.4ºC (C) 37ºC (D) 37.5ºC
Sol. We have – = (
– s
) e–kt
where = Initial temperature of body = 40°C
= temperature of body after time t.
Since body cools from 40 to 38 in 10min, we have 38 – 30 = (40 – 30) e– k 10 (1)
Let after 10 min, The body temp. be
– 30 = (38 – 30) e–k 10 (2)
(1)
gives
8 10 , – 30 = 6.4 = 36.4 °C
(2) 30 8
gy. ge tkurs gS – = ( – ) e–kt
s s
tgk¡ = oLrq dk izkjfEHkdrkieku =40°C
= oLrq dk t le; i'pkr~ rkieku .
pwafd oLrq 40ls38°C rd 10feuVesa B.Mhgksrh gS, vr% 38 – 30 = (40 – 30) e– k 10 (1)
ekuk 10feuVi'pkr~ ,oLrq dkrkieku gkstkrk gS
– 30 = (38 – 30) e–k 10 (2)
(1) ls
8 10 , – 30 = 6.4 = 36.4 °C
(2) 30 8
14.41 The temperature of the body in ºC denoted by the variation of versus time t is best denoted as
oLrqdkºCesarkieku}kjkçnf'kZrgSdkle;t dslkFkifjorZudkloksZfprvkjs[kgS
(A*)
(B)
(C)
(D)
Sol. Temperature will decrease exponentially
rki?k rakdh:ils?kVsxkA
14.42 When the body temperature has reached 38 ºC, it is heated again so that it reaches to 40ºC in 10 minutes .The total heat required from a heater by the body is: tc oLrqdk rki38ºCigq¡prk gSrks bldksnqckjk blçdkj xeZ djrsgS fdbldk rkieku10feuVesa40ºCgks tkrk gS rks oLrq ds fy, ghVj ls çkIr dqy Å"ek gksxh &
(A) 3.6J (B) 0.364J (C*) 8 J (D) 4 J
Sol. During heating process from 38 to 40 in 10 min. The body will lose heat in the surrounding which will be exactly equal to the heat lost when it cooled from 40 to 38 in 10 min, which is equal to ms = 2 × 2
= 4 J.
During heating process heat required by the body = m s = 4 J.
Total heat required = 8 J.
gy. 38ls40°Crd10feuVesaxeZdjusdsnkSjku, oLrqokrkoj.kesaÅ"ek{k;djsxhtksfd10feuVesa40ls38°C
rd B.Mh gkus esa mRlftZr Å"ek ds cjkcj gksxh] tks fd ms = 2× 2 = 4J ds cjkcj gSA xeZ djus ds nkSjku oLrq ds fy, vko';d Å"ek = m s = 4 J.
dqy vko';d Å"ek = 8 J.
SECTION - V : MATRIX - MATCH TYPE
[k.M-V:eSfVªDl&lqesyizdkj
14.43 An ideal monoatomic gas undergoes different types of processes which are described in column-I. Match the corresponding effects in column-II. The letters have usual meaning.
,dvkn'kZ,dijek.kqdxSltksfdLrEHk-Iesafn[k ;sx,fofHkUuizØeksadkikyudjrhgSALrEHk-IIesalaxrizHk o fn;s x;s gSA inksa dk lkekU; vFkZ gSA
Column-I Column-II
(A) P = 2V2 (p) If volume increases then temperature will also increase.
(B) PV2 = constant (q) If volume increases then temperature will decrease.
(C) C = CV + 2R (r) For expansion, heat will have to be supplied to the gas.
(D) C = CV – 2R (s) If temperature increases then work done by gas is positive.
(t) If temperature decreases then work done by gas is positive
dkWye-I dkWye-II
(A) P = 2V2 (p) ;fnvk;ruc<+rkgSrksrkiekuHkhc<+sxkA
(B) PV2 = fu;r (q);fnvk;ruc<+rkgSrksrki?kVsxkA
(C) C = CV + 2R (r) izlkjds fy,]xSl dksÅ"ek iznkudhtkuhvko';dgSA
(D) C = CV – 2R (s) ;fnrkieku c<+rkgS rc xSl }kjk fd;k x;k dk;Z/kukRedgSA
(t) ;fnrki?kVrkgSrksxSl}kjkfd;kx;kdk;Z/kukRed gksxkA
Ans. (A) p,r,s (B) q,t ; (C) p,r,s (D) q,r,t
Sol. (A) If P = 2V2 , from ideal gas equation we get
2V3 = nRT
with increase in volume
(A) Temperature increases implies dU = +ve
dW = +ve
Hence dQ = dU + W = +ve
(B) If PV2 = constant, from ideal gas equation we get VT = K (constant) On decreasing temperature, as volume of an ideal gas increases
dW = +ve
Now dQ = dU + PdV = nCvdT –
PK K
T2 dT [ dV = – T 2 dT]
= nC dT – PV dT = n(C
v T v
– R) dT
with increase in temperature dT = +ve
and since Cv > R for monoatomic gas. Hence dQ = +ve as temperature is increased
(C) dQ = nC dT = nCv dT + PdV
n (C + 2R) dT = nC dT + PdV
v
2nRdT = PdV
dV
dT = +ve
Hence with increase in temperature volume increases and vice versa.
dQ = dU + W = +ve
(D) dQ = nC dT = nCv dT + PdV
or n (Cv – 2R)dT = nCrdT + PdV
dV
or – 2nRdT = PdV dT = –ve
On decreasing temperature, as volume of an ideal gas increases
dW = +ve
Also dQ = n(Cv – 2R)dT
with increase in temperature dT = +ve but Cv < 2R for monoatomic gas. Therefore dQ = –ve with increase in temperature.
(A) ;fn P = 2V2, vkn'kZ xSl lehdj.k ls
2V3 = nRT
vk;ruc<+kusij
(A) rkiekuc<+rkgSvFkkZr~dU=+ve
W = +ve
vr% dQ = dU + W = +ve
(B) ;fn PV2 = fu;r] vkn'kZ xSl lehdj.k ls VT = K (fu;r)
rkieku?kVkusijvkn'kZxSldkvk;ruc<+rkgSA
dW = +ve
vc dQ = dU + PdV = nCvdT –
PK K
T2 dT [ dV = – T 2 dT]
= nC dT – PV dT = n(C
v T v
– R) dT
rki c<+us ij dT =+ve
rFkk pwafd Cv >R ,dijek.kqd xSl ds fy, vr% dQ= +vetc rkieku c<+rk gSA
(C) dQ = nC dT = nCv dT + PdV
n (C + 2R) dT = nC dT + PdV
v
2nRdT = PdV
dV
dT = +ve
vr% rkieku c<+us ijvk;ru c<+rk gS oblh rjg vk;ru c<+usij rkic<+rk gSA
dQ = dU + W = +ve
(D) dQ = nC dT = nCv dT + PdV
;k n (Cv – 2R)dT = nCrdT + PdV
dV
;k – 2nRdT = PdV dT = –ve
rkieku?kVkusijvkn'kZxSldkvk;ruc<+rkgSA
dW = +ve
vr% dQ = n(Cv – 2R)dT
rki c<+us ij dT = +ve ysfduCv < 2R ,d ijek.kqd xSl ds fy, vr% dQ = –ve tc rki c<+rk gSA
14.44 A sample of gas goes from state A to state B in four different manners, as shown by the graphs. Let W be the work done by the gas and U be change in internal energy along the path AB. Correctly match the graphs with the statements provided.
fp=k esa n'kkZ;s vuqlkj] ,d xSl dk uewuk pkj fofHkUu rjhdksa ls voLFkk A ls B rd igq¡prk gSA ;fn xSl }kjk ekxZ AB esa fd;k x;k dk;Z W
rFkk vkUrfjd ÅtkZ esa ifjorZu U gks] rks xzkQksa dks fn;s x;s dFkuksa ls lqesfyr dhft,A &
V
(A)
(p) Both W and U are positive
P
WrFk Unksuksa/kukRedgSaA
(B) (q) Both W and U are negative
T
W rFk Unksuksa_.k RedgSaA
T
(C) (r) W is positive whereas U is negative
V
W /kukRed, ijUrqU_.kkRedgSaA
V
(D) (D)
P
Ans. (A) – s ; (B) – q,t ; (C) – r,t ; (D) – q,t
Sol. in (A), V is on vertical axis. (a)esa,Vm/okZ/kjv{kijgSA
(s) W is negative whereas U is positive
W _.kkRed, ijUrq U/kukRedgSA
(t) Final temperature of an ideal gas is less than its initial temperature.
(t) vkn'kZxSldkvfUre rki]çkjfEHkdrkilsdegSA
As V is icreasing, W is positive. pqafdV c<+rk gS, W /kukRed gSA
V
V is decreasing, W is negative.V ?kVus ij, W _.kkRed gSA
P
As negative work in part-II is greater than positive work in part-I, net work during the process is negative. Using PV = nRT and as Vremains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive. Hence option (S) is connected with (A).
As T = PV
nR
From graph we can say Tfinal < Tinitial in part B,C & D
Similar arguments can be applied to other graphs.
pwafdHk x-IIesa_.k Reddk;ZHk x-I,lsT;knkgS]vr%çfØ;kesadqydk;Z_.kkRedgSA
PV= nRT dkç;ksx djusij vkSjV çkjEHk,oa vfUrefcUnqvksa ijleku jgrkgS, vr% Li"VgS fdvfUre rki] çkjfEHkdrkilsT;knkgSpwafdnkcc<+rkgSAvr%dU/kukRedgS]fodYi(S),(A)lslEcfU/krgSA
pwafd T = PV
vkjs[kls gedg ldrsgS fdHkkx B,CoDesa TvfUre
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