DC Pandey Waves And Thermodynamics

390 — Waves and Thermodynamics Match the Columns 4. I ∝ A2 or A ∝ I 1. (a) f = v Due to a point source, 2l I = P (P = power of source) 4 πr2 fc = v = v = f 4 (2l) 8l 4 or I ∝ 1 and A∝1 (b) Second overtone frequency r2 r = 5 fc = 5 ( f /4) = 1.25 f Due to a line source (c) Third harmonic frequency = 3 fc I = P or I ∝ 1 = 3 ( f /4) = 0.75 f 2πl r (d) First overtone frequency = 3 fc ∴ A∝ 1 = 3 ( f /4) = 0.75 f r 5. k =π Q O O λ = 2π = 2 m O2 k 3 1 l = 5λ = 5 (2) T' 44 2. T v v 4 4 (a) fb = fT ′ − fT = 2.5 m  v   v  Distance of displacement node from the closed end  −v   +v  = f v / 4 − f  v / 4 = λ or λ 2 =8f 15 = 1 m or 2m (b) fb = fT ′ − fT = f  v + v / 4  − f  v − v/4  =2f 3 (c) T and T ′ both are approaching towards O3. Pressure node means displacement antinode. Its ∴ fT = fT ′ or fb = 0 distance from closed end (d) Not T and T ′ both are receding from O3. = λ or 3λ or 5λ Hence, again fT = fT ′ or fb = 0 44 4 3. Let f1 = frequency of tuning fork and = 0.5 m,1.5 m and 2.5 m f2 = frequency of streamed wire Subjective Questions Given, f1 > f2 1. By definition sound level = 10 log I = 60 ∴ fb = f1 − f2 I0 or I = 106 (a) By loading wax on turning fork f1 will I0 decrease. Hence, f1 − f2 may be less than fb ⇒ I = 10−12 × 106 = 1µW/m2 or f2 − f1 may be greater than fb. (b) If prongs are filed, f1 will increase. Hence, Power entering the room = 1 × 10−6 × 2 = 2 µW f1 − f2 > fb Energy collected in a day = 2 × 10−6 × 86400 (c) If tension in stretched wire is increased, f2 will increase. Hence, f1 − f2 may be less than fb = 0.173 J Ans. or f2 − f1 may be greater than fb. (d) If tension in stretched wire is decreased, f2 2. (a) At a distance r from a point source of power P, will decrease. Hence, the intensity of the sound is f1 − f2 > fb

Chapter 19 Sound Waves — 391 I = P = 0.8 5. (a) λ = v = 340 = 0.57 m 4 πr2 (4π ) (1.5)2 f 600 or I = 2.83 × 10−2 W/m2 …(i) S1 dθ Further, the intensity of sound in terms of d = 2m (∆p)m, ρ and v is given by S2 I = (∆p)2m …(ii) d sin θ = λ 2ρv 2 From Eqs. (i) and (ii), ∴ θ = sin− 1  2λd (∆p)m = 2 × 2.83 × 10−2 × 1.29 × 340 = sin− 1  0.457 = 4.98 N/m2 Ans. (b) Pressure oscillation amplitude (∆p)m and = 8.14° displacement oscillation amplitude A are (b) d sin θ = λ or θ = sin − 1  λ  related by the equation d (∆p)m = BAk = sin− 1  0.257 Substituting B = ρv2, k = ω v and ω = 2πf = 16.5° We get , (∆p)m = 2πAρvf (c) d sin θ = nλ ∴ A = (∆p)m = 4.98 ∴ nmax = d (When sin θ = 1) λ 2πρvf (2π ) (1.29) (340) (600) = 3.0 × 10−6 m Ans. = 2 = 3.5 0.57 3. (a) Fundamental frequency when the pipe is open at ∴ Maximum 3 maximas can be heard both ends is corresponding to n = 1, 2 and 3 (beyond θ = 0°) f1 = v = 340 6. ∆φ1 =  2π  ∆x =  2π  λ =π 2l 2 × 0.6 λ λ 2 = 283.33 Hz Ans. (a) ∆φnet = ∆φ1 + ∆φ2 (b) Suppose the hole is uncovered at a length l =π + 0 from the mouthpiece, the fundamental =π frequency will be f1 = v So, they will interfere destructively and 2l Inet = 0 ∴ l = v = 340 (b) No interference will take place. Hence, 2 f1 2 × 330 Inet = I1 + I2 = 2I0 = 0.515 m (c) ∆φnet = ∆φ1 ± ∆φ2 = 51.5 cm Ans. = 2π or 0 (as ∆φ2 = π) In both conditions, they interfere Note Opening holes in the side effectively shortens the constructively. Hence, length of the resonance column, thus increasing I = Imax = 4I0 the frequency. 7. ∆φ = 2π  2πf  4. 2 H 2 + d2/4 − d = nλ …(i) (a) λ (∆x) = v (∆x) Now, 2 (H + h)2 + d2/4 − d = nλ + λ …(ii) = (2π ) (170) (11 − 8) 2 340 Solving these two equations, we get = 3π = φ (say ) λ = 2 4 (H + h)2 + d2 − 2 4H 2 + d2 I = 4I0 cos2 θ …(i) 2

392 — Waves and Thermodynamics Substituting value of φ, we get 10. Given length of pipe, l = 3 m x = 0 I =0 Third harmonic implies that 3  λ2 = l (b) Now, φ = 3π ± π = 4π or 2π or λ = 2l = 2 × 3 = 2 m Hence, from Eq. (i) 33 l I = 4I0 Now, L1 − L2 = 10 log10 I1 The angular frequency is I2 ω = 2πf . ∴ L1 − 60 = 10 log10  4 I 0 = 6 = 2πv = (2π )(332) or x=l   I0  λ2 ω = 332π rad/s ∴ L1 = 66 dB (c) Sources are incoherent. The particle displacement y(x, t) can be written as Hence, y(x, t) = A cos kx sin ωt k = 2π = 2π = 3π I = I0 + I0 = 2I0 λ (2l/3) l L1 − L2 = 10 log10 I1 and ω = kv = 3πv  v = ωk  I2 l = 10log10 2I 0 ∴ y(x, t) = A cos  3πl x ⋅ sin  3πl v t I0 ∴ L1 − 60 = 10 log10 2 = 3 or L1 = 63 dB The longitudinal oscillations of an air column can 8. (a) I = P be viewed as oscillations of particle displacement 4 πr2 or pressure wave or density wave. Pressure 1 × 10− 3 variation is related to particle displacement as 4π (2)2 ∴ I1 = = 19.9 × 10− 6 W/m2 p (x, t) = − B ∂y (B = Bulk modulus) ∂x = 19.9 µW/m2 =  3 BAπ  sin  3πl x sin  3πl v t l 1 × 10− 3 I2 = 4π (3)2 = 8.84 × 10− 6 W/m2 The amplitude of pressure variation is = 8.84 µW/m2 pmax = 3BAπ ⇒ v= B or B = ρv2 l ρ (b) Ip = Imax = ( I1 + I2 )2 ∴ pmax = 3 ρv2Aπ or A = pmax l l 3 ρv2π = 55.3 µW/m2 Here, pmax = 1% or p0 = 103 N/m2 (c) Ip = Imin = ( I1 − I2 )2 Substituting the values, = 2.2 µW/m2 A = (103)(3) = 0.0028 m (d) Ip = I1 + I2 (3)(1.03)(332)2π = 28.7 µW/m2 or A = 0.28 cm Ans. 9. vs = v0 = v θ S v0 According to definition of Bulk modulus (B) θ Let u be the speed of sound. vs B = − dP …(i) (dV /V ) Then, f ′ = f  u + v0 cos θ O Volume = Mass or V = m  u + vs cos θ Density ρ  u + v cos θθ or dV = − m dρ = − Vdρ  u + v cos ρ2 ρ = f  or f ′ = f Ans. or dV = − dρ Vρ

Chapter 19 Sound Waves — 393 Substituting in Eq. (i), we get (b) vs = 30 m/min dρ = ρ (dP) Spacing between the pies will be B = 300 − 30 or 13.5 m or amplitude of density oscillation is 20 dρmax = ρ pmax = pmax  ρ = 1  and f′ = f  v  = (20)  300  B v2 B v2      v − vs   300 − 30 103 = (332)2 = 9 × 10−3 kg/m3 = 22.22 min−1 Ans. 11. Sound level (in dB) 14. (a) Comparing with the equation of a travelling wave L = 10 log10  I  y = a sin (kx − ωt)  I0 k = 15π and ω = 6000π where, I0 = 10−12 W/m2 ∴ Velocity of the sound, v = ω L = 60 dB k Hence, I = (106)I0 = 10−6 W/m2 = 6000π = 400 ms−1 as v = B 15π ρ Intensity, I = Power = P Hence, ρ = B = 1.6 × 105 =1 kg/ m 3 Ans. Area 4 πr2 v2 (400)2 ⇒ P = I (4πr2) (b) Pressure amplitude, p0 = BAk P = (10−6)(4π )(500)2 = 3.14 W Hence, A = p0 = 24 π Bk 1.6 × 105 × 15π ∴ Time, t = (1.0 × 103)  13000 3.14 = 10−5 m = 10µm Ans. t = 95.5 s (c) Intensity received by the person, 12. Let n1 harmonic of the chamber containing H2 is I = W = W 4 πR 2 4π (10)2 equal to n2 harmonic of the chamber containing O2. Then, Also, I = p02 ⇒ W = (24π )2 2ρv 4π (10)2 2 × 1 × 400 AN A W = 288π 3 W Ans. 0.5 m 0.5 m 15. (a) Path difference and hence phase difference at P n1  vH 2  = n2  vO 2  from both the sources is 0°, whether θ = 45° , or     θ = 60°. So, both the wave will interfere  4l   4l  constructively. Or maximum intensity will be obtained. From ∴ n1 = vO2 = 300 = 3 n2 vH2 1100 11 Imax = ( I1 + I2 )2 we have I0 = ( I + I )2 = 4I (I1 = I2 = I , say ) ∴ fmin = 3  vH2  = 3  41×1000.5 ∴ I = I0/4  4l   When one source is switched off, no interference will be obtained. Intensity will = 1650 Hz Ans. be due to a single source or I0/4. (b) At θ = 60° , also maximum intensity or I0 will 13. This problem is a Doppler’s effect analogy. be observed. (a) Here, f = 20 min−1 16. (a) Frequency of second harmonic in pipe v = 300 m/min Ans. vs = 0 and v0 = 0 A = frequency of third harmonic in pipe B Spacing between the pies = 300 = 15 m ∴ 2  vA  = 3  vB  20  2lA  4lB  and f ′ = f = 20 min−1

394 — Waves and Thermodynamics γ ARTA (b) Velocity of sound in air is or vA = 3 ⇒ MA = 3 Air speed vB 4 γBRTB 4 va = 5 m/s MB Source Observer at (rest) or γA MB = 3 (as TA = TB ) vs = 10 m/s γB MA 4 γRT = (1.4)(8.31)(20 + 273) MA γA  196  5/3  196 va = M 28.8 × 10−3 MB γB  7/5 ∴ = = = 344 m/s or MA =  2251  196 = 400 Frequency does not depend on the medium. MB 189 Therefore, frequency in air is also f0 = 105 Hz. (b) Ratio of fundamental frequency in pipe A and ∴ Frequency of sound detected by receiver (observer) in air would be in pipe B is fA = vA/2lA = vA (as lA = lB ) f2 = f0  va va − vw vs  fB vB/2lB vB  − vw −    γ ARTA = 105  344 − 5  Hz  344 − 5 −  = MA = γA .MB (as TA = TB )  10  γ BRTB γB MA f2 = 1.0304 × 105 Hz Ans. MB 18. Frequency of fundamental mode of closed pipe, Substituting MB = 189 from part (a), we get MA 400 f1 = v = 200 Hz 4l fA = 25 × 189 = 3 Ans. Decreasing the tension in the string decrease the fB 21 400 4 beat frequency. 17. Velocity of sound in water is Hence, the first overtone frequency of the string should be 208 Hz (not 192 Hz) vw = B= 2.088 × 109 = 1445 m/s ∴ 208 = 1 T ρ 103 lµ Frequency of sound in water will be ∴ T = µ ⋅ l2 (208) f0 = vw = 1445 Hz =  2.5 × 10−3 (0.25)2(208)2 λw 14.45 × 10−3   0.25  f0 = 105 Hz = 27.04 N Ans. (a) Frequency of sound detected by receiver 19. (a) Wavelength of sound ahead of the source is (observer) at rest would be λ′ = v − vs = 332 − 32 = 0.3 m f 1000 Ans. source observer(At rest) vs = 10m/s (b) f ′ = f  v + v0 = 1000  332 + 64  v − vs   332 − 32 vr = 2 m/s = 1320 Hz Ans. f1 = f0  vw + vr vs  (c) Speed of reflected wave will remain 332 m/s.  vw + vr −  Ans.   (d) Wavelength of reflected wave. = 105  1445 + −210 Hz λ′′ = v − v0 = 332 − 64  1445 + 2 f ′ 1320  f1 = 1.0069 × 105 Hz Ans. = 0.2 m Ans.

20. Thermometry, Thermal Expansion and Kinetic Theory of Gases INTRODUCTORY EXERCISE 20.1 2. Density of water will increase by increasing the temperature from 0 to 4° C, then it will decrease. 1. (a) TC − 0 = TF − 32 Fraction of volume immersed is given by 100 180 fi = ρs ρl Putting TF = 0, we get TC = − 17.8° C (b) TC − 0 = TF − 32 or TK − 273 = TF − 32 ρl is first increased the decreased first decreased 100 180 100 180 the increased. Or percentage of volume above Putting TK = 0, we get TF = − 459.67° F water level will first increased then decreased. 2. (a) TC − 0 = TF − 32 3. fi = ρs ⇒ fi = ρ′s ρl ρl′ 100 180 ∴ fi′ =  ρ′s   ρl  =  1 + γl∆T  =  1 + γ2∆T  Putting TF = 2TC, we get TC = 160° C fi  ρs   ρ′l   1 + γs ∆T   1 + γ1 ∆T  (b) Putting TF = TC , we get TC = − 24.6° C 1 2 γ1 ∆T Now, ≈ (1 − γ1 ∆T ) 3. TC = 100 1+ 52− 5 99 − 5 ∴ fi′ = (1 + γ2∆T ) (1 − γ1∆T ) fi ∴ TC = 50° C Neglecting γ1γ2 (∆T )2 term, we get Correct reading (°C) 100 fi′ fi TC Wrong reading ≈ (γ 2 − γ1) ∆T 0 5 52 99 (°C) 4. Steel Now, using the equation, TC − 0 = TF − 32 Brass 100 180 Putting TC = 50° C , we get TF = 122° F Since, α B > α Fe On cooling brass will contract more. 4. TC − 0 = TF − 32 or TK − 273 = TF − 32 100 180 100 180 5. ∆d = d (α B −α Fe ) ∆θ Putting TF = TK , ∴ ∆θ = ∆d We get, TF or TK = 574.25 d (α B − α Fe ) 5. TC − 0 = TF − 32 = 60 × 0.01 10− 5 = 20.83° C 100 180 0.8 × Putting TC = TF, we get ∴ New temperature = (30 + 20.83) = 50.83°C TC or TF = − 40° C or − 40° F 6. (a) ∆l = l α∆θ INTRODUCTORY EXERCISE 20.2 = (88.42) (2.4 × 10− 5) (35 − 5) 1. ∆T = 1 T α∆θ = 0.064 cm 2 (b) At higher temperature, it measures less. Hence, Temperature is decreased. Hence, l or T will l = l′ + ∆l decrease. So, it will gain time. = (88.42) + (0.064) = 88.484 cm Time gained = ∆T t ≈ ∆T t (as T ′ ≈ T ) 7. ∆l × 100 = − l α ∆θ × 100 T′ T ll = 1 α∆θt = 1 (1.2 × 10− 5) (30) (24 × 60 × 60) 22 = − (α∆θ) × 100 = − (1.2 × 10− 5) (35) × 100 = 15.55 s = − 0.042%

396 — Waves and Thermodynamics INTRODUCTORY EXERCISE 20.3 5. V = nRT =  nR T 1. pV = nRT =  m  RT p  p M For given mass, V - T graph is a straight line ∴ T =  VM  p passing through origin having mR slope = nR ∝ 1 At constant volume T - p graph is a straight line of pp slope, = VM ∝ 1 Slope of p2 is more. Hence, p2 < p1. mR m 6. p = (nRT ) 1 Slope of m1 is less. Hence, m1 is greater. V 2. p= nRT or p ∝ nT i.e. p versus 1 graph is a straight line passing V V through origin of slope nRT . ∴ p2 = n2T2 p1 n1T1 7. pV and T both are constants.  n2  T2 INTRODUCTORY EXERCISE 20.4  n1  T1 or p2 = p1 1. Internal energy of n moles of an ideal gas at  12  87 + 273 temperature T is given by  27 + 273 = (20 atm) U = n  f RT  2 = 12 atm where, f = degrees of freedom 7 1 Hence, 3. n1 = number of moles of nitrogen = 28 = 4 = 5 for O2 and 3 for Ar n2 = number of moles of CO2 = 11 = 1 U = U O2 + U Ar 44 4 = 2  5 RT  + 4  3 RT  2 2 M = n1M1 + n2M2 n1 + n2 = 11RT =  41 (28) +  41 (44) = 36 2. Average translational kinetic energy of an ideal  41 +  41 gas molecule is 3 kT which depends on 2 temperature only. Therefore, if temperature is ρ = RM same, translational kinetic energy of O2 and N2 RT both will be equal. Now, = (1.01× 105) (36 × 10− 3) 3. fT = 3 = KT = KT (8.31) (290) fR 2 KR KO ∴ KT = 3 K O 2 = 1.5 kg/m3 4. pV = nRT INTRODUCTORY EXERCISE 20.5 ∴ n = pV = (ρgh) V 1. vrms = 3RT = 3 × 8.31 × 373.15 RT RT M 2 × 10− 3 = (13.6 × 103) (9.8) (10− 6)(250 × 10− 6) = 2156 m/s = 2.15 km/s 8.31 × 300 2. vav = 500 + 600 + 700 + 800 + 900 = 700 m/s = 1.33 × 10− 8 5 ∴ Number of molecules = (n) N A (500)2 + (600)2 + (700)2 = (1.33 × 10−8 ) (6.02 × 1023) vrms = + (800)2 + (900)2 = 714 m/s = 8 × 1015 5

Chapter 20 Thermometry, Thermal Expansion and Kinetic Theory of Gases — 397 4. Helium gas is monoatomic. So, its degree of = f kT = 3 kT 22 freedom f = 3. Average kinetic energy of 1 molecule of gas =  23 (1.38 × 10− 23) (300) = 6.21 × 10− 21 J = f kT = 3 kT 22 6. vrms = 3RT M = 3 (1.38 × 10− 23) (300) ∴ 2 T = mv2rms = (4 × 10− 3) (103)2 3R 3 × 8.31 = 6.21 × 10−21 J 5. (a) vrms = 3RT M 3 × 8.31× 300 = 160 K 4 ×10− 3 For He gas : vrms = 7. Suppose n1 molecules have v1 velocity and n2 = 1368 m/s molecules have v2 velocity. Then, For Ne gas : vrms = 3 × 8.31 × 300 vav = n1 v1 + n2 v2 20.2 × 10− 3 n2 + n2 = 609 m/s But, vrms = n1 v12 + n2 v22 n1 + n2 (b) Each gas is monoatomic for which degree of freedom f = 3. Hence, average kinetic energy Now, vrms ≥ vav because v1 and v2 may be in of one atom opposite directions also. Exercises LEVEL 1 Objective Questions Assertion and Reason 2. vrms = (1)2 + (0)2 + (2)2 + (3)4 4 1. Straight line passing through origin represents = 3.5 m/s isochoric process. 3. In monoatomic gas, there is no rotational degree of 3. T1 = 273 + 27 = 300 K freedom. T2 = 273 + 927 = 1200 K vrms ∝ T 4. Temperature of the gas is associated with random T has become four times. or disordered motion of the gas molecules. By the Therefore, vrms will become two times. motion of container total kinetic energy of gas molecules will increase. But temperature will not 4. Hydrogen and oxygen both are diatomic gases. increase. Therefore, average kinetic energy per molecule is 5. Internal energy is equally distributed in all degrees f kT 2 of freedom. For example KT = 3 in diatomic gas or 5 kT (as f = 5) KR 2 2 As fT = 3 and fR = 2 5. E ∝ T 7. Density of water is maximum at 4°C. Hence, ∴ T2 = E2 = 2E = 2 T1 E1 E volume is minimum at 4°C. 8. U ∝ T ∝ pV 9. p= 1  mN  vr2ms ∴ T2 = 2T1 = 2 (273 + 10) 3 V = 566 K = 293° C mN = Total mass of gas = density of gas 7. If p = constant, then V ∝ T V Volume of gas

398 — Waves and Thermodynamics 8. It means rod is compressed from its natural length 6. The temperature on the platinum scale is by ∆l. t = Rt − R0 × 100° ∴ Strain = − ∆l = − l∆α∆θ R100 − R0 ll = 86 − 80 × 100°C = (− αdθ) = − (12 × 10− 6) (50) 90 − 80 = − 6 × 10− 4 = 60°C Ans. 9. ∆l1 = ∆l2 7. Let the relation between the thermometer reading ∴ l1α 1 ∆θ = l2α 2 ∆θ and centigrade be y = ax + b or l1 = α 2 = 9 Given, at x = 100, y = 80 and at x = 0, y = 10 l2 α 1 11 In option (a), ratio is 19. ∴ 80 = 100 a + b, 10 = b 11 ⇒ a = 0.7 10. V = nRT = mR (t + 273) Now, we have to find x when y = 59 p Mp ∴ 59 = 0.7x + b ⇒ x = 70 m ratio is same. Therefore, slope and intercept of ∴ The answer is 70° C. Ans. p 8. If T be the corresponding temperature, straight line between V and t should also remain same. 3RT = 3R (300) MO MH Subjective Questions ∴ T = (300)  MO   MH 1. For Q.Nos. 1 and 2 Ans. = 4800 K Apply the formula, 9. Mass of 6.02 × 1023 molecules is 17 g or 0.017 kg. TC − 0 = TK − 273 = TF − 32 = TR − 0 100 100 180 80 ∴ Mass of one NH3 molecule 3. In the above equation, putting TF = TC, = 0.017 kg 6.02 × 1023 We get TF = − 40° F or TC = − 40° C = 2.82 × 10− 26 kg 4. Initially TC =  TF18−032 (100) 10. n1 + n2 = n1 + n2 =  68.2 − 32 (100) 180 γ − 1 γ1 − 1 γ2 − 1 ∴ 3+ 2= 3 + 2 γ − 1 1.67 − 1 1.4 − 1 Finally = 20.11°C 5 = 9.48 ∴ TC = 20.11 + 40 = 60.11° C γ −1 TF = 32 +  110800 TC Solving, we get γ = 1.53 = 32 +  110800 (60.11) 11. v = γ p ρ = 140.2° F ∴ γ = ρv2 = 1.3 × (330)2 = 1.4 p 1.013 × 105 0°C 100° γ =1+ 2 ( f = degree of freedom) f 5. 1.4 = 1 + 2 TC f 0° Thermometer 20° 32° 80° TC = 100 or TC = 20° C Solving, we get 32 − 20 80 − 20 f =5

Chapter 20 Thermometry, Thermal Expansion and Kinetic Theory of Gases — 399 12. 4 g hydrogen = 2 moles hydrogen f has become 5 time (3 for He and 5 for O2). p has 3 11.2 litre He at STP = 1 mole of He 2 become 3 times and V has become 3 times. 2 p= pH + pHe = (nH + nHe ) RT V ∴ U will become 5 × 3 × 3 or 7.5 times. 32 =  2 + 12 8.31 × (300K) (20 × 10−3)m3 17. vrms = 3RT ∝ T M M = 3.12 × 105 N/m2 Ans. vrms is same. Hence, 13. Since, pressure is constant.  T   T  ∴ V ∝T M = M ∴ Vi = Vf H 2 O 2 Ti Tf or TH 2 =  M H 2  TO 2 M O 2 = Vf ∴ Tf Vi Ti =  322 (273 + 47) ⇒ Tf = 2Ti = 600K 18. (i) = 20 K = − 253°C ∴ ∆U = f nR∆T ∴ 3RT = 11.2 × 103 m/s 2 M = 3 R (600 − 300)= 450R Ans. T = (11.2 × 103)2 M 2 3R …(i) 14. From energy conservation principle, = (11.2 × 103)2 × 2 × 10− 3 3 × 8.31 Ei = Ef or E1 + E2 = E ∴ n1  3 RT1 + n2  3 RT2 = (n1 + n2 )  3 RT  = 10059 K 2 2 2 (ii) Escape velocity from moon ∴ T = n1T1 + n2T2 Ans. = 2gm Rm n1 + n2 = 1.6 × 2 × 1750 × 103 15. Number of gram moles = 1 = n = 2366 m/s Substituting in Eq. (i), we have 18 T = (2366)2 × 2 × 10− 3 Avogadro number, NA = 6.02 × 1023/g-mol 3 × 8.31 ∴ Total number of molecules, = 449 K N = nN A = 3.34 × 1022 19. At constant volume, ∴ Number of molecules per cm 2 = N (R → in cm) p∝T 4 πR 2 ∴ p1 = p2 = 3.34 × 1022 = 6500 T1 T2 (4π ) (6400 × 105)2  p2  16. pV = nRT ∴ T2 =  p1  T1 ∴ nRT = pV =  18600 (273) = 546 K U = f (nRT ) ( f = degree of freedom) 2 20. t =  Rt − T0  × 100 =  6.5 − 2.5 × 100  R100 − R0  3.5 − 2.5 ∴ U = f (pV ) 2 = 400° or U ∝ fpV

400 — Waves and Thermodynamics 21. At constant volume, 26. n1 = n2 p∝T ∴ p1V1 = p2V2 RT1 RT2 ∴ p2 = T2 or T2 =  p2 T1 p1 T1  p1  p1  T2   01.5  237000  45 + 75 ∴ V2 =  p2  T1  V1 = (500)  5+ 75  = (273 + 30) = 900 m3 = 450 K = 450 − 273 27. p = constant = 177°C ∴ V ∝T 22. Strain = ∆l = (α∆θ) or h ∝ T ∴ hf = hi l Stress = Y × strain = γα∆θ Tf Ti ∴ Tension = (A) (stress) = YAα∆θ or hf =  Tf  hi Substituting the value, we get  Ti  T = YAα∆θ  227733 ++ 12000 = (2.0 × 1011) (2 × 10− 6) (1.2 × 10− 5) (40)  = 192 N = (4) = 5.09 cm 23. Change in weight = upthrust on 100% volume 0.025 10− 4 28 ∴ ∆W ′ = F′ =  1 + γs ∆θ 28. n1 = = 8.93 × ∆W F  1 + γl ∆θ n2 = 0.04 = 0.01 or (50 − 45.1) = 1 + (12 × 10− 6) (75) 4 (50 − 45) 1 + γl (75) ∴ n1 = 0.089 n2 Solving this equation, we get γl = 3.1 × 10− 4 per° C In equilibrium p and T are same. 24. (a) n = pV = (1.52 × 106) (10−2) ∴ pV = nRT RT (8.31) (298.15) ∴ V ∝ n or L ∝ n = 6.135 ∴ L1 = n1 = 0.089 (b) ρ = pM = (1.52 × 106) (2 × 10−3) L2 n2 RT 8.31 × 298.15 29. n = n1 + n2 = 1.24 kg/m3 pV = p1V1 + p2V2 RT RT RT (c) ρ ∝ M (if P and T are same) ∴ p = p1V1 + p2V2 V ∴ ρO2 = MO2 ρH2 MH2 = (0.11) (1.38) + (0.16) (0.69) 0.11 + 0.16  MO  ∴ ρO2 = MH 2 ρH2 = 0.97 MPa 2 30. (n1 + n2 )1 = (n1 + n2) f =  322 (1.23) ∴ piV1 + piV2 = pV1 + pV2 = 19.68 kg/m3 RTi RTi RT1 RT2 25. If T = constant, then ∴ P = P (V1 + V2) Ti (V1 / T1 + V2 /T2) p1V1 = p2V2 = (1 atm) (600) ∴ p2 = p1  V1 = (1 atm) =  760 (293)  400 + 220703  V2 373 = 11.7 atm = 1.13 atm

Chapter 20 Thermometry, Thermal Expansion and Kinetic Theory of Gases — 401 31. V = nRT = (1) (8.31) (273.15) 36. (a) Average translational kinetic energy of any type p 1.013 × 105 of molecules is = 0.0224 m3 = 22.4 L 3 kT = 3 (1.38 × 10− 23) (300) 22 32. n1 = n2 = 6.21 × 10− 21J ∴ p1V1 = p2V2 or p2 =  V1  T2 p1 (b) 2 energy is associated with rotation. Hence, it RT1 RT2  V2  T1 5 is diatomic gas. =  00..4785  273 + 157 (1.5 × 105)  273 + 27  (1)  5 R Q = nCV ∆T = 2 (1) = 3.36 ×105 Pa = 5 × 8.31 = 20.8 J 3 33. n1= number of moles of N2 = 1.4 = 0.05 28 n1C p1 + n2C p2 30% of the gas dissociates into atoms. 37. Cp = n1 + n2 ∴ n1′ = 2 (0.3) (0.05) + (0.7) (0.05) = 0.065 = 2.5R + 3.5R = 3R 1+ 1 n2 = number of moles of He = 0.4 = 0.1 4 n1CV1 + ∴ n = n1′ + n2 = 0.165 CV = n1 + n2 CV2 n2 Now, P= nRT = (0.165) (8.31) (1500) V (5 × 10− 3) = 1.5R + 2.5R = 2R 2 = 4.1 × 105 N/m2 γ = Cp = 3 = 1.5 34. In diatomic gas rotational degree of freedom is CV 2 two. ∴ Rotational KE of one molecule 38. pV − b = constant 1 I ω2 = f kT = kT 22 (as f = 2) C= R + R γ − 1 1 − (−b) ∴ ω = 2kT C = 0 , if γ −1 = − (1 + b) I ∴ b=−γ = 2 × 1.38 × 10− 23 × 300 8.28 × 10− 38 × 10− 9 39. pV − 1 = constant = 1013 rad /s ∴ C = CV + R 1− x 35. (a) Cp = (1 + f /2) R Here x = − 1 29 = (1 + f /2) 8.31 ∴ C = CV + R Solving, we get 2 f =5 40. K = 3 kT ⇒ T = 2K (b) pT = constant 2 3k ∴ p (pV ) = constant or pV 1/ 2 = constant Now, n = pV = 3kpV RT 2KR C in the process pV x = constant is given by = 3 (200 × 103) (0.1 × 10− 3) (1.38 × 10− 23 ) 2 × 6 × 10− 21 × 8.31 R f R+ R C = CV + 1− x = 2 1− x = 8.3 × 10− 3 mol ∴ 29 = 8.31  f + 1 1  Now, N = nN A  2 − 1/ 2 = (8.3 × 10− 3) (6.02 × 1023) Solving, we get f = 3 = 5 × 1021

402 — Waves and Thermodynamics 41. vrms = 3RT = 3 × 8.31 × (273 + 57) (b) KT in one mole = 3 RT M 46 × 10− 3 2 = 423 m/s = 3 × 8.31 × 300 2 vrms = 3740 J vrms (c) vrms = 3RT = 3 × 8.31 × 300 M 32 × 10− 3 ∆p = 2 mvrms = 484 m/s Mass of one gas molecule, 45. Let mass of nitrogen = (m) g. Then, mass of 46 oxygen = (100 − m) g. Number of moles of 6.02 × 1023 m = g nitrogen, n1 = m and number of moles of oxygen 28 = 7.64 × 10−26 kg n2 =  100 − m 32 Let n molecules strike per second per unit area. Then, For air Pressure = F = ∆P / ∆t = 2 mn vrms ρ = pM = p  m1 + m2 A A RT RT  n1 + n2  ∴ n = Pressure = 2 × 1.013 × 105 ∴ 1.284 = 1.01 × 105  100 × 10− 3  (2m) vrms × 7.64 × 10− 26 × 423 8.31 × 273  32  2  (m / 28) + (100 − m) / = 3.1 × 1027 Solving this equation, we get 42. KT = 3 kT = 3 k  pV  m = 76.5 g 2 2 nR This is also percentage of N2 by mass on air as =3k pV total mass we have taken is 100 g. 2 (M /mN A) R 46. n1 = n2 = 3kPV m N A ∴ p1V1 = p2V2 2MR RT1 RT2 (3 × 1.38 × 10− 23) (100 × 103) ∴ V2 = p1 V1 T2 = ( p0 + ρgh) V1 T2 p2 T1 p0 T1 = (2 × 10− 6) (8.0 ×10− 23) (6.02 × 1023 ) 2 × (50 × 10− 3) (8.31) = (1.01 × 105 + 103 × 10 × 40) (20) (273 + 20) 1.01 × 105 (273 + 4) = 4.8 × 10−22J ≈ 105 cm3 43. (a) vrms ∝ T Q  T ′   573  47. (a) Molar heat capacity Cp = n∆T  T  293  ∴ v′0 = v0 = v0 = 1.40 v0 and heat capacity Cp = Q ∆T (b) vrms does not depend on pressure, as long as temperature remains constant. ∴ Cp = nCP Similarly, Cv = nCV (c) vrms = 3RT or vrms ∝ 1 Now, Cp − Cv = n (CP − CV ) − nR M M ∴ n = Cp − Cv = 29.1 = 3.5 44. (a) KT = 3 kT R 8.31 2 (b) Cp = nC P = n  5 R = 3 (1.38 × 10− 23) (300) 2 2 = (3.5) (2.5) (8.31) = 6.21 × 10− 21J = 72.75 J/ K

Chapter 20 Thermometry, Thermal Expansion and Kinetic Theory of Gases — 403 Cv = nCV = n  3 R 3. T = mV + C 2 ∴ pV = mV + C = (3.5) (1.5) (8.31) nR = 43.65 J/K or p = nR  m + C  V (c) Cp = nC P = n  7 R 2 ∴ p -V graph is as shown in graph (a). = (3.5) (3.5) (8.31) 4. p2V = constant = 101.8 J/ K  nRT  2 V Cv = nCV = n  5 R ∴ V = constant 2 ∴ T2 ∝V = (3.5) (2.5) (8.31) = 72.75 J/K or T ∝ V 48. (a) As discussed in the above problem, V is made three times. So, T will become 3 Cv = nCV times. ∴ n = Cv = Cv 5. n = pV n ∝ 1 or m ∝ 1 CV (3/2)R TT = 35 = 2.81 RT ∴ 1.5 × 8.31 m1 = T2 = 273 + 37 (b) Internal energy, m2 T1 273 + 7 U = n  f RT  = 310 = 1.1 2 280 f = degree of freedom = 3 6. vav = 8RT ∴ U = (2.81)  23 (8.31) (273) πM = 9562 J = 9.56 kJ T and M are same. Hence, average speed of O2 molecules in both vessels will remain same or v1. (c) CP = 5R=  25 (8.31) 2 7. n = pV RT = 20.8 J/mol-K ∴ N = nN A = pVN A RT LEVEL 2 = (1.013 × 105 × 10− 13) (10− 6) (6.02 × 1023) (8.31) (300) Single Correct Option = 2.4 × 106 1. (n1 + n2)i = (n1 + n2)f 8. vrms = 3RT = 3 × 8.31 × 273 M 28 × 10− 3 ∴ p1V1 + p2V2 = p (V1 + V2) RT1 RT2 RT = 493 m/s T = T1T2p (V1 + V2) Solving, we get p1V1T2 + p2V2T1 Now, increase in gravitation PE = decrease in kinetic energy 2. ∆l = lα∆θ ∴ (mg h) = 1 m vr2ms or 1 2gh = vr2ms 1+ h 2 + (h/R) ∴ α = ∆l = 0.008 = 8 × 10− 6 per °C l ∆θ (10) (100) R γ = 3α = 2.4 × 10− 5 per° C or 1+ 2× 9.81 × h = (493)2 (h / 6400 × 103) ∆V = Vγ ∆θ = (100) (2.4 × 10− 5) (100) Solving this equation, we get h ≈ 12000 m = 0.24 cc ∴ V ′ = V + ∆V = 100.24 cc or 12 km

404 — Waves and Thermodynamics 9. U ∝ T 2. p ∝ 1 According to given graph, V p∝V If volume becomes 4 times than p will remain half ∴ p∝T V ∝T Hence, V = constant p 10. 28 g of N2 mean 1 g-mol ∴ Vp2 = constant Now, n = pV or n ∝ p ∴  TP p2 = constant RT T nf =  pf   Ti  =  21  273 + 57 = 11 or pT = constant ni  pi  Tf  273 + 27 20 or p ∝ 1 or nf =  2101 ni = 11 mol T 20 i.e. p - T graph is a rectangular hyperbola. ∆n = ni − nf =1− 11 = 9 g - mol 20 20 If p is halved than T will becomes two times. ∆m = 9 × 28 g = 63 g 4. ρ ∝ 1 20 5 V 11. 4 g H2 means 2 g-moles and 8 g He means V is doubled so ρ will remain half. 2 g-moles. U ∝ T ∝ pV According to given graph, p ∝ V ∴ U ∝ T ∝ p2 or V 2 Now, M = n1M1 + n2 M2 = (2) (2) + (2) (4) V is doubled, soU and T both will become four n1 + n2 2+ 2 times. = 3 g/mol p∝V ρ = PM = (1.013 × 105) (3 × 10− 3) = 0.13 kg/m3 ∴ T ∝V  as p ∝ T  RT (8.31) (273) V V ∴ T ∝V2 12. ρ = 1 p ∝ Kρ or T -V graph is a parabola passing through origin. ρ ∝1 V 6. V = nRT = mRT =  mR T Given, V p Mp  Mp ∴ pV = constant T = constant i.e. V - T graph is a straight line passing through or ∴ origin of slope =  mR ⇒ Slope ∝ m  Mp p More than One Correct Options Hence, slope depends on both m and p. 1. p ∝ ρ as per process. If ρ becomes half, then p 7. (a) p = nRT = (m/V )RT will becomes 2 times. VM Further, ρ∝ 1 So, given m in the question is mass of gas per V unit volume. (b) m = total mass of gas ∴ p2 = constant (c) m = mass of one molecule of gas. (1/V ) or (pV ) p = constant Match the Columns Hence, Tp = constant p∝ 1 (as pV ∝ T ) 1. (a) KT = n fT RT = (2) (3) RT = 3RT ∴ T 2 2 or P - T graph is a rectangular hyperbola. (b) KR = n fRRT = (2)(2) RT = 2RT 2 2

Chapter 20 Thermometry, Thermal Expansion and Kinetic Theory of Gases — 405 (c) Potential energy = 0 i.e. T - V graph is a straight line passing through origin. (d) U = KR + KT = 5RT (d) ρ = pM 2. U ∝ T and ρ ∝ 1 RT If p = constant, then ρ ∝ 1 V T According to given graph, i.e. ρ -T graph is rectangular hyperbola. U ∝ρ ⇒ T ∝ 1 V Subjective Questions ∴ pV ∝ 1 or p ∝ 1 1. pV = nRT V V2 Density is increasing. So, volume is decreasing. pdV = nRdT Hence, pressure will increase. dV = nR , dV = VγdT dT P U is increasing. So, temperature will also increase. ∴ γ = dV /dT = nR = 1 Now, T ∝ 1 V PV T V V2 Volume is decreasing. Hence, T will increase. 2. In the process pV x = constant C = R + R , V γ −1 1− x 3. Speed of sound = γ RT Given, C = R and γ = 7/5 M Substituting these values, we get x = 5 3 4. (a) Density of water is maximum at 4°C. From 0°C Now, pV 5/ 3 = constant or p ∝ 1 to 4°C, density will first increase, then decrease. (V )5/ 3 (b) and (c) : Same logic as given in option (a). (d) T = 2π l or T ∝ l By increasing volume to two times pressure will g decrease (2)5/ 3 times. As temperature is increased, length will vrms ∝ T or vrms ∝ pV increase. So, time period will also increase. or vrms will become (2) times 5. (a) p = constant. Hence p - T graph is as shown (2)5/ 3 below. or vrms will become (2)−1/ 3 times p or 1 3 times. (2)1/ Now, p ∝ (no. of collisions) (vrms) T 1 ∝ (no. of collisions) 1 (2)5/ 3 (2)1/ 3 (b) ρ = pM RT or number of collisions will decrease (2)4/ 3 times. If p = constant, then Ans. ρ ∝ 1 or ρ ∝ 1 TU 3. p1V = n1RT U (as U ∝ T ) p2V = n2RT (T = 273 K) ∴ ( p1 − p2)V = (n1 − n2)RT =  m1 − m2  RT M (∆p)V = ∆m RT …(i) M In the initial condition (at STP) P RT = p0 …(ii) Mρ (c) If p = constant, then T ∝V (T = 273 K)

406 — Waves and Thermodynamics From Eqs. (i) and (ii), we get ∴ ρ0 = ρ50(1 + 50γ) ∆m = ρV∆p = 1.2 × 0.4 × 0.24 Substituting in Eq. (i), we have P0 1.0 = 0.1152 kg = 115.2 g % of fraction submerged 4. p2 = pressure at depth x Ans. =  1 + 50γ s  × 100 =  1 + 0.3 × 10−5  × 100 1+ 50γ l   1 + 8 × 10−5  = 99.99 % Ans. p p 7. ρ100gh100 = ρ0gh0 or ρ100 = h0 ρ0 h100 1 2 x x ∆h  1100γ  39.2  40 1 + = p1V1 = p2V2 Solving this equation we get γ = 2.0 × 10−4 per °C or p1(Ah) = p2(A)(h − ∆h) Ans. ∴ p1 = p2(h − ∆h) …(i) 8. Let ∆l be the change in length. (let ∆ls > ∆l > ∆la) h Ans. Strain in steel = ∆ls − ∆l Initially, mg = ρgxA …(i) l0 Now, p2 = p0 + ρgx = p0 + mg and strain in aluminium = ∆l − ∆la A l0 Substituting in Eq. (i), we have p1 =  p0 + mAg 1 − ∆hh In equilibrium, 2Fs = Fa 5. p = RT 2  ∆ls − ∆l  Ys A =  ∆l − ∆la  Ya A  l0   l0  V (n = 1) or p = R (T0 + αV 2) or 2(l0α sθ − ∆l)Ys = (∆l − l0α aθ)Ya V Solving this equation, we get For minimum attainable pressure ∆l =  2α sYs + α aYa  l0θ  2Ys + Ya  dp = 0 or − RT0 + αR = 0 dV V2 ∴ Total length or V = T0 α = l0 + ∆l = l0   α aYa + 2α sYs  θ  Ans. 1 +  2Ys + Ya   d2p   At this volume we can see that dV 2 is positive or p 9. From ∆l = lα∆θ we have, is minimum. 0.05 = 25α A(100) From Eq. (i) ∴ α A = 0.00002 per° C 0.04 = 40 α B (100) pmin = RT0 + αR T0/α α B = 0.00001 per° C T0/α In third case let l is the length of rod A. Then = 2R α T0 Ans. length of rod B will be (50 − l) cm. 6. At 50°C, density of solid = density of liquid. ∆l = ∆l1 + ∆l2 or 0.03 = l(0.00002) (50) + (50 − l) (0.00001) (50) At 0°C, fraction submerged  ρs  × 100 …(i) Solving we get, l = 10 cm and 50 − l = 40 cm  ρl  0°C Ans. (ρ)50 = 1 ρ0 + 50γ

21. Laws of Thermodynamics INTRODUCTORY EXERCISE 21.1 6. VT = constant 1. ∆U will be same along all paths. Then, we can or V (pV ) = constant (as T ∝ pV ) apply Q = W + ∆U for all. ∴ pV 2 = constant 2. ∆U = Q − W Comparing with pV x = constant, we have = 254 − (−73) = 327 J x=2 INTRODUCTORY EXERCISE 21.2 W = nR∆T (Table 21.3) 1− x 1. (a) Work done by gas under constant pressure is = (2)(R)(300) 1− 2 W = p(V f + Vi ) = (1.7 × 105)(0.8 − 1.2) = − 600 R 7. pV 2 = K = p0V02 = − 6.8 × 104 J ∴ p = K V2 (b) Q = W + ∆U 2V0 K = − 6.8 × 104 − 1.1 × 105 ∫ ∫W = = V0 V 2 dV pdV or |Q | = 1.78 × 105J − K 2V0 V V0 As Q is negative. So, net heat flow is out of = the gas. =K − K 2. (a) W1 = + ve, as cycle is clockwise V0 2V0 W2 = − ve as cycle is anti-clockwise =K Since, |W1| > |W2| 2V0 ∴ Net work done by the system is positive. Substituting, K = p0V02, we have (b) In cyclic process, Qnet = Wnet W = 1 Since, Wnet is positive. So, Qnet is also 2 p0V0 positive. So, heat flows into the system. (c) W1 = Q1 = + ve , so into the system. INTRODUCTORY EXERCISE 21.3 W2 = Q2 = − ve , so out of the system. 1. (a) Q = nCV ∆T 3. Helium is monoatomic. So, degree of freedom 200 = (1)  3 R (T f − 300) f = 3. 2 U = nf RT = 3n RT 22 or 200 = 1.5 × 8.31(Tf − 300) Solving we get, or n = 2U = 2 × 100 3RT 3 × 8.31 × 300 Tf = 316 K (b) Q = nCp ∆T = 2.67 × 10−2mol Ans. 4. In isochoric process, W =0 200 = (1)  5 R (Tf − 300) 2 Q = ∆U = nCV ∆T = 4  3 R (300) 2 or 200 = 2.5 × 8.31 (Tf − 300) Solving we get, = 1800 R Tf = 309.6 K 5. W = Area under p-V diagram 2. p = constant Further p ∝ V along AB. Therefore, pB = 2pA = 2p0 ∴ T ∝V

408 — Waves and Thermodynamics As the gas expands V increases. So, T also U is a state function. Hence, ∆U depends only on increases. Hence, ∆T is positive. Therefore, in the the initial and final positions. Therefore, expression, ∆U 1 = ∆U 2 Q = nCp ∆T But, W1 > W2 Q is positive. as the area under 1 is greater than area under 2. 3. AB Hence, V is increasing, so W = + ve . Product of pV and Q1 > Q2 therefore T and therefore U is increasing. ∫ ∫7. W = Vf pdV = 2 (αV 2) dV So, ∆U is also + ve. Vi 1 Q = W + ∆U ∫= 2 × 1.01 × 105) V 2dV ∴ Q is also positive. BC V = constant (5 1 ∴ W =0 Solving we get, ∴ Q = ∆U = − ve W = 1.18 × 106 J CA = 1.18 MJ 8. W = nR ∆T (in isobaric process) V is decreasing. = (1)(8.31)(72) ∴ W = − ve = 600 J = 0.6 kJ ∆U = Q − W = 1.0 kJ ∆U is also negative. γ = Cp = n Cp ∆T = Q ∴ Q = W + ∆U is also negative. CV n CV ∆T ∆U 4. Q = 0 as the box is insulated. = 1.6 = 1.6 Wgas = 0 as on the right hand side there is vacuum. 1.0 ∴ ∆U = Q − W = 0 ∴ ∆T is also zero, INTRODUCTORY EXERCISE 21.4 as U ∝ T 1. Given, Q1 = 106 cal 5. In an isobaric process p = constant. and Therefore, C = Cp . as, T1 = (827 + 273) = 1100K Now, ∆Q = nCp ∆T = Cp = γ ∴ T2 = (27 + 273) = 300 K ∆U nCV∆T CV Q2 = T2 and ∆Q = ∆Q Q1 T1 (in Carnot engine) ∆W ∆Q – ∆U T2  1310000 = nCp ∆T Q2 = T1 ⋅ Q1 = (106) nCp ∆T – nCV∆T = 2.72 × 105 cal Ans. = Cp = Cp /CV Efficiency of the cycle, Cp – CV Cp /CV – 1 η=  − T2  × 100 =γ 1 T1  γ –1 or η = 1 − 1310000 × 100 6. Q1 = W1 + ∆U 1 and Q2 = W2 + ∆U 2 pp = 72.72% Ans. 1 2. T1 = 2100 K, T2 = 700 K, ηactual = 40% AB A2 B Maximum possible efficiency of Carnot engine is W1 W2 V ηmax = 1− T2 = 1− 700 V T1 2100 = 0.666 = 66.6%

Chapter 21 Laws of Thermodynamics — 409 Actual efficiency as the percentage of the βB = (−27 + 273) = 5.6 273) − (−27 + maximum possible efficiency is (17 + 273) ηactual × 100 = 40 × 100 βA > βB ηmax 66.6 Therefore, refrigerator A is better. = 60% Ans. 7. Here, T1 = 25 + 273 = 298 K 3. (a) η = 1 − T2 = 1− 375 = 0.25 = 25% T2 = − 10 + 273 = 263 K Q2 = 263 J/s T1 500 Coefficient of performance is given by (b) W = ηQ1 = 0.25 × 25 × 105 β = Q2 = T2 = 6.25 × 105 J Q1 − Q2 T1 − T2 (c) Q2 = Q1 − W = (25 − 6.25) × 105 ∴ Q1 = T1 = 18.75 × 105J Q2 T2 4. T1 = 627 + 273 = 900 K T1 298 T2 263 T2 = 27 + 273 = 300 K ∴ Q1 = × Q2 = × 263 Q1 = 3 × 106 cal = 298 Js−1  T2  1 − Q2  η= 1 − T1  = Q1  Average power consumed, = 298 − 263 = 35 J s−1 = 35 W ∴ Q2 = T2 Q1 T1 8. η = Wnet × 100 T2 ΣQ+ ve T1 ∴ Q2 = × Q1 Wnet = Area under the cycle = p0V0 = 3 × 106 × 300 = 106 cal 2 900 ΣQ+ve = QABC Work done by the engine, = WABC + ∆U ABC W = Q1 − Q2 = 3 × 106 − 106 cal = 2 × 106 cal = (Area under ABC) + nCV ∆T =  5 p0V0 + n  3 R (TC − TA) 4 2 5. T2 1 = 1− T2 η1 = 1− T1 or 6 T1 ...(i) p In the second case, the temperature of the sink is 3p0 B 2 C reduced by 65°C. Hence, p0 A p0 D η2 = 1− T2 − 65 2 T1 or 1 = 1− T2 − 65 ...(ii) 3 T1 V Solving Eqs. (i) and (ii), we get V0 V0 3V 0 22 T1 = 390 K = 117° C, T2 = 325 K = 5 p0V0 + 3 ( pC VC − pAVA ) 4 2 = 52°C 6. Coefficient of performance is given by = 5 p0V0 + 3  3 p0V0 − p0V0  4 2 2 2 β = T2 T1 − T2 = 11 p0V0 4 βA = (−10 + 273) (27 + 273) − (−10 + 273) ∴ η = (p0V0/2) × 100 = 18.18 % (11p0V0 /4) = 7.1

Exercises LEVEL 1 2. Wgas = nRT  Vf  = nRT ln  12  Vi  ln Assertion and Reason = − nRT ln 2 1. Q = 0, ∆U = − W ∴ Work done on the gas = + nRT ln 2 In expansion, W is positive. So, ∆U is negative. 3. ∆U is same p. Hence, T also decreases. 2. p p Q = W + ∆U V V W1 = + ve,W2 = 0 (i) (ii) and W3 = − ve 4. In adiabatic compression temperature increases. Hence, pV increases, as T ∝ pV 5. W = − π (Radius)2 In both figures, Vi = V f = − π (Radius) (Radius) but W1 = Hatched area = − π  p2 − p1  V2 − V1  W2 = 0 2 2 3. First law is just energy conservation law, which 6. p can be applied for any system. 1 2 5. Q − W = ∆U is state function. So, it will remain constant for any two 3 thermodynamic states. V 6. For given temperatures T1 and T2 efficiency of a 1 → Isobaric heat engine can’t be greater than efficiency of Carnot engine. 7. pT = constant 2 → Isothermal  T  3 → Adiabatic V ∴ T = constant Minimum area is under graph-3. or V ∝ T 2 7. W1 > W2 as area under graph-a is more. Hence, If temperature is increased, then volume will also ∆Q1 > ∆Q2, as increase. So, work done is positive. ∆Q = W + ∆U and ∆U 1 = ∆U 2 8. Q = 0 as chamber is adiabatic 8. η =  − T2  × 100 1 T1  W = 0 as expansion is taking place in vacuum. ∴ ∆U = Q − W = 0 T2 = 300 K and T1 = 600 K ∴ U or T = constant ∴ η = 50% or pV = constant ∴ p∝ 1 9. p = n RT V V V = constant, but temperature will decrease with time. So, p will decrease. Objective Questions 10. In adiabatic compression, internal energy of gas 1. p ∝ V 2 increases. So, temperature increases. ∴ T ∝ V or T ∝ V 3 ⇒ V ∝ T 1/ 3 V γγ T is increasing. So, V will also increase. Hence, 11. pT 1 − γ = constant or p ∝ T γ − 1 work done will be positive. ∴ α = γ = 1.4 = 3.5 γ − 1 1.4 − 1

Chapter 21 Laws of Thermodynamics — 411 12. C = CV +R 4. Qnet = Wnet = area under the cycle 1− x = (2 × 105)(2.5 × 10−6) =5R+ R =5R− R 2 1− x 2 x −1 = 0.5 J C is negative if 5. (a) WI = + ve (as cycle is clockwise) x −1< 2 5 WII = − ve (as cycle is anti-clockwise) But, ∴ x < 1.4 |WI | > |WII | If x lies between 1 and 1.4, then also C is negative. ∴ Wnet = + ve (c) Qnet = Wnet = + ve 13. CV = n1CV1 + n2CV2 (d) In loop-I, n1 + n2 QI = WI = + ve In option (c), In loop-II, (2)  3 R + (4 )  5 R QII = WII = − ve 2 6 2 CV = 6. Work done per cycle = area under the cycle = 13R = 1 (3 − 2)(10−3)(30 − 10)(1.01 × 105) 6 2 14. Qnet = Wnet = area under the graph = 1010 J = π (Radius)(Radius) Total work done per second = π (10 × 103 Pa)(10 × 10−3 m3) = 1010 × 100 60 = 102π J = 1683 J/s 15. Area is same, signs are opposite. 7. (a) Q = nCV ∆T = (1)  3 R (T f − Ti ) 2 Subjective Questions ∴ 200 = 1.5 × 8.31(Tf − 300) 1. Helium is monoatomic. So, degree of freedom Solving we get, f = 3. Tf = 316 K U = nf RT = 3n RT 22 (b) Q = nC p ∆T = (1)  5 R (T f − Ti ) 2 or n = 2U 3RT ∴ 200 = 2.5 × 8.31 (Tf − 300) ∴ Tf ≈ 310 K = 2 × 100 3 × 8.31 × 300 8. V = constant = 2.67 × 10−2 mol From pV = nRT , 2. For all process, V∆p = nR(∆T ) ∆U = nCV∆T Q = nCV ∆T = nR∆T γ −1 (as CV = R) = n  5 R ∆T = 5 (V∆p) γ −1 2 2 3. (a) In a cyclic process, = 5 × (10 × 10−3)(4 × 105) 2 Qnet = Wnet cyclic is clockwise. Hence, work done is = 104 J Ans. positive. So, Qnet is also positive. (b) Wnet = Qnet = 7200 J 9. (a) W = + (Area of the cycle) (c) In counter-clockwise direction, = + 1 (2V0 − V0 )(3 p0 − p0 ) 2 Wnet = Qnet = − ve |Qnet | = 7200 J = p0V0 (b) QCA = nCp ∆T

412 — Waves and Thermodynamics = n  5 R (TA − TC ) Therefore, temperature also becomes three times. 2 (Tf = 3Ti ) = 5 ( pAVA − pCVC ) Second process 2 p = constant = 5 ( p0V0 − 2 p0V0 ) = − 5 p0V0 ∴ V ∝T 2 2 Volume is doubled. So, temperature also becomes QAB = n CV ∆T two times. (Tf ′ = 6Ti ) Now, C = Q = Q1 + Q2 = n  3 R (TB − TA) 2 n ∆T n ∆T = 3 − pAVA ) = nCV ∆T1 + n Cp ∆T2 2 (pBVB n ∆T = 3p0V0  5 R (3Ti − Ti ) +  7 R (6Ti − 3Ti ) 2 2 (c) WBC = + (Area under the graph) = = 2p0V0 (6Ti − Ti ) ∆U BC = n CV ∆T = 3.1 R = n  3 R (TC − TB ) 11. First Process 2 V = constant = 3 ( pCVC − pBVB ) = − 3 p0V0 ∴ T∝p 2 2 Pressure becomes half. So, temperature also ∴ QBC = WBC + ∆U BC = p0V0 becomes half. 2 Q1 = n1CV∆T = 2CV (150 − 300) (d) p - V equation along path BC is = − 300 CV p= −  2 p0  V + 5 p0 Second Process  V0  p = constant or pV = −  2 p0  V2 + 5 p0V Q2 = nCp ∆T  V0  = 2 Cp (300 − 150) = 300 Cp or RT = −  2 p0  V 2 + 5 p0V  V0  Q = Q1 + Q2 = 300(Cp − CV ) = 300 R ∴ T = 1  − 2 p0 ⋅ V 2  …(i) = 300 × 8.31 = 2493 J R 5 p0V V0   3 R For T to be maximum, 12. ∆U = Q − W = nCV ∆T = n 2 (T f − Ti ) dT = 0 ∴ 1200 − 2100 = 5  3 × 8.31 (Tf − 400) dV 2 ∴ 5 p0 − 4 p0 V =0 ∴ Tf = 385 K V0 ∴ V = 5 ≈ 113 °C 4 V0 13. Vi = m = 2 = 2.0002 × 10−3 m3 ρi 999.9 So at this volume, temperature is maximum. Substituting this value of V in Eq. (i), we get Vf =m = 2 = 2 × 10−3m3 ρf 1000 Tmax = 25 p0V0 8R ∆U = Q − W 10. First process = ms ∆θ − pa∆V V = constant = (2)(4200)(4) − (10)5(−0.002 × 10−3) ∴ p∝T Pressure becomes three times. = 33600.2 J

Chapter 21 Laws of Thermodynamics — 413 14. Vi = m = 0.01 = 10−5m3 18. (a) WAC is less than WABC as area under graph is ρi 1000 less. Vf =m = 0.01 = 0.0167 m3 ρf 0.6 p ∆U = Q − W 8 Pa B C = ms∆θ + mL − p0∆V = (0.01 × 4200 × 100) + (0.01)(2.5 × 10)6 4 Pa − 105(0.0167 − 10−5) A = 27530 J 5 m3 15 m3 V 15. (a) W = p0∆V (b) For process A to C, = (1.013 × 105)(1671 − 1) × 10−6 Q = 200 J = 169 J Work done (b) Q = mL = (10−3)(2.256 × 106) WAC = area under AC = 2256 J = 1 (8 + 4) × 10 ∴ ∆U = Q − W 2 = 2087 J = 60 J 16. (a) W = p∆V = (2.30 × 105)(1.2 − 1.7) From first law of thermodynamics, = − 1.15 × 105 J ∆U = Q − WAC UC − U A = 200 − 60 (b) Q = W + ∆U ∴ UC = U A + 140 = (−1.15 × 105) + (−1.40 × 105) = 10 + 140 = 150 J Ans. = − 2.55 × 105J (c) From A to B, 17. (a) and (b) (pV )A = (pV )C = 3p0V0 U B = 20 J ∴ ∆U = Q − WAB ∴ TA = TC or ∆T = 0 UB −UA = Q − 0 ∴ ∆U ABC = 0 20 − 10 = Q QABC = WABC = Area under the graph ∴ Q = 10 J Ans. = − 2p0V0 Ans. 19. (a)UC − U A along two paths should be same. i.e. Heat is released during the process. ∴ (QAB + QBC ) − (WAB + WBC ) = QAC − WAC ∴ (250 + QBC ) − (500 + 0) = 300 − 700 (c) Process AB Solving, we get ρ - p graph will be a straight line parallel to QBC = − 150 J ρ - axis because pressure is constant. Further, V (b) QCDA = (WCD + WDA) + (U A − UC ) is decreasing. Therefore, ρ will increase. Process BC V = constant ⇒ ρ = constant = (−800 + 0) − (UC − U A) = − 800 − (300 − 700) Therefore, ρ- p graph is a straight line parallel to p - axis. = − 400 J ρ - p graph is as shown below. 20. V = m = 1.0 ρ 8.92 × 103 ρ 2ρ0 B C = 1.12 × 10−4m3 γ = 3α = 2.1 × 10−5per ° C ρ0 A ∆V = Vγ ∆θ p = (1.12 × 10−4)(2.1 × 10−5)(30) p0 2p0 = 7.056 × 10−8 m3

414 — Waves and Thermodynamics W = p∆V 23. Along the process CD apply, = (1.01 × 105)(7.056 × 10−8 ) pV = constant and find VC = 7.13 × 10−3J Similarly along path AB, again apply, Q = mc ∆Q pV = constant and find VB. = (1)(387)(30) Now, Wnet = WAB + WBC + WCD + WDA ∆U = Q − W = nRTA ln  VB  + pB (VC − VB)  VA  = 11609.99287 J + nRTC ln  VD  + pD (VA − VD )  VC  21. (a) T = pV = (2 × 105)(1.0 × 10−2) Further, nRTA = pAVA nR (1)(8.31) and nRTC = pCVC = 240.7 K 24. pV = nRT (b) In adiabatic process, ∴ p∆V = nR∆T TV γ − 1 = constant Work done under constant pressure is W = p∆V = nR∆T ∴ Tf V γ − 1 = TiViγ − 1 = (0.2)(8.31)(100) = 166.2 J f   γ − 1   ∴ Tf = Ti Vi Vf 1.0 × 10−2 5/ 3−1 LEVEL 2 × 10−3  = (240.7)  5.0   Single Correct Option ≈ 383 K ∫1. W = 2V pdV V (c) Work done on the gas = ∆U 2V  n RT  = nCV ∆T ∫= V  V − b dV = (1)  3 R (T f − Ti ) = nRT ln  2VV−−bb 2  = 1.5 × 8.31(383 − 240.7) Put n = 1, ∴ ≈ 1770 J W = RT ln  2VV−−bb  (d) ∆U ≈ 1770 J 2. For AB V = constant 22. (a) From conservation of linear momentum, 0.01 × 200 = (2 + 0.01) v ∴ W =0 For BC ∴ v ≈ 1m/s = n RT2  Vf  For CA  Vi  Mechanical energy dissipated in collision, W ln = Ki − Kf = RT2 ln  V2  = 1 × (0.01)(200)2 − 1 (2.01)(1)2  V1  22 V ∝T ≈ 199 J ∴ p = constant (b) n = m = 2000 + 10 = 10.05 3. W = nR∆T M 200 Q = R(T1 − T2) Using, Q = n Cp ∆T Q = nCV ∆T ∴ ∆T = Q = 199 7 RT0 = (10)  7 R (T f − T0) 2 2 nCV 10.05 × 3 × 8.31 = 0.80°C Tf = 1.1T0

Chapter 21 Laws of Thermodynamics — 415 If p = constant, then V ∝ T 9. T ∝ U If temperature becomes 1.1 times, then volume will also become 1.1 times. TC = TD = T0 = 300 K 4. Wnet = area under the cycle TA = TB = 2T0 = 600 K = 2p0V0 Qnet = QAB + QBC + QCD + QDA Q+ve = QABC = WABC + ∆U ABC = nRTA ln  Vf  + nCV (TC − TB)  Vi  = area under the graph + n CV ∆T  Vf  = (3 p0V0 ) + n  3 R (TC − TA ) + nRTC ln  Vi  + nCV (TA − TD ) 2 = (3p0V0) + 3 − pAVA ) = (1)(R)(600) ln (2) + (1)(CV )(−300) 2 (pCVC + (1)(R)(300)ln  12 + (1)(CV )(300) = 10.5 p0V0 10. 1-2 = 300 R ln 2 η = Wnet = 4 T ∝V Q+ve 21 5. Q = W + ∆U ∴ p = constant = area under the graph + ∆U Qnet = Wnet = W12 + W23 + W31 = − p0V0 − 3 p0V0 ∴ W23 = Qnet − W12 (as W31 = 0) 2 = − 300 − nR∆T 5 = − 2 p0V0 = − 300 − 2 × 8.31 × 300 6. pV −1 = constant = − 5286 J In the process, pV x = constant 11. QA = QB R nCp ∆TA = nCV ∆TB 1− x C = CV + ∴ ∆TB = Cp ⋅ ∆TA = γ∆TA CV Here, C = 3 R + R = 2R 2 1+ 1 = (1.4)(30) = 42 K Q = n C∆T = (1)(2R)(2T0 − T0) 12. TV n − 1 = constant = 2RT0 ∴ (pV )V n − 1 = constant (as T ∝ pV ) 7. p ∴ pV n = constant Bulk modulus in the process pV γ = constant is γp. 3 Hence, in the given process Bulk modulus is np. 1 13. pV −1 = constant 2 In the process, pV x = constant V Vi Vf   W =  1 R x ∆T In process 3 − 1 − V = constant Here, x = −1 ∴ T∝p R ∆T R 2 2 If p is decreasing, then T will also decrease. ∴ W = = (T2 − T1) 8. TV γ − 1 = constant 14. ∆U = Q − W = 450 J = nCV∆T ∴ T1  VV12 γ − 1 T2  = = n  3 R ∆T 2  L2  5/ 3 − 1  L2  2/ 3 ∴ n ∆T =  3R00  L1   L1  = =

416 — Waves and Thermodynamics Now, C= Q 20. In second process, p ∝ V i.e. p-V graph is straight n ∆T line passing through origin. = 600 = 2R 300/ R p 15. W = p∆V = p0(2V0 − V0) = p0V0 2 ∆U = U f − U i 1 = 2p0(2V0) − 2p0V0 V = 2p0V0 Area under graph-2 is more. ∴ Q = W + ∆U = 3p0V0 ∴ W2 > W1 16. C = Q Further, T ∝ pV n ∆T or C ∝ Q (pV )2 > (pV )1 T2 > T1 ∆U is same. But W1 < W2 as area under p-V graph ∴ is less. or dU + dW = − dU 21. dQ = − dU Hence, Q1 < Q2 or 2dU + dW = 0 or C1 < C2. ∴ C1 < 1 ∴ 2 [n CVdT ] + pdV = 0 C2 or 2   γ R  dT  + pdV = 0 n  − 1  17. Qnet = Wnet   QAB = QCD = 0 or 2nRdT +  nRT  dV =0 γ −1 V  5 R QBC = n Cp ∆T = (1) 2 (4T −T)  2  dT dV  − 1 T V or γ + =0 = 7.5RT QDA = n Cp ∆T Integrating we get, = (1)  5 R (3T − 5T )  2 ln (T ) + ln (V ) = ln (C ) 2  γ − 1 = − 5RT Solving we get ∴ Wnet = Qnet = 7.5RT − 5RT γ−1 = 2.5RT TV 2 = constant 18. Uρ = constant 22. In a cyclic process, ∴ T  1  = constant Qnet = Wnet V ∴ QAB + QBC + QCA = area under the graph 1 (3 or T ∝ V ∴ 600 + 200 + QCA = 2 × 10−4 )(5 × 105) ∴ p = constant = 75 J ∆U = ∆U W Q − ∆U ∴ QCA = − 725 J = WCA + ∆UCA = n CV ∆T = − 725 = (−Area under the graph) + ∆UCA n Cp∆T − n CV∆T = −  1 × 11 × 105 (3 × 10−4 ) + ∆UCA 2 1= 1 3 = γ −1 5 − = 2 Solving we get, ∆UCA = − 560 J 3 1 23. T2 = T3 (lying on same isotherm ) 19. W = Area under F -x graph ∴ ∆U 1 = ∆U 2 W = Hatched area = 17.5 J = 1 ( pi + pf )(V f − Vi ) Now, Q = W + ∆U = 20 J 2

Chapter 21 Laws of Thermodynamics — 417 = 1 ( piV f − piVi + pfV f − pfVi ) 26. Pressure becomes half. So, temperature is doubled. 2  ∝ 1p p  as T ∆U = n CV ∆T pf Ti = p0V0 pi nR Tf = 2 p0V0 nR Vi Vf V ∴ ∆T = Tf − Ti = p0V0 nR Now, pi Vi and pf V f are same for both processes. ∴ ∆U = n  3 R  p0V0  = 3 p0V0 2 nR 2 Further pi and Vi , area also same for both processes. 27. W = nRT0 ln  pi   pf  For 1→ 3 Vf is more and pf is less. Hence, W will be more. WBC = 2 WAB Now, Q = W + ∆U ∴ nRT0 ln  pp0C/ 2 = 2nRT0 ln  pp0 0/2 W13 > W12   ∴ Q13 > Q12 (as ∆U is same) Solving this equation we get, 24. ∆U net = 0 pC = p0 ∴ ∆U 1 + ∆U 2 + ∆U 3 = 0 8 ∴ Q1 + (Q2 − W2) + ∆U 3 = 0 More than One Correct Options (200 kJ) + (−100 + 50) kJ + ∆U 3 = 0 1. p ∴ ∆U 3 = − 150 kJ pA In adiabatic process, W = − ∆U p 2B ∴ W3 = − ∆U 3 = 150 kJ V 25. QBC = 0 QCD = n Cp ∆T V 2V = n  5 R (TD − TC ) Now, see the hint of Q-No 9 (d) of subjective 2 questions for Level 1. = 5 − pCVC ) 2. Temperature is increased. So, internal energy will 2 (pDVD also increase. = 5 (105 − 2 × 105) 2 ∴ ∆U = + ve Further, = − 25 × 104 J pV 2 = constant QDA = n CV ∆T ∴  T  V 2 = constant or V∝1 V T = n  3 R (TA − TD ) 2 Temperature is increased. So, volume will = 3 − pDVD ) decrease and work done will be negative. 2 (pAVA In the process pV x = constant, molar heat capacity = 3 (2.4 × 105 − 105) is given by 2 R = 21 × 104 J C = CV + 1− x Now, Wnet = Qnet Here, x = 2 = (9 − 25 + 21) × 104 J = 5 × 104J ∴ C = CV − R

418 — Waves and Thermodynamics CV of any gas is greater than R. Here, x = 2 So, C is positive. Hence, from the equation, ∴ C = CV − R Q = nC∆T For any of the gas, CV ≠ R. ∴ C ≠0 Q is positive if T is increased. or ∆T is positive. ∴ Q = nC∆T ≠ 0 as ∆U ≠ 0 and ∆T ≠ 0 3. ab W =0 (as V = constant) ca Q1 = ∆U 1 = nCV∆T ρ is increasing. Hence, V is decreasing. So, work done is negative. =  3 R − = (2) 2 (2T0 T0 ) 3RT0 5. Q1 = nCV∆T bc ∆U = 0 (as T = constant) Q2 = nCp ∆T Q3 = 0 ∴ Q2 = W2 = nRT ln  Vf  Cp > CV  Vi  Q2 > Q1 > Q3 = (2)(R)(2T0)ln (2) ∴ V = constant = 4RT0 ln (2) cd W = 0 6. 1-2 Q3 = ∆U 3 = nCV∆T ∴ p∝T T is increasing. So, p will also increase. = (2)  3 R (T0 − 2T0) 2-3 V ∝ T 2 ∴ p = constant. = − 3RT0 V and T both are increasing. da ∆U = 0 Process 4-1 is reverse of 2-3. Q4 = W4 = nRT ln  Vf  Comprehension Based Questions  Vi  1. Qnet = Wnet = area under the cycle = pV 2 = (2)(R)(T0) ln 21 2. CV = 1 = 5 = − 2RT0 ln(2) Now in complete cycle, Cp γ 3 ∆U net = 0 3. In adiabatic process, γ −1 Qnet = Wnet = 2RT0 ln(2) = + ve p1 − γT γ = constant or T ∝ p γ ρ = constant 4. ab γ−1 ∴ V = constant ∴ TB =  pB  γ ∴ W =0 TA  pA  Q = ∆U 5/ 3 − 1 ∆U is positive, as U is increasing. ∴ TB = (1000)  32 5/ 3 Hence, Q is also positive.  23 0.4 bc ρ ∝ U ⇒ ∴ 1 ∝ T = (1000) 850 K V = ρ is decreasing, so V is increasing. Hence, work 4. In adiabatic process, done is positive. WAB = − ∆U AB (as Q = 0 ) Further, 1 ∝ T (T ∝ pV ) = nCV (TA − TB ) V = (1) 3 R (TA − TB ) ∴ pV 2 = constant 2 In the process, pV x = constant,  3 235 2 Molar heat capacity is given by = (1) × (1000 − 850) C = CV + R = 1875 J 1− x

Chapter 21 Laws of Thermodynamics — 419 5. WBC = 0 (as V = constant) (b) W = 0 ∴ T∝p (as V = constant) ∆U = − ve Q = ∆U = − ve pC = pB/ 2 ∴ (c) Q = 0 ∴ TC = TB/ 2 = 425 K Q = ∆U = nCV (∆T ) W = + ve (as V is increasing) = (1)  3 R (TC − TB ) ∴ ∆U = − W = − ve 2 (d) ∆T = 0 = 3 × 25 × (425 − 850) ⇒ ∆U = 0 23  Vf  = 5312.5 J and Q =W = nRT ln  Vi  =+ ve Match the Columns 5. (a) Wab = area under p-V graph 1. (a) W = nR∆T 3 p0V0 = (2)(R)(T ) = 2RT = 2 = 1.5 p0V0 (b) ∆U = nCV∆T (b) ∆U ab = Qab − Wab = (2)  3 R (T ) = 3RT = 6 p0V0 − 3 p0V0 = 4.5P0V0 2 2 (c) W = − ∆U = − 3RT (c) C = Q = W + ∆U n∆T n∆T (d) ∆U = 3RT (in all processes) = 6p0V0 = 2R 2. For both processes,  pbVb paVa  nR − nR p = constant (as V ∝ T ) n ∴ C = Cp for same gases. (d) CV = ∆U For W , ∆U and Q we will require number of n∆T moles also. = 4.5p0V0 = 1.5R 3. p2V = constant or pV 1/ 2 = constant  pbVb paVa  nR nR Comparing with pV x = constant, we get n − x=1 Subjective Questions 2 1. In a cyclic process, ∴ Molar heat capacity, ∆U = 0 C = CV + R 1− x Qnet = Wnet  3 R QAB = n Cp ∆T 2 = + 2R = 3.5 R = (2)  5 R (400 − 300) 2 Q = nC∆T = n(3.5R) (∆T ) = 500R = 3.5 nR∆T  pi  ∆U = nCV∆T QBC = n R TB ln  pf  = n  3 R ∆T = 1.5 nR∆T = (2)(R)(400) ln  21 2 ∴ W by the gas = Q − ∆U = 2nR∆T and work = 800R ln (2) done on the gas = − 2nR∆T 4. (a) Q = nCp ∆T QCD = nCp ∆T = − 500R ∆U = nCV∆T W = Q − ∆U = nR∆T QDA = n RTD ln  pi  W < ∆U as CV > R  pf 

420 — Waves and Thermodynamics = (2)(R)(300) ln  12 Substituting the values like above we get, = − 600R ln (2) ∴ Qnet = Wnet = (200R) ln (2) ∆U 2 = 1.5p0 M = (200)(8.31)(0.693) ρ0 W3 = Q2 − ∆U 2 = p0 M ρ0 ≈ 1153 J For Process 3-1 (1.013 × 105)(22.4) Density is constant. Hence, (103)(8.31) 2. TA = pAVA = volume is constant. nR ∴ W3 = 0 = 273 K ∴ Q3 = ∆U 3 = nCV∆T Along path AB = (1)  3 R (T1 − T3) 2 V = constant ∴ T∝p = 3  p0 M − 2 p0 M  2  ρ0 ρ0  p is doubled. So, T is also doubled. ∴ TB = 2TA = 546 K. = − 1.5p0M Further, ρ0 ∴ TB = TC ∴ pB VB = pCVC (b) ΣQ−ve = |Q1 + Q3| VC = (pB )(VA) = (2)(22.4 ) = p0 M  3 + ln 2 ( pA ) 1 ρ0 2 = 44.8 m3 (+ p0 M /ρ0) +  − p0 M ln 2  ρ0 3. For Process 1 - 2 ρ ∝ p (c) η = Wnet = ΣQ+ ve (2.5 p0 M /ρ0) ∴ 1∝p V = 2 (1 − ln 2) 5 ∴ Process is isothermal. 4 (a) p . ∆U 1 = 0 c  pi  1.5p Q1 = W1 = nRT ln  pf  0 pa b = p0M ln  12 = PM ) 0 ρ0 ρ (as n = 1and RT V V0 1.5V0 = − p0M ln (2) Wnet = Qnet in a cycle. ρ0 Since, Qnet is negative (flows out of the gas). For Process 2-3 Hence, Wnet should also be negative. Or, cycle should be anti-clockwise as shown in figure. Q2 = Qp = nCp ∆T = (1) 5 R (T3 − T2 ) From a to b 2 p = constant 5  p3 M p2 M  ∴ V ∝T 2  ρ3 ρ2  = − T has become 1.5 times. Therefore, V will also become 1.5 times. = 5  2 p0M − 2 p0M  From c to a 2  ρ0 2ρ0  p∝V = 2.5 p0M V has become 1 times. Therefore, ρ0 1.5 ∆U 2 = nCV∆T p will also become 1 times. 1.5

Chapter 21 Laws of Thermodynamics — 421 In a cycle ∴ WTotal = 3RT0 ln (2) − 3 RT0 2 |Qnet | = |Wnet | = Area of cycle ∴ 800 = 1 (c) QTotal = 3RT0 ln (2) − 21 2 (0.5p0)(0.5V0) 4 RT0 ∴ p0V0 = 6400 J 7. (a) p(× 105 N/m2) Wca = − Area under the graph = − 1 (2.5 p0 )(0.5V0 ) (p0, V0) (p1, V1) 2 2.5 A B = − 0.625p0V0 0.44 C (p2 , V2) = − 0.625 × 6400 = − 4000 J 20 40 113.1 V (L) 5. First process is isobaric. ∴ ∆Q1 = nCV∆T + p∆V 2 × 8.31 × 300 Second process is isochoric 20 × 10–3 ∴ ∆Q2 = nCV∆T (b) p0 = nRT0 = V0 ∆Q1 − ∆Q2 = p∆V =  p0 + mg  [ Ax ]  A  = 2.5 × 105 N/m2 = (p0A + mg)x p1 = p0 = 2.5 × 105 N/m2 = [105 × 60 × 10−4 + 8 × 10](0.2) V1 = 2V = 40 × 10−3 m3 = 136 J Ans. Process AB : V ∝ T 6. (a) p A p A ∴ T1 = 2T0 = 600 K Process BC CB Using T1V1γ−1 = T2V2γ−1, we get CB V2 = 2 2V1 = 113.1 × 10−3 m3 Ans. Ans. VT nRT2 (2)(8.31)(300) V2 113.1 × 10−3 Ans. and p2 = = …(i) (b) For the process AB, p0V0 = pB (2V0) = 0.44 × 105 N/m2 T0 T0 (c) WTotal = W1 + W2 ∴ pB = p0 = p0 (V1 − V0) + nR (T1 − T2) 2 γ −1 ∆U = 0 Substituting the values, we get Q = W + ∆U WTotal = 12479 J = nRT ln VB 8. (a) p1V1 = p2V2 VA T1 T2 = 3RT0 ln 2 T2 =  p2V2T1   p1V1  For the process BC, 2V0 = V0 T0 TC ∴ TC = T0 2 W = nR∆T = 3R  T0 − T0 =− 3 RT0 Here, 2 2 p1 = 1.0 × 105 N/m2 Q = nCp ∆T = − 21 RT0 V1 = 2.4 × 10−3 m3 4

422 — Waves and Thermodynamics T1 = 300 K ∴ W = Q − ∆U = n(C − CV )∆T W = C − CV p2 = p1 + kx ∆U CV A = 1.0 × 105 + 8000 × 0.1 ∴ W =  C − CV  ∆U 8 × 10−3  CV  = 2.0 × 105 N/m2 =  9/2 − 5/ 2 (100) = 80 J Ans.  5/2  V2 = V1 + Ax Ans. = 2.4 × 10−3 + 8 × 10−3 × 0.1 10. dQ = dU + dW Ans. = 3.2 × 10–3 m3 CdT = CV dT + pdV F (CV + 3aT 2) dT = CV dT + pdV A Substituting in Eq. (i), we get Ans. T2 = 800 K Ans.  RT  V (b) Heat supplied by the heater, ∴ 3aT 2 dT = pdV = dV Q = W + ∆U  3Ra dV V Here, ∴ TdT = ∆U = nCV ∆T Integrating, we get =  p1V1   3 R (800 − 300)  3aT 2   RT1  2   = ln V − ln C  2R  = (1.0 × 105)(2.4 × 10–3)(1.5)(500) (300) 3aT 2 − 3aT 2 = 600 J V = Ce 2R or Ve 2R = constant W = 1 kx2 + p1∆V 11. (a) p = αT 1/ 2 2 or pT −1/ 2 = constant = 1 × (8000)(0.1)2 + (1.0 × 105)(0.1)(8 × 10–3) p1/ 2V −1/ 2 = constant 2 pV −1 = constant = (40 + 80) J = 120 J ∴ x = –1 ∴ Q = 600 + 120 = 720 J Ans. 9. U ∝ V ∆W =  1 R  (∆T )  − x As U ∝ T ∴ T ∝ V 1/ 2 =  8.231 (50) or TV −1/ 2 = constant = 207.75 J or pV 1/ 2 = constant R 3 R 1− x 2 2 Comparing with pV x = constant, we have (b) C = CV + = R + = 2R x=1 12. F + pA = p0A 2 F = (p0 − p)A p0 ∴ Molar specific heat p C= R + R 2V γ −1 1− x ∫W = V (p0 − p)A dx =R+R (γ = 7/5) 2V 2V 7/5 − 1 1 − 1/2 ∫= V p0dV − ∫V pdV = 5 R + 2R = 9R ∫= p0V− 2V  dV  22 V Ans. RT Q = nC∆T ∆U = nCV∆T V = p0V − RT ln (2) = RT − RT ln (2) = RT (1 − ln 2)

Chapter 21 Laws of Thermodynamics — 423 13. (a) p ∝ 1 or pT = constant ∆WDA = − 1  p0 + p0 (2V0 − V0) 2 2 T ∴ p(pV )) = constant = − 3 p0V0 or pV 1/ 2 = constant 4 In the process, pV x = constant, ∆U AB = nCV ∆T = (2)  3 R (TB − TA) 2 Molar heat capacity is  n 3 R C = CV + R = 2, CV = 2 1− x 3R  p0V0 p40VR0 or C = 3R + R = 2R − 2 1− 1 3 T pV  2 = 4 p0V0 = ∆QAB = nR = 3 R + 2R = 7 R Ans.  3 R 22 2 ∆U BCD = nCV ∆T = (2) (TD − TB ) (b) W = Q − ∆U = nC∆T − nCV∆T  2 p0V0 p0V0  3 2R 2R 2 p0V0 = n(C − CV )∆T = (3R ) − = = 2 7 R − 3 R(T2 − T1 ) Hence, ∆QBCD = ∆U BCD + ∆WBCD  2 2 = 4R(T2 − T1) Ans. =  π + 25 p0V0 4 14. V = a ∆UDA = nCV∆T T 3 or VT = constant = (2)  2 R (TA − TD ) or V (pV ) = constant = (3R)  p0V0 − 2 2p0RV0  ∴ pV 2 = constant 4R In the process pV x = constant, molar heat capacity 9 4 is = − p0V0 C= R + R γ −1 1− x ∴ ∆QDA = ∆U DA + ∆WDA Here, x=2 = − 9 p0V0 − 3 p0V0 4 4 2 1γ  ∴ C = R + R =  γ − R = − 3p0V0 γ −1 1− 2 − Net work done is, Now, Q = nC∆T Wnet =  π + 1− 43 p0V0 4 2 − γ1 = (1)  γ − R∆T = 1.04 p0V0 and heat absorbed is  2 − γ1 =  γ − R∆T Ans. Qab = ∆Q+ ve 15. Process AB is isochoric (V = constant). =  3 + π + 52 p0V0 = 4.03 p0V0 4 4 Hence, Hence, efficiency of the cycle is η = Wnet × 100 ∆WAB = 0 Qab = 1.04 p0V0 × 100 ∆WBCD = p0V0 + π ( p0 )  V0  4.03 p0V0 2 2 = 25.8% =  π + 1 p0V0 Ans. 4

424 — Waves and Thermodynamics 16. p = αT − βT 2 (p = constant) or p -V graph will be a rectangular hyperbola V with increasing p and decreasing V . Hence V = αT − βT 2 ρ ∝ 1 . Hence, ρ -V graph is also a rectangular p V hyperbola with decreasing V and hence or dV =  α − 2βT  dT increasing ρ.  p  ρ∝ p ρ = pM  RT  T2 α − 2βT  ∫ ∫W = = T1  p  pdV p dT Hence, ρ - p graph will be a straight line passing through origin, with increasing ρ and p. or W = [αT − βT 2 ]TT12 Process B -C is an isochoric process, because = α (T2 − T1) − β(T22 − T12) Ans. p -T graph is a straight line passing through origin 17. (a) First law of thermodynamics for the given i.e. V = constant process from state 1 to state 2 Hence, p -V graph will be a straight line parallel to p - axis with increasing p. Q12 − W12 = U 2 − U 1 Here, Q12 = + 10 p0V0 joule Since, V = constant hence ρ will also be constant W12 = 0 (Volume remains constant) Hence ρ -V graph will be a dot. U 2 − U 1 = nCV (T2 − T1) nCV (T2 − T1) = 10 p0V0 ρ- p graph will be a straight line parallel to p -axis For an ideal gas, with increasing p, because ρ = constant Process C-D is inverse of A- B and D - A is inverse p0V0 = nRT0 of B-C. Cp − CV = R and Different values of p, V , T and ρ in tabular form ∴ CV = Cp − R are shown below = 5R − R = 3R pV T ρ 22 A p0 V0 T0 ρ0 ∴ n  32R (T2 − T0) = 10nRT0 B 2 p0 T0 2ρ0 V0 T2 = 23 C 4 p0 2 2T0 2ρ0 3 T0 D 2 p0 V0 2T0 ρ0 As p ∝ T for constant volume 2 23 V0 3 p2 = p0  T0  P0M  P0  RT0 (b) Here, V0 = nR and ρ0 = 2 The corresponding graphs are as follows ρ p 2ρ0 B,C 1 V 18. Process A -B is an isothermal process ρ0 A,D i.e. T = constant V Hence, p∝ 1 V0 V0 2 V

Chapter 21 Laws of Thermodynamics — 425 p D 19. First process 4p0 C A T = constant 2p0 B V ∴ p∝ 1 p0 V0 V C V is made 5 times. Therefore, p will become 1 th. 4p0 p 5 V0 Second process 2 V = constant ⇒ T ∝ p ρ Pressure again becomes first times. So, 2ρ0 B temperature will also become 5 times. Q1 + Q2 = 80 × 103 ∴ nRT ln  Vf  + nCV ∆T = 80 × 103  Vi  ρ0 A ∴ (3)(8.31)(273) ln(5) + (3)  8.31  (5 × 273 − 273) D  γ − 1 = 80 × 103 p0 2p0 Solving we get, γ = 1.4

22. Calorimetry and Heat Transfer INTRODUCTORY EXERCISE 22.1 Q = m1s∆θ = 10 × 1 × 40 1. Q = mLf + msw ∆θ + mLf = 400 cal = m (Lf + sw∆θ + Lf ) Mass of ice melted by this heat, = 10 [80 + 1 × 100 + 540] = 7200 cal m2 = Q = 400 = 5 g < 15 g L 80 2. Let θ (> 60°) is the mixture temperature in Therefore, whole ice is not melted. equilibrium. Temperature of mixture is 0° C. Heat given = Heat taken 0 = m(1)(θ − 20) + m(0.5)(θ − 40) Mass of water = m1 + m2 = 15 g + m (0.25)(θ − 60) Mass of ice = 15 − m2 = 10 g Solving the equation, we get 7. Similar to Example 22.4 θ = 31.43°C INTRODUCTORY EXERCISE 22.2 3. mL = ms (θ − 0°) 2. Q = KA (θ1 − θ2)t ∴ θ = L = 80 = 80°C l s1 ∴ K = Ql = J-m A (θ1 − θ2) t m2 K - s 4. 75% heat is retained by bullet 3 1 mv 2 = ms∆θ + mL =W (as J/s = W) 4  2  m-K or v = (8s∆θ + 8L) 3. R = l = m m2 3 KA (W/m -K) Substituting the values, we have = K = KW−1 W v = (8 × 0.03 × 4.2 × 300) + (8 × 6 × 4.2) 3 4. In convection, liquid is heated from the bottom. = 12.96 m/s 5. 200°C K 2K 3K 5. Let m be the mass of the steam required to raise H θ1 H θ2 H 100°C the temperature of 100 g of water from 24°C to 1 2 3 90°C. H1 = H2 = H3 Heat lost by steam = Heat gained by water ∴ 200 − θ1 = θ1 − θ2 = θ2 − 100 ∴ m (L + s∆θ1) = 100s∆θ2 or m = (100)(s)(∆θ2) (l/KA) (l/2KA) (l/3KA) L + s(∆θ1) Solving these equations we get, Here, s = specific heat of water = 1 cal/g-°C, and θ1 = 145.5° C 0°C L = latent heat of vaporization = 540 cal/g θ2 = 118.2° C ∆θ1 = (100 − 90) = 10° C 6. R/2 θ R/2 and ∆θ2 = (90 − 24) = 66° C 100°C H1 H2 H3 Substituting the values, we have R m = (100)(1)(66) = 12 g (540) + (1)(10) 25°C ∴ m = 12 g H1 + H2 = H3 100 − θ + 25 − θ = θ − 0° 6. Heat liberated by 10 g water (at 40°C) when it (R/2) R R/2 converts into water at 0° C

Chapter 22 Calorimetry and Heat Transfer — 427 Solving we get, θ = 45° C ∴  80 − 50 =α  80 + 50 − 20 5  2 HCD = TD = 45° C − 25° C R 5  60 − 30 =α  60 + 30 − 20 = 4W and t  2 7. Let us start with the assumption that Solving these two equations, we get ∆θ ∝ Temperature difference t = 9 min ∆ Exercises LEVEL 1 5. λm ∝ 1 T Assertion and Reason ∴ (λ m)1 = T2 or T1 = (λ m)2 1. Specific heat is a function of temperature. But for (λ m)2 T1 T2 (λ m)1 most of the substances variation with temperature = 350 = 0.69 510 is almost negligible. 4. R1 = 2R and R2 = R 6. dQ = CdT 2 Thermal resistance becomes 1 th. Therefore heat ∴ dT = 1 = Slope of T - Q graph. 4 dQ C current becomes four times. 7. = l = l ∝ l KA K (πR2) R2 5. If temperature of a normal body is more than a Thermal resistance perfectly black body, it can radiate more energy. l R2 6. This law was given by Kirchhoff. The rod for which is minimum will conduct 9. In steady state, temperature of different sections maximum heat. (perpendicular to the direction of heat flow) 8. Heat taken by 1 g ice in transformation from ice at becomes constant but not same. 0° C to water at 100° C is 10. − dT = eAσ (T 4 − T04) or − dT ∝ 1 Q = mL + ms∆θ dt mc dt m = (1) (80) + (1) (1) (100) = 180 cal Objective Questions 1. λm ∝ 1 Mass of steam condensed to give this much heat is T m = Q = 180 = 1 g L 540 3 ∴ (λ m)2 = T1 = 2000 = 2 (λ m)1 T2 3000 3 This is less than 1g or total mass of steam. Therefore, whole steam is not condensed and ∴ (λ m)2 = 2 (λ m)1 mixture temperature is 100° C. 3 9. H A = HTotal 2. At higher temperature radiation is more. So, (TD)A = TD ∴ cooling is fast. RA RA + RB 3. Q1 = Q2 ∴ ms1 (32 − 20) = ms2 (40 − 32) ∴ (TD)A =  RA RA RB  TD ∴ s1 = 8 = 2  +  s2 12 3 =  1 TD ...(i)  1 + RB /RA 4. 2 Q = ∫ dQ = ∫1msdT RB  lB   KA A  1200 RA  lA   KB A ∫= 2 (1)(aT 3) dT = 15a = = (3) = 6 14

428 — Waves and Thermodynamics Substituting in Eq. (i), we get 5. P = er σ A (T 4 − T04) (TD)A =  1 × 35 = 5° C = (0.4)(5.67 × 10−8 )(4π )(4 × 10−2)2[(3000)4  1 + 6 − (300)4 ] 10. Thermal resistance, = 3.7 × 104 W R= l ∝ 1 ⇒ KA = 2 KB 6. L  ddmt  = dQ = H = TD = θ − 100 KA K dt R (l / KA ) ∴ RA = RB ∴ θ = 100 +  KLAl   ddmt  2 So, let RA = R then RB = 2R Now, HA = HB  2.256 × 106 × 1.2 × 10−2  0.44    60 × 5 ∴ (TD)A = (TD)B =  RA  = 100 +  50.2 × 0.15  RA RB  RB  or (TD)A (TD)B =  2RR (36°C) = 18°C = 105° C 7. (a) H 1 = H 2 11. R = l θ KA –10°C H2 H1 R2 19°C R2 =  ll12  K 1 ⋅ A1  R1   K 2 A2  R1 = (2)  12  12 = 1 Wood Insulation 2 ∴ 19 − θ = θ + 10  H = TRD Therefore, thermal resistance of 2nd rod is half. R1 R Hence, rate of heat flow will be twice. or 19 − θ = θ + 10 Subjective Questions (l1 /K1A) (l2 /K2A) 1. Total rate of radiation of energy ∴ K1(19 − θ) = K2(θ + 10) l1 l2 = er σT 4A = (0.6)(5.67 × 10−8 )(800 + 273)4(0.1)2(2) ∴ 0.01 (19 − θ) = 0.08 (θ + 10) 3.5 2.0 ≈ 900 W Solving, we get 1  1 mv2 2. 2 2 = ms∆θ + mL θ = − 8.1°C ∴ v = 2 s∆θ + L (b) H = H1 = 19 − θ (l1 /K1A) = 2 (125) (327 − 27) + 2.5 × 104 = K1A (19 − θ) l1 = 500 m/s = (0.01) (19 + 8.1) (1) 3. 0.4 [mg ∆h] = ms∆θ 3.5 × 10−2 ∴ ∆θ = 0.4 g∆h = 0.4 × 9.8 × 0.5 = 7.7 W/ m2 s 800 8. Hint is already given in the question = 2.5 × 10−3° C 9. Let mixture is water at θ°C (where 0°C< θ <40°C) 4. P =  ddmt  s ⋅ ∆θ Heat given by water = Heat taken by ice ∴ (200) (1) (40 − θ) = (140) (0.53) (15) ∴ dm = P = 500 × 106 dt s∆θ (4200) × 10 + (140) (80) + (140) (1) (θ − 0) Solving we get, = 11904 kg/s = 1.2 × 104 kg/s θ = − 12.7° C

Chapter 22 Calorimetry and Heat Transfer — 429 Since, θ < 0° C and we have assumed the mixture Solving this equation, we get to be water whose temperature can't be less than θ ≈ 206° C 0° C. Hence, mixture temperature θ = 0° C. Now, temperature gradient on Heat given by water in reaching upto 0° C is, LHS = θ − 100 = 212°C/m 0.5 θ = (200) (1) (40 − 0) = 8000 cal. and on RHS = θ − 0 = 424°C/m Let m mass of ice melts by this heat, then 0.5 8000 = (140) (0.53) (15) + (m) (80) 13. Net power = erσ (T 4 − T04)A Solving we get m = 86 g In first case, 210 = erσ [(500)4 − (300)4 ] A ∴ Mass of water = 200 + 86 = 286 g ...(i) In second case, ...(ii) Mass of ice = 140 − 86 = 54 g 700 = σ [(500)4 − (300)4 ] A 10. Let heat is supplied at a constant rate of (α ) J . Dividing Eq. (i) by Eq. (ii), we get min In 4 minutes, (when temperature remains constant) 210 700 ice will be melting. er = = 0.3 ∴} mL = (4) α ∴ α = L = 336 × 103 LEVEL 2 m4 4 = 84 × 103 J/min - kg Single Correct Option Now in last two minutes, 1. R = l Q = ms∆θ KA l is halved, A is four times and K is 1 times. ∴ (α ) (2) = (m) (4200) (θ − 0° C) Substituting the value of α we get, 4 m ∴ R will become half. Hence, heat current will become two times. Therefore, rate of melting θ = 40° C of ice will also become two times or 0.2 g/s. 11. T3 T2 2. RR K1 TT R/2 T R/2 H2 T 12 3 H1 K2 H3 d 3a R T1 H1 = H2 H1 + H2 = H3 ∴ T1 − T2 = T2 − T3 (d /K1A) (3d /K2A)  T3 − T  +  T2 − T  = T − T1 ∴ K1 =  T2 − TT23 ⋅ 1  3R / 2   3R / 2  R K2  T1 − 3  ∴ T = 3T1 + 2 (T2 + T3) But T1 : T2 and T3 are in AP 7 ∴ T2 − T3 = T1 − T2 12. 100°C θ 0°C ∴ K1 = 1 K2 3 H1 25 W H2 3. H = TD H 1 + H 2 = 25 R = TD = 100 − 0 R H1 ∴ θ − 100 + θ−0 = 25 ∴ 0.5/(400 × 10−4) 0.5/(400 × 10−4) = 100 kW−1

430 — Waves and Thermodynamics Now, ∫ ∫x x dx More than One Correct Options R = dR = 0 0 K0 (1 + ax) A 1. Total radiation per second is given by ∫or x dx P = er σT 4A 100 = 0 102 (1 + x) (10−4) Here, er : σ : T and A all are same. Solving this equation we get, Hence, P will be same. x = 1.7 m Same is the case with absorption per second. 4. Let R = thermal resistance of each rod. Now, P = ms  − dT  dt In first case, Rnet = R + R = 2R (in series) (in parallel)  dT  P 1 In second case, Rnet = R or Rate of cooling − dt = ms ∝ m 2 In second case thermal resistance has become 1 th. Mass of hollow sphere is less. So, its initial rate of 4 cooling will be more. So, heat current will become 4 times. So time 2. 40°C taken to flow same amount of heat will be reduced to 1 th. H1 4 θ 5. AT H2 H3 20°C 30°C H 2 B H C H1 = H2 + H3 √2T ∴ 40 − θ = θ − 20 + θ − 30 1 RR R H1 = H2 ∴ 2T − TC = TC − T where, R = thermal resistance of each rod (l/KA) ( 2l/KA) Solving this equation, we get Solving we get, θ = 30° C Heat flows from higher temperature to lower temperature. TC = 3T 3. ms (2θ − θ′ ) = m (2s) (θ′ − θ) 2+1 Solving this equation we get, 6. Net power available per second, θ′ = 4 θ 3 P = 1000 − 160 = 840 J/s Heat required, Further, heat capacity Q = ms∆θ = (2) (4200) (77 − 27) C = ms or C ∝ s (as m is same) ∴ C1 = s1 = 1 = 420000 J C2 s2 2 ∴ Time required = Q = 500 s = 8 min 20 s 4. q1 = TD P R1 7. Rnet = R1 + R2 =  KXA + 4X = 3X ∴ R1 = TD 2KA KA Similarly, q1 Now, H = dQ = TD R2 = TD dt Rnet q2 = (T2 − T1) =  KA (T2 − T1 )   31 In series, (3X /KA)  x  qs = TD = TD ∴ f =1 R1 + R2 TD + TD 3 q1 q2

Chapter 22 Calorimetry and Heat Transfer — 431 = q1 q2 3. If R = thermal resistance of one rod q1 + q2 c In parallel, qp = TD = TD  1  H/2 H/2 Rnet  Rnet  ab de 1 1 100°C H H –80°C  R1 R2 H/2 = TD + H/2 f = TD  q1 + TqD2  = q1 + q2 Then, Rnet = 3R TD ∴ H = 100 − (−80) = 60 5. (b) If temperature difference is small, the rate of 3R R cooling is proportional to TD. (a) ab H = 60 = 100 − θb RR Match the Columns ∴ θb = 40° C 1. (a) E = σT 4 = energy radiated per unit surface are (b) bc H = 30 = 40 − θc per unit time by a black body 2R R ∴ σ = E  =  ML2T−2  ∴ θC = 10° C T 4    (c) Same as above  L2T θ 4  = [MT−3 θ−4 ] (d) de (b) b = λmT H = 60 = θd − (−80) ∴ [b] = [Lθ ] RR (c) Emissive power is energy radiated per unit ∴ θd = − 20° C time per unit surface area. 4. (a) m (s)(θ′ − θ) = 2m (s) (2θ − θ′ ) ∴ [E ] =  ML2 T−2  = [ MT −3 ] Solving we get,  L2 T  θ′ = 5 θ   3 (d) H = d Q = TD (b) ms (θ′ − θ) = 3m (s) (3θ − θ′ ) dt R Solving we get, ∴ R = TD = θ θ′ = 5 θ (d Q /dt) [ML2T−2 /T ] 2 = [M −1 L−2 T3θ ] (c) 2m (s) (θ′ − 2θ) = 3m (s) (3θ − θ′ ) 2. (a) dQ = ms dθ Solving we get, dt dt θ′ = 13 θ ∴ 5  ddθt  = dQ /dt ms (d) ms (θ′ − θ) + 2 ms (θ′ − 2θ) + 3ms (θ′ − 3θ) ∴ Slope of θ - t graph ∝ 1 =0 m Solving we get, (b) QTotal = mL θ′ = 7 θ or 3  dQ  (time) = mL dt 5. Unit of heat capacity is J / °C. or time ∝ m or length of line bc ∝ m (d) ab : only solid state Unit of latent heat is J/kg. bc : solid + liquid state Subjective Questions cd : only liquid state de : liquid + gaseous state 1. (a) Between t = 1 min to t = 3 min, there is no rise ef : only gaseous state. in the temperature of substance. Therefore, solid melts in this time.

432 — Waves and Thermodynamics L = Q = Ht = 10 × 2 4. Let heat capacity of flask is C and latent heat of m m 0.5 fusion of ice is L. = 40 kJ/kg (b) From Q = ms∆T or s = Q Then, m∆T C (70 − 40) + 200 × 1 × (70 − 40) = 50 L Specific heat in solid state + 50 × 1 × (40 − 0) s = 10 × 1 = 1.33 kJ/kg-° C 0.5 × 15 or 3C − 5L = − 400 …(i) Further, C (40 − 10) + 250 × 1 × (40 − 10) 2. Let T0 be the temperature of surrounding and T be = 80L + 80 × 1 × (10 − 0) the temperature of hot body at some instant. Then, or 3C − 8L = − 670 …(ii) dT Solving Eq. (i) and (ii), we have dt − = K (T − T0) L = 90 cal/g Ans. T dT t 5. ms∆θ = work done against friction = − K dT ∫ ∫or = (µmg cos θ)d (but µmg cos θ = mg sin θ) Tm T − T0 0 ∆θ = (µg cos θ)d = (g sin θ)d ss (Tm = temperature at t = 0) = (10) 53 (0.6) Solving this equation, we get 420 T = T0 + (Tm − T0)e−Kt (i) = 8.57 × 10−3 ° C Maximum temperature it can lose is (Tm − T0) ≈ 8.6 × 10−3 ° C From Eq. (i), T − T0 = (Tm − T0)e−Kt Given that Ans. T − T0 = Tm − T0 = (Tm − T0)e−Kt 6. Using dQ = mc dθ = KA(θ1 − θ2) 2 dt dt l Solving this equation we get, Area = 0.4 m2 Temp = 400 K t = ln (2) K 3. Let θ be the temperature of junction. Then, 0.4 m Cylinder 100°C 80°C H1 H2 C B θ Disc H3 S Mass = 0.4 kg c = 600 J/kg-K We have, 60°C mc  dθ  = KA(400 − θ) dt 0.4 H1 + H2 = H3 ∴ 100 − θ + 80 − θ = θ − 60 (0.4)(600) dθ = (10)(0.04)(400 − θ) dt 0.4 (46/0.92A) (13/0.26A) (12/0.12A) Solving this equation we have, ∴  dθ  = 1 dt  400 − θ 240 θ = 84° C (b) Heat current in copper rod, t 350 dθ dt = 240 100 − θ ∫ ∫∴ H1 = (46/0.92 × 0 300 400 − θ 4) Solving this, we get t ≈ 166 s = 1.28 cal/s Ans. Ans.

Chapter 22 Calorimetry and Heat Transfer — 433 7. Consider a differential cylinder 10. Let T1 be the temperature of C1 and T2 the dr temperature of C2 at some instant of time. Further let T be the temperature difference at that instant. C1 H C2 T1 T2 r Then, C1  − dT1  = H = T = KA (T ) dt l/KA l H = dQ = KA dθ = (2πKrl) dθ (Q T = T1 − T2) dt dr dr  dT2  T KA (T ) H r 2 dr = 50 and C2 + dt = H = l/KA = l ∫ ∫∴ 2πKl r 1 r 0 dθ or H ln  r2  = 50 ∴ − dT = − dT1 + dT2 2πKl  r1  dt dt dt ∴ H = 100πKl ∴ − dT1 = KA (T ) ln (r2 /r1) dt lC1 Now, Ht = mL and + dT2 = KA dt lC2 t = mL = mL ln (r2/r1) ∴ H 100πKl Ans. − dT = − dT1 + dT2 dt dt dt Further, 8. Three thermal resistances are in series.  R = KlA = KA(C1 + C2) T l C1C1 ∴ R = R1 + R2 + R3 = 2.5 × 10−2 + 1.0 × 10−2 + 2.5 × 10−2 T dT = − KA(C1 + C2) t 0.125 × 137 1.5 × 137 1.0 × 137 ∫ ∫or ∆T0 T l C1C2 dt 0 = 0.0017 °C-s /J Solving, we get T = ∆T0e−αt Now, heat current H = Temperature difference where, α = KA(C1 + C2) Net thermal resistance lC 1C 2 = 30 = 17647 W 11. x T 0.0017 H 9. Thermal resistance of plastic coating  Rt = KlA T1 dx T2 Rt = t (d = diameter) H = TD = (− dT ) K (πd)l R (dx)/KA = (0.16 × 10−2 × 0.06 × 10−3 × 10−3)(2) = −  dT   a A = constant …(i) 4.18 × 102)(π )(0.64 dx T = 0.0223 ° C-s /J T2 − dT = H l Now, i2Re = TD ∫ ∫∴ dx ∴ Rt T1 T aA 0 TD = i2ReRt ln  TT21 = Hl  aA = (5)2(4)(0.0223) aA = 2.23°C Ans. or H = l ln (T1/T2 )

434 — Waves and Thermodynamics Substituting in Eq. (i), we have R2 dr = − 4πK θ2 dθ R1 r2 H aA  T1   dT  aA ∫ ∫or θ1 l  T2  dx T ln = − R2 − R1 4πK R1R2 H or = (θ1 − θ2) T dT ln  T1  ∴ H = 4πK (R1R2)(θ1 − θ2)  T2  R2 − R1 =− x ∫ ∫or dx T1 T l 0 T  x  T1  − x Substituting this value of H in Eq. (i), we have  T1 l  T2  l ∴ ln = − ln (T1/ T2) = ln R1R2 (θ1 − θ2) = − dθ (r2) R2 − R1 dr −x or T =  TT21 l  (θ1 + θ2)  R1R2 r dr T1  2 R2 − R1 R1 r2 ∫ ∫∴ = θ1 dθ − (θ1 − θ2) − x x l or T = T1  T1  =  T2  l Ans. or θ1 + θ2 − θ1 = R1R2 (θ1 − θ ) 1 − 1  T2  T1 T1  2 R2 − R1 r R1  2 12. L  ddmt  = TD or θ1 − θ2 = R1R2 (θ1 − θ2) 1 − 1 R1 + R2 2 R2 − R1  R1 r 100°C H ice at 0°C 1 − 1 = R2 − R1 R1 R2 R1 r 2R1R2 or 80 × 360 = 100 1 = 1 − R2 − R1 = R1 + R2 3600 20 + 10 r R1 2R1R2 2R1R2 0.25 × 10 K × 10 or r = 2R1R2 Ans. R1 + R2 Solving this equation, we get K = 0.222 cal/cm-s-°C Ans. 14. See the extra points just before solved examples. …(i) 13. H = − dθ =  − dθ  (4 πKr2 ) Growth of ice on ponds. We have already derived dr/K (4πr2) dr that H t = 1 ρL y2 r 2 Kθ ∴ dt = ρLy dy K θ dr ∴ dy = K θ dt Lρy

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. Two moles of an ideal monoatomic gas Cp − CV = a for hydrogen gas Cp − CV = b occupies a volume V at 27°C. The gas for nitrogen gas. The correct relation expands adiabatically to a volume 2 V . Calculate (i) the final temperature of the between a and b is (2017) gas and (ii) change in its internal energy. (a) a = b (b) a = 14b (2018) (c) a = 28b (d) a = 1 b 14 (a) (i) 189 K (ii) 2.7 kJ (b) (i) 195 K (ii) −2.7 kJ 6. A copper ball of mass 100 g is at a (c) (i) 189 K (ii) −2.7 kJ (d) (i) 195 K (ii) 2.7 kJ temperature T . It is dropped in a copper 2. A silver atom in a solid oscillates in calorimeter of mass 100 g, filled with 170 g simple harmonic motion in some direction of water at room temperature. with a frequency of 1012 per second. What is Subsequently, the temperature of the the force constant of the bonds connecting system is found to be 75°C. T is one atom with the other? (Take, molecular (Given, room temperature = 30°C, specific weight of silver = 108 and Avogadro heat of copper = 0.1 cal/g°C) (2017) number = 6.02 × 1023 g mol−1) (2018) (a) 885°C (b) 1250°C (a) 6.4 N/m (b) 7.1 N/m (c) 825°C (d) 800°C (c) 2.2 N/m (d) 5.5 N/m 7. An external pressure p is applied on a 3. A granite rod of 60 cm length is clamped cube at 0°C so that it is equally at its middle point and is set into compressed from all sides. K is the bulk longitudinal vibrations. The density of modulus of the material of the cube and α granite is 2.7 × 103 kg/m3 and its Young’s is its coefficient of linear expansion. modulus is 9.27 × 1010 Pa. What will be the Suppose we want to bring the cube to its fundamental frequency of the original size by heating. The temperature longitudinal vibrations? (2018) should be raised by (2017) (a) 5 kHz (b) 2.5 kHz p 3α (a) αK (b) (c) 10 kHz (d) 7.5 kHz (c) 3pKα pK 4. An observer is moving with half the speed p of light towards a stationary microwave (d) 3αK source emitting waves at frequency 8. The temperature of an open room of 10 GHz. What is the frequency of the volume 30 m3 increases from 17°C to 27°C microwave measured by the observer? due to the sunshine. The atmospheric (speed of light = 3 × 108 ms−1) (2017) pressure in the room remains 1 × 105 Pa. (a) 12.1 GHz (b) 17.3 GHz If ni and nf are the number of molecules in the room before and after heating, then (c) 15.3 GHz (d) 10.1 GHz nf − ni will be (2017) 5. Cp and CV are specific heats at constant (a) 1.38 × 23 (b) 2.5 × 25 pressure and constant volume, 10 10 respectively. It is observed that (c) −2.5 × 25 (d) −1.61 × 23 10 10

2 Waves & Thermodynamics 9. A uniform string of length 20 m is 13. n moles of an ideal gas undergoes a suspended from a rigid support. A short process A and B as shown in the figure. wave pulse is introduced at its lowest end. The maximum temperature of the gas It starts moving up the string. The time during the process will be (2016) taken to reach the support is (Take, p g = 10 ms−2) (2016) (a) 2 π 2 s (b) 2 s 2 p0 A p0 B (c) 2 2 s (d) 2 s 10. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water, so that half V of it is in water. The fundamental V0 2V0 9p V frequency of the air column is now (2016) pV pV pV 00 9 3 9 f 3f (d) (a) (b) 00 00 00 nR (a) (b) (c) nR nR nR 4 2 2 2 4 14. A train is moving on a straight track with (c) 2f (d) f speed 20 ms−1. It is blowing its whistle at 11. A pendulum clock loses 12 s a day if the the frequency of 1000 Hz. The percentage temperature is 40°C and gains 4 s a day if change in the frequency heard by a the temperature is 20°C. The temperature person standing near the track as the at which the clock will show correct time, train passes him is close to (speed of and the coefficient of linear expansion α of sound = 320 ms−1) (2015) the metal of the pendulum shaft are, (a) 12% (b) 6% (c) 18% (d) 24% respectively. (2016) 15. Consider a spherical shell of radius R at (a) 25°C, α = 1.85 × −5 /° C temperature T. The black body radiation 10 (b) 60°C, α = 1.85 × −4 /° C inside it can be considered as an ideal gas 10 (c) 30°C, α = 1.85 × −3 /° C of photons with internal energy per unit volume u = U ∝ T 4 and pressure 10 V (d) 55°C, α = 1.85 × −2 /° C 10 12. An ideal gas undergoes a quasi static, p = 1 U  . If the shell now undergoes an reversible process in which its molar heat 3  V  capacity C remains constant. If during this process the relation of pressure p and adiabatic expansion, the relation between volume V is given by pV n = constant, then n is given by (Here, Cp and CV are T and R is (2015) molar specific heat at constant pressure and constant volume, respectively) (2016) (a) T ∝ e− R (b) T ∝ 1 (c) T ∝ e− 3R (d) T ∝ 1 R R3 (a) n = Cp CV 16. Consider an ideal gas confined in an (b) n = C − Cp isolated closed chamber. As the gas C − CV undergoes an adiabatic expansion, the (c) n = Cp − C C − CV average time of collision between molecules increases as V q, where V is the (d) n = C − CV C − Cp volume of the gas. The value of q is  γ = Cp     CV  (2015) 3γ + 5 γ+1 3γ − 5 γ −1 (a) (b) (c) (d) 62 62

Previous Years’ Questions (2018-13) 3 17. A solid body of constant heat capacity 21. One mole of diatomic ideal gas undergoes 1 J/°C is being heated by keeping it in contact with reservoirs in two ways a cyclic process ABC as shown in figure. (i) Sequentially keeping in contact with 2 The process BC is adiabatic. The reservoirs such that each reservoir supplies same amount of heat. temperatures at A, B and C are 400 K, (ii) Sequentially keeping in contact with 8 800 K and 600 K, respectively. Choose the reservoirs such that each reservoir supplies same amount of heat. correct statement. (2014) In both the cases, body is brought from p initial temperature 100°C to final temperature 200°C. Entropy change of B the body in the two cases respectively, is 800 K (2015) A 600 K 400 K C (a) ln2, ln2 (b) ln2, 2ln2 V (c) 2ln2, 8ln2 (d) ln2, 4ln2 18. A pipe of length 85 cm is closed from one (a) The change in internal energy in whole cyclic process is 250 R end. Find the number of possible natural (b) The change in internal energy in the process CA is 700 R oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The (c) The change in internal energy in the process AB is −350 R velocity of sound in air is 340 m/s. (2014) (a) 12 (b) 8 (c) 6 (d) 4 (d) The change in internal energy in the process BC is −500 R 19. Parallel rays of light of intensity I = 912 Wm−2 are incident on a spherical 22. A sonometer wire of length 1.5 m is made black body kept in surroundings of of steel. The tension in it produces an temperature 300 K. Take Stefan constant σ = 5.7 × 10−8 Wm−2 K−4 and assume that elastic strain of 1%. What is the the energy exchange with the fundamental frequency of steel if density surroundings is only through radiation. and elasticity of steel are 7.7 × 103 kg/m 3 and 2.2 × 1011 N/m2 respectively? The final steady state temperature of the (a) 188.5 Hz (2013) black body is close to (2014) (a) 330 K (b) 660 K (c) 990 K (d) 1550 (b) 178.2 Hz 20. Three rods of copper, brass and steel are (c) 200.5 Hz welded together to form a Y-shaped structure. Area of cross-section of each rod is 4 cm2. (d) 770 Hz End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 23. If a piece of metal is heated to 0°C. Lengths of the copper, brass and steel temperature θ and then allowed to cool in rods are 46, 13 and 12 cm respectively. a room which is at temperature θ0. The graph between the temperature T of the metal and time t will be closed to (2013) The rods are thermally insulated from T T (a) (b) θ0 surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 in CGS units, O tO t T T respectively. Rate of heat flow through (c) θ0 (d) θ0 copper rod is (2014) (a) 1.2 cal/s (b) 2.4 cal/s Ot Ot (c) 4.8 cal/s (d) 6.0 cal/s

4 Waves & Thermodynamics 24. The shown p-V diagram represents the 25. An ideal gas enclosed in a vertical thermodynamic cycle of an engine, cylindrical container supports a freely operating with an ideal monoatomic gas. moving piston of mass M. The piston and The amount of heat, extracted from the the cylinder have equal cross-sectional source in a single cycle is (2013) area A. When the piston is in equilibrium, 2p0 BC the volume of the gas is V0 and its p0 pressure is p0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system AD is completely isolated from its surrounding, the piston executes a simple V0 2V0 harmonic motion with frequency (2013) 1 Aγ p V Mp 1 0 00  13  (a) 2π V M (b) (a) p V (b) pV 2 π A2γ 00 00 0 2 1 A2 γp MV 0  11 1 (c) pV (d) 4p V 2 π MV 0 00 00 0 (d) 2π Aγ p (c) 0 2 Answer with Explanations 1. (c) For adiabatic process relation of temperature and ∴Wave speed, v = Y ρ volume is, T2V2γ −1 = T1V1γ −1 So, frequency f = v T2(2 V )2/ 3 300(V )2/ 3 λ ⇒ = [γ = 5 3 for monoatomic gases] ⇒ f = 1 Y 2L ρ 300 189 K ⇒ T2 = 2 2/ 3 ≈ 1 9.27 × 1010 60 × 2.7 × 103 Also, in adiabatic process, = 2 × 10−2 ∆Q = 0, ∆U = − ∆W 3 25 (300 ≈ 5000 Hz nR(∆T ) 2 3 f = 5 kHz or ∆U = − γ −1 = − 2 × × − 189) ≈ − 2.7 kJ 4. (b) As observer is moving with relativistic speed; formula ∆f = vradial , does not apply here. T2 ≈ 189 K, ∆U ≈ 2.7 kJ fc 2. (b) For a harmonic oscillator, Relativistic doppler’s formula is T = 2 π m, where k = force constant and T = 1 fobserved = factual ⋅  1 + v / c 1/ 2 kf  1 −    ∴ k = 4π2f2m v / c   22  2 108 × 10−3 Here, v 1 7 6.02 × 1023 c 2 = 4 × × (1012 )2 × = ⇒ k = 7.1 N / m  3 / 2  1 / 2 1 / 2 3. (a) λ So, fobserved = factual 4 ∴ fobserved = 10 × 3 = 17.3 GHz 5. (b) By Mayor’s relation, for 1 g mole of a gas, L Cp − CV = R So, when n gram moles are given, From vibration mode, Cp − CV = R λ = L ⇒ λ = 2L n 2

Previous Years’ Questions (2018-13) 5 As per given question, =2 π L(1+ α ∆θ) = 2 π L (1 + α 1 g  g  R ; for H2 ∆θ) 2 2 a = Cp − CV = b = Cp − CV = R ; for N2 ≈ T0 1+ α ∆θ  28 2 a = 14b ∴ ∆T = T ′−T0 = α ∆θT0 …(i) 2 6. (a) Heat gained (water + calorimeter) = Heat lost by copper ball or ∆T1 = α ∆θ1T0 ⇒ 12 = 40 − θ ∆T2 α ∆θ2T0 4 θ − 20 ⇒ mwsw∆T + mc sc ∆T = mBsB∆T ⇒ 3(θ − 20) = 40 − θ ⇒ 4θ = 100 ⇒ 170 × 1 × 45 + 100 × 0.1 × 45 ⇒ θ = 25°C = 100 × 0.1 × (T − 75) Time gained or lost is given by ∴ T = 885°C ∆T =  ∆T  t ≈ ∆t t   p ∆V p  T0 + ∆T  T0 7. (d) K = (− ∆V / ) ⇒ V = K V From Eq. (i), ⇒ −∆V = pV ⇒ pV = V(3α) ∆T ⇒ ∆T = p ∆T = α∆θ ⇒ ∆t = α (∆θ)t K K 3αK T0 2 2 8. (c) From pV = nRT = N RT 12 = α (40 − 25)(24 × 3600) ⇒ α = 1.85 × 10−5 /°C NA 2 We have, nf − ni = pVNA − pVNA 12. (b) ∆Q = ∆U + ∆W RTf RTi In the process pV n = constant, molar heat capacity is ⇒ nf − ni = 105 × 30 × 6.02 × 1023 .  1 − 1  given by 8.3 300 290 C = R 1+ R = CV + R = − 2.5 × 1025 γ− 1− n 1− n ∴ ∆n = − 2.5 × 1025 C −CV = R ⇒ 1− n = Cp − CV 9. (c) At distance x from the bottom 1− n C − CV  mgx  ⇒n = 1−  C p − C V  = (C − C V )− (C p − C V ) = C −Cp L   T= gx ⇒ dx =  C −CV  C −CV C −CV v= = xg µ  mL  dt 13. (a) p-V equation for path AB L x−1/ 2dx = t  x1 /2 L p= −  p0  V + 3 p0 ⇒ pV = 3 p0V − p0 V2  (1/ =  V0  V0 0 dt  2 0    ∫ ∫⇒ g ⇒ ) g ⋅t 0 1 or T = pV = nR  3 p0 V − p0 V 2  nR  V0  t =2 L ⇒t =2 20 2 2s  ⇒ g 10 = For maximum temperature, − 2 p0V 10. (d) Fundamental frequency of open pipe. dT = 0 ⇒ 3 p0 V0 = 0 dV f = v 2l 3 and p = 3 p0 − p0 3 p0 ⇒ V = 2 V0 V0 = 2 Now, after half filled with water it becomes a closed pipe Therefore, at these values : of length l .  3 p0   3V0  9 p0V0 2 2 2 4nR Fundamental frequency of this closed pipe, ∴ Tmax = = f′ = v = v = f nR 4(l /2 ) 2l 14. (a) Observer is stationary and source is moving. 11. (a) T0 = 2 π L During approach, f1 = f  v g   v − vs T ′ = T0 + ∆T = 2 π L + ∆L ⇒ T ′ = T0 + ∆T = 1000  320  = 1066.67 Hz g  20   320 −