Title 13. KINETIC THEORY - Explanations.pdf ✅

1 (b) vrms ∝ √T; To double the rms velocity temperature should be made four times, i. e., T2 = 4T1 = 4(273 + 0) = 1092K = 819°C 2 (a) Rate of cooling proportional to (T 4 − T0 4 ), as per Stefan’s Law. ∴ R ′ R = (900) 4 − (300) 4 (600) 4 − (300) 4 = 9 4 − 3 4 6 4 − 3 4 = 3 4(3 4 − 1) 3 4(2 4 − 1) = 80 15 = 16 3 R ′ = 16 3 R 3 (b) Thermal energy corresponds to internal energy Mass=1 kg Density = 4 kg m−3 Volume = Mass Density = 1 4 m3 Pressure = 8 × 104 Nm−2 ∴ Internal energy = 5 2 p × V = 5 × 104 J 4 (c) CV = R (γ − 1) ⇒ γ = 1 + R CV = 1 + R 3 2 R = 5 3 5 (d) Let initial conditions = V, T And final conditions = V ′ , T′ By Charle’s law V ∝ T [P remains constant] V T = V ′ T ′ ⇒ V T = V ′ 1.2T ′ ⇒ V ′ = 1.2V But as per question, volume is reduced by 10% means V ′ = 0.9V So percentage of volume leaked out = (1.2 − 0.9)V 1.2V × 100 = 25% 6 (a) ∵ θ1 < θ2 ⇒ tan θ1 < tan θ2 ⇒ ( V T ) 1 < ( V T ) 2 Form PV = μRT; V T ∝ 1 P Hence ( 1 P ) 1 < ( 1 P ) 2 ⇒ P1 > P2 7 (c) vr.m.s. is independent of pressure but depends upon temperature as vrms ∝ √T 8 (c) In the absence of intermolecular forces, there will be no stickness of molecules. Hence, pressure will increase. 9 (d) Oxygen being a diatomic gas possesses 5 degrees of freedom, 3 translational and 2 rotational. Argon being monoatomic has 3 translational degrees of freedom. Total energy of the system = Eoxygen + Eargon = n1f1 ( 1 2 RT) + n2f2( 1 2 RT) = 2 × 5 × 1 2 RT + 4 × 3 × 1 2 RT = 5RT + 6RT = 11RT 10 (c) In Vander Waal’s equation (P + a V2 ) (V − b) = RT a represents intermolecular attractive force and b represents volume correction 11 (d) ∆Q = KA ( ∆T ∆x ) ∆t, where A = 4 π r 2 = 0.008× 4 × 22 7 (6 × 108) 2 × ( 32 105 ) × 86400 = 1018 cal 12 (c) Let for mixture of gases, specific heat at constant volume be CV CV = n1 (CV)1+n2 (CV)2 n1+n2 where for oxygen; CV1 = 5R 2 , n1 = 2 mol For helium; CV2 = 3R 2 , n2 = 8 mol Therefore, CV = 2×5R 2 +8× 3R 2 2+8 = 17R 10 = 1.7 R 13 (c) From the Mayer’s formula Cp − CV = R ...(i) and γ = Cp CV ⇒ γCV = Cp ...(ii) Substituting Eq. (ii) in Eq. (i) we get ⇒ γCV − CV = R CV(γ − 1) = R CV = R γ−1 14 (b) vrms ∝ √T 15 (b) Average speed or mean speed of gas molecules v̅= √ 8RT πM or v̅∝ 1 √M or v̅H v̅He = √ MHe MH Here, MHe = 4 MH ∴ v̅H v̅He = √ 4 1 = 2 or v̅He = 1 2 v̅H 16 (c) At constant temperature; PV = constant
⇒ P × ( m D ) = constant ⇒ P D = constant = K. [D = Density] 17 (a) For a gas, PV = μRT = m M RT For graph A, PV = m M RT Slope of graph A, ( P T ) = m M R V ...(i) For graph B, PV = 3m M RT Slope of graph B, ( P T ) = 3m M R V ...(ii) Slope of curve B Slope of curve A = 3m M R V m M R V = 3 1 18 (d) Mean kinetic energy of any ideal gas is given by E = f 2 RT which is different gases. (f is not same for all gases) 19 (c) At constant volume P1 T1 = P2 T2 ⇒ T2 = ( P2 P1 ) T1 ⇒ T2 = ( 3P P ) × (273 + 35) = 3 × 308 = 924K = 651°C 20 (d) Vrms = √ 3RT M % increase in Vrms = √ 3RT2 M −√ 3RT1 M √ 3RT1 M × 100% = 20 − 17.32 17.32 × 100 = 15.5% 21 (b) P = μRT V = mRT MV (μ = m M ) So, at constant volume pressure-versus temperature graph is a straight line passing through origin with slope mR MV . As the mass is doubled and volume is halved slope becomes four times. Therefore, pressure versus temperature graph will be shown by the line B 22 (b) Thermal capacity = Mass × Specific heat Due to same material both spheres will have same specific heat. Also mass = Volume (V) × Density (ρ) ∴ Ratio of thermal capacity = m1 m2 = V1ρ V2ρ = 4 3 πr1 3 3 4 πr2 3 = ( r1 r2 ) 3 = ( 1 2 ) 3 = 1 8 23 (b) In isothermal changes, temperature remains constant 24 (b) RMS velocity is given by v = √ 3kT m or v 2 = 3kT m For a gas, k and m are constants. ∴ v 2 T =constant 25 (d) Root mean square velocity of gas molecules vrms = √ 3RT M or vrms ∝ 1 √M or vO3 vO2 = √ MO2 MO3 Here, MO2 = 32, MO3 = 48 ∴ vO3 vO2 = √ 32 48 = √2 √3 26 (c) For carbon dioxide, number of moles (n1 ) = 22 44 = 1 2 ; molar specific heat of CO2 at constant volume CV1 = 3 R For oxygen, number of moles (n2 ) = 16 32 = 1 2 ; molar specific heat of O2 at constant volume CV2 = 5R 2 . Let TK be the temperature of mixture. Heat lost by O2 = Heat gained by CO2. n2CV2 ∆T2 = n1CV1 ∆T1 1 2 ( 5 2 R) (310 − T) = 1 2 × (3R)(T − 300) Or 1550−5T = 6T − 1800 Or T = 304.54K = 31.5°C 27 (d) We know vs = √ γP ρ and vrms = √ 3P ρ ∴ vrms vs = √ γ 3 28 (a) PV = μRT = m M RT ⇒ PV T ∝ 1 M [∵ M = molecule mass]
From graph ( PV T ) A < ( PV T ) B < ( PV T ) C ⇒ MA > MB > MC 29 (a) For one g mole; average kinetic energy = 3 2 RT 30 (d) P ∝ T ⇒ P2 P1 = T2 T1 = (273 + 100) (273 + 0) = 373 273 ⇒ P2 = 760 × 373 273 = 1038mm 31 (a) According to Avogadro’s hypothesis 34 (b) Vander Waal’s gas equation for μ mole of real gas (P + μ 2a V2 ) (V − μb) = μRT P = ( μRT V − μb − μ 2a V2 ) Given equation, P = ( RT 2V − b = a 4b 2 ) On comparing the given equation with this standard equation, we get μ = 1 2 Hence , μ = m M ⇒ mass of gas, m = μM = 1 2 × 44 = 22g 35 (b) Here , m = 0.1 kg, h1 = 10m, h2 = 5.4 m c = 460 J-kg −1°C−1 , g = 10ms −2 , θ =? Energy dissipated, Q = mg(h1 − h2 ) = 0.1 × 10(10 − 5.4) = 4.6j J From Q = c m θ θ = Q cm = 4.6 460 × 0.1 = 0.1°C 36 (c) Since P and V are not changing, so temperature remains same 37 (d) (∆Q)P = μCP∆T ⇒ 207 = 1 × CP × 10 ⇒ CP = 20.7 Joule mol − K . Also CP − CV = R ⇒ CV = CP − R = 20.7 − 8.3 = 12.4 Joule mole − K So, (∆Q)V = μCV∆T = 1 × 12.4 × 10 = 124 J 38 (a) 22 g of CO2 is half mole of CO2 ie, n1 = 0.5 16 g of O2 is half mole of O2 ie, n2 = 0.5 ∴ T = n1T1 + n2T2 n1 + n2 = 0.5×(27+273)+0.5(37+273) 0.5+0.5 = 305 K = 305−273 = 32°C 39 (d) Let T0 be the initial temperature of the black body ∴ λ0T0 = b (Wien’s law) Power radiated, P0 = CT0 4 , where, C is constant. If T is new temperature of black body, then 3λ0 4 T = b = λ0T0 or T = 4 3 T0 Power radiated, P = CT 4 = CT0 4 ( 4 3 ) 4 P = P0 × 256 81 or P P0 = 256 81 40 (c) ( ∆Q ∆t ) inner + ( ∆Q ∆t ) outer = ( ∆Q ∆t ) total K1πr 2 (T2 − T1) l + K2π[(2r) 2 − r 2 ](T2 − T1) l = Kπ(2r) 2(T2 − T1 ) l or (K1 + 3K2 ) πr 2(T2−T1 ) l = Kπ 4 r 2(T2−T1) l ∴ K = K1 + 3K2 4 41 (a) vrms = √ 3RT M ⇒ T ∝ M ⇒ THe TH = MHe MH ⇒ (273 + 0) THe = 2 4 ⇒ THe = 546K = 273°C 42 (d) Average kinetic energy E = f 2 kT Sinec f and T are same for both the gases so they will have equal energies also 43 (a) Average kinetic energy E = f 2 kT = 3 2 kT ⇒ E = 3 2 × (1.38 × 10−23)(273 + 30) = 6.27 × 10−21J = 0.039eV < 1 eV 44 (a) From Boyle’s law pV = constant ∴ p1V1 = p2V2 Here, p1 = (h + l), V1 = 4 3 πr 3 p2 = l, V2 = 4 3 π(3r) 3 ∴ (h + l) 4 3 πr 3 = l × 4 3 π(3r) 3 or h + l = 27l ∴ h = 26l 45 (a) CP − CV = R = 2. cal g − mol − K
Which is correct for option (a) and (b). Further the ratio CP CV (= γ) should be equal to some standard value corresponding to that of either, mono, di, or triatomic gases. From this point of view option (a) is correct because ( CP CV ) mono = 5 3 46 (a) E = 3 2 RT ⇒ E ′ E = T ′ T = 400 300 = 4 3 = 1.33 47 (c) Eav = f 2 kT = 3 2 × 1.38 × 10−23 × 273 = 0.56 × 10−20J 48 (b) CP − CV = R = Universal gas constant 49 (b) E = f 2 RT; f = 5 for diatomis gas ⇒ E = 5 2 RT 50 (c) Molar specific heat at constant pressure CP = 7 2 R Since, CP − CV = R ⇒ CV = CP − R = 7 2 R − R = 5 2 R ∴ CP CV = (7/2)R (5/2)R = 7 5 51 (a) From ideal gas equation, we have pV = nRT ∴ n = pV RT Given, p = 22.4 atm pressure = 22.4 × 1.01 × 105 Nm−2 , V = 2L = 2 × 10−3 m3 , R = 8.31 J mol −1 − K −1 , T = 273 K ∴ n = 22.4 × 1.01 × 105 × 2 × 10−3 8.31 × 273 n = 1.99 ≈ 2 Since, n = Mass Atomic weight We have, mass = n × atomic weight = 2 × 14 = 28 g 52 (d) Vander Waal’s gas constant b = 4 × total volume of all the molecules of the gas in the enclosure Or b = 4 × N × 4 3 π ( d 2 ) 3 = 2 3 πNd 3 = 2 3 × 3.14 × 6.02 × 1023 × (2.94 × 10−10) 3 = 32 × 10−6 m3 mol 53 (d) vrms = √ 3RT M ⇒ vrms ∝ 1 √M 54 (b) Temperature becomes 1 4 th of initial value [1200K = 927°C → 300K = 27°C] So, using vrms ∝ √T. r. m. s. velocity will be half of the initial value 55 (b) Vrms = √ 3RT M ⇒ (Vrms )O2 (Vrms )H2 = √ TO2 TH2 × MH2 MO2 ⇒ (Vrms )O2 (Vrms )H2 = √ 900 300 × 2 32 = √3 4 ⇒ (vrms )O2 = 836m/s 56 (c) Here, ∆l = 80.3 − 80.0 = 0.3 cm l = 80 cm, α = 12 × 10−6°C−1 Rise in temperature ∆T = ∆l lα ∆T = 0.3 80 × 12 × 10−6 = 312.5°C 57 (c) vrms = √ 3RT M ⇒ T ∝ M [∵ vrms , R → constant] ⇒ TO2 TN2 = MO2 MN2 ⇒ TO2 (273 + 0) = 32 28 ⇒ TO2 = 312K = 39°C 58 (d) From Maxwell’s velocity distribution law, we infer that vrms > v > vmp ie, most probable velocity is less than the root mean square velocity. 59 (c) PV = m M RT ⇒ V ∝ mT ⇒ V1 V2 = m1 m2 . T1 T2 = 2V V = m m2 × 100 200 ⇒ m2 = m 4 60 (a) We treat water like a solid. For each atom average energy is 3kBT. Water molecule has three atoms, two hydrogen and one oxygen. The total energy of one mole of water is U = 3 × 3kBT × NA = 9RT [∵ kB = R NA ] ∴ Heat capacity per mole of water is C = ∆Q ∆T = ∆U ∆T = 9R 61 (b) As the temperature increases, the average velocity increases. So the collisions are faster 62 (d) We have vrms = √ 3RT M ; at T = T0(NTP)