Tiêu đề 12.THERMODYNAMICS - Explanations.pdf ✅

1 (c) The efficiency of Carnot engine η = Work output Heat input = W QH or η = QH−QL QH = 1 − QL QH Also we can show that QL QH = TL TH ∴ η = 1 − TL TH where TL is temperature of sink and TH is temperature of hot reservoir. According to question 1 5 = 1 − TL TH ...(i) and 1 3 = 1 − TL−50 TH ...(ii) From Eq. (i) TL TH = 4 5 ⇒ TH = 5 4 TL Substituting value of TH in Eq. (ii), we get 1 3 = 1 − TL − 50 5 4 TL or 4(TL − 50) 5TL = 2 3 or TL − 50 = 2 3 × 5 4 TL or TL − 5 6 TL = 50 ∴ TL = 50 × 6 = 300 K 2 (a) Q2 = 2000 cal. As COP = Q2 W ∴ 4 = 2000/W W = 500 cal = 500 × 4.2 = 2100 J 3 (c) ∆Q = ∆U + ∆W = 167 + 333 = 500 cal 4 (c) Work done by the gas (as cyclic process is clockwise) ∴ ∆W = Area ABCD So from the first law of thermodynamics ∆Q (net heat absorbed) = ∆W =Area ABCD As change in internal energy in cycle ∆U = 0 5 (a) Heat absorbed by the system at constant pressure Q = ncp∆T Change in internal energy ∆U = ncv∆T W = Q − ∆U ∴ W Q = Q − ∆U Q = 1 − ∆U Q = 1 − ncv∆T ncp∆T = 1 − cv cp = (1 − 1 γ ) 6 (c) PV γ = constant : Differentiating both sides PγV γ−1dV + V γdP = 0 ⇒ dP P = −γ dV V 7 (a) From symmetry considerations and also from theory, Va Vd = Vb Vc 8 (b) ∆W = P∆V = 103 × 0.25 = 250 J 9 (d) Work done at constant temperature (ie, isothermal process), W = 2.3nRT log10 ( V2 V1 ) = 2.3 × 10 × 8.31 × 500 × log10 ( 5 50) = −9.6 × 104 J 10 (b) In isothermal process, heat is released by the gas to maintain the constant temperature 11 (c) Work done during isothermal process in expanding volume of gas fromV1 to V2 is given by W = ∫ p dV V2 V1 = ∫ ( nRT V ) dV V2 V1 (as p = nRT V ) = nRT ∫ dV V (as T = V2 V1 constant) = nRT loge ( V2 V1 ) For expansion of 1 mole of gas, ie, n=1 W = RT loge ( V2 V1 ) 12 (a)
As isothermal at T1 is farther from the origin than the isothermal at T2, therefore, T1 > T2 14 (b) ∆Q = ∆U + ∆W ⇒ ∆U = ∆Q − ∆W = Q − W [using proper sign] 15 (c) dU = dQ − dW = mL − p(dV) = 1 × 540 − 1.013 × 105(1671 − 1)10−6 4.2 = 540 − 40 = 500 cal 16 (d) Eφ = γP = 1.4 × (1 × 105) = 1.4 × 105N/m2 17 (c) Internal energy (∆U) does not depend upon path. It depends only on initial and final states 18 (a) Ideal gas possess only kinetic energy 19 (d) TV γ−1 = constant ⇒ T2 = T1 ( V1 V2 ) γ−1 = 927°C 20 (c) In isothermal process temperature remains constant 21 (b) Input energy = 1g s × 2kcal g = 2kcal/s Output energy = 10 kW = 10 K J/S = 10 4.2 kcal/ s ⇒ η = output energy input energy = 10 4.2×2 > 1, it is impossible 22 (d) (i)Curve OA represents isobaric process (since pressure is constant). Since, the slope of adiabatic process is more steeper than isothermal process. (ii)Curve OB represents isothermal process. (iii)Curve OC represents adiabatic process. (iv)Curve OD represents isochoric process. (since volume is constant). 23 (a) For every gas, Cp − Cv = R ∴ x = y 24 (b) From FLOT ∆Q = ∆U + ∆W ∵ Heat supplied to the system so ∆Q → Positive and work is done on the system so ∆W → Negative Hence +∆Q = ∆U − ∆W 25 (a) In isothermal process, compressibility Eθ = P 26 (c) Ka = γp = − ∆p ∆V/V ∴ ∆V V = − ∆p γp 27 (a) For adiabatic process P1 V1 γ = p2V2 RT1 V1 V1 γ = RT2 V2 V2 γ T1V1 γ−1 = T2V2 γ−1 28 (a) In isothermal process temperature remains constant i. e. , ∆T = 0. Hence according to C = Q m∆T ⇒ Ciso = ∞ 29 (c) Given that, the temperature of freezer, T2 = −13°C ⟹ T2 = −13 + 273 = 260 K Coefficient of performance, β=5 The coefficient of performance is defined as, β = T2 T1 − T2 or 5 = 260 T1 − 260 ∴ T1 − 260 = 260 5 or T1 − 260 = 52 or T1 = (52 + 260)K = 312 K or T1 = (312 − 273)°C ⇒ T1 = 39°C 30 (a) Here, for hydrogen Cp − Cv = m = R 2 Or R = 2m And for nitrogen, Cp − Cv = n = R 28 or R = 28 n ∴ 2m = 28n m = 14n 31 (d) As indicator diagram if all the three cases are closed curves, representing cyclic changes, therefore, U = const and ∆U = 0 in all the cases 32 (c) PV γ = K or PγV γ−1dV + dP. V γ = 0
⇒ dP P = −γ dV V or dP P × 100 = −γ ( dV V × 100) = −1.4 × 5 = 7% 33 (a) η = T1 − T2 T1 = (127 + 273) − (87 + 273) (127 + 273) = 400 − 360 400 = 0.1 → 10% 34 (b) Q1 Q2 = CVdT p dT = 3 2 R R = 3 2 that is 60: 40. 36 (a) Here, Q1 = 200cal,Q2 = 150cal, T1 = 400K As Q1 Q2 = T1 T2 ∴ T2 = Q2 Q1 × T1 = 150 200 × 400 = 300K 37 (b) Gain of entropy of ice S1 = ∆Q T = mL T = 80 × 100 (0 + 273) = 8 × 103 273 cal/K Loss of entropy of water = S2 = − ∆Q T = − mL T = 80 × 100 (273 + 50) = 8 × 103 323 cal/K Total change of entropy S1 + S2 = 8 × 103 273 − 8 × 103 323 = +4.5 cal/K 38 (b) Efficiency of a heat engine η = 1 − T2 T1 or 30 100 = 1 − 77 + 273 T1 or 350 T1 = 1 − 30 100 = 7 10 ⇒ T1 = 500K or 227°C 39 (b) As work done in process = area under the curve, which increases continuously 40 (a) ∆S = ∆Q T = 80 × 1000 273 = 293 cal/K 41 (b) Work done at constant pressure is W = p ∆V = nR ∆T Where p is presure, ∆V the volume change, R the gas constant, ∆T the change in temperature and n the number of moles. Given, n=1, T2 = 127°C = 400K, T1 = 27°C = 300K, R = 8.14 J/mol − K ∴ W = 1 × 8.14 × (400 − 300) W=814 J 42 (a) TV γ−1 = constant ∴ T1 T2 = ( V2 V1 ) γ−1 or ( 1 2 ) γ−1 = √ 1 2 ∴ γ − 1 = 1 2 or γ = 3 2 ∴ PV 3/2 = constant 43 (b) Let the initial pressure of the three samples be PA, PB and PC, then PA (V) 3/2 = (2V) 3/2P, PB = P and PC (V) = P(2V) ⇒ PA: PB:PC = (2) 3/2 : 1: 2 = 2√2: 1: 2 44 (d) Work done = Area under curve = 6P1×3V1 2 = 9 P1V1 45 (a) Here, T1 = 927°C = (927 + 273)K = 1200K T2 = 27°C = (27 + 273)K = 300K As U ∝ T ∴ ∆U U2 = U1 − U2 U2 = 1200 − 300 300 × 100 = 300% 46 (d) In adiabatic process ∆Q = 0 Therefore, first law of thermodynamics becomes dU+dW=0 47 (a) For heat engine, Q2 Q1 = T2 T1 ⇒ Q2 = T2Q1 T1 = 375 × 600 500 = 450 J 48 (a) ∆Q = ∆U + ∆W ⇒ ∆U = ∆Q − ∆W = 6 × 4.18 − 6 = 19.08kJ ≈ 19.1kJ 50 (a) The efficiency of heat engine is given by η = W Q = 1 − Q2 Q1 = 1 − T2 T1 where T1 is temperature of source and T2 is temperature of sink. Given, η1 = 1 6 , η2 = 1 3 ∴ 1 6 = T1−T2 T1 ...(i) and 1 3 = T1−(T2−62) T1 ...(ii)
Solving Eqs. (i) and (ii), we get T1 = 372 K and T2 = 310 K 51 (a) dQ = 2kcal = 200cal = 2000 × 4.2 J = 8400 J dW = 500 J, dU = dQ − dW = 8400 − 500 = 7900 J 52 (c) Work done in expansion = Cp − Cv = R Joule 53 (d) The change in entropy of an ideal gas ∆S = ∆Q T ...(i) In isothermal process, there is no change in internal energy of gas ie, ∆U = 0 ∴ ∆U = ∆Q − W ⇒ 0 = ∆Q − W ⇒ ∆Q = W ∴ ∆Q =work done by gas in isothermal process which went through from (p1,V1, T) to (p2,V2, T) or ∆Q = nRTloge ( V2 V1 ) ...(ii) For 1 mole of an ideal gas, n=1 So, from Eqs.(i) and (ii), we get ∆S = Rloge ( V2 V1 ) = R In ( V2 V1 ) 54 (d) For adiabatic process pV γ = constant ⇒ P2 P1 = ( V1 V2 ) γ ∴ P2 P = ( 800 100) 5/3 ⇒ P2 = 32 P 55 (a) Curve IV is parallel to volume axis. It represents isobaric curve. Out of II and III, slope of III is smaller. Therefore, III curve represents an isothermal curve 57 (d) Volume of the ideal gas is constant so W = P∆V = 0 Using FLOT ∆Q = ∆U ⇒ ∆U = i 2Rt = 1 2 × 100 × 5 × 60 = 30 × 103 = 30kJ 58 (b) Volume of the gas is constant V = constant ∴ P ∝ T i. e., pressure will be doubled if temperature is doubled ∴ P = 2P0 Now let F be the tension in the wire. Then equilibrium of any one piston gives F = (P − P0 )A = (2P0 − P0 )A = P0A 59 (a) As gas is suddenly expanded so it is an adiabatic process, ie, pV γ = constant or p1V1 γ = p2V2 γ Given, V2 = 3V1, CV = 2R ∴ CP = 2R + R = 3R ⇒ γ = CP CV = 3R 2R = 1.5 ∴ P1 P2 = ( V2 V1 ) γ = (3) 1.5 = 5.1 ≈ 5 60 (a) T2 = 27°C = (27 + 273)K = 300 K, η = 25% = 1 4 We know that, η = 1 − T2 T1 ⇒ 1 4 = 1 − 300 T1 or 300 T1 = 1 − 1 4 ⇒ 300 T1 = 3 4 or T1 = 300 × 4 3 ⇒ T1 = 400 k or T1 = (400 − 273)°C = 127°C 61 (b) η = 1 − T2 T1 = 1 − 500 800 = 1 − 600 x ∴ 3 8 = 1 − 600 x 600 x = 1 − 3 8 = 5 8 PA P0A F