Tiêu đề 7-gaseous-state--thermodynamics-.pdf ✅

GASEOUS STATE & THERMODYNAMICS 1. (B) Apply gas equation PV = nRT to identify decrease or increase in temp and volume for each process 2. (A) As the system is thermally insulated, ΔQ = 0 Further as here the gas is expanding against vacuum (surroundings), the process is called free expansion and for it, ΔW = ∫ PdV = 0 [ As for vacuum P = 0] So in accordance with first law of thermodynamics, i.e. ΔQ = ΔU + ΔW, we have 0 = ΔU + 0, i.e. ΔU = 0 or U = constant 3. (A) ab and cd are adiabatic bc and da are isothermal. Hence PbVb = PcVc PdVd = PaVa PcVc γ = PdVd γ PaVa γ = PbVb γ } ⇒ Pb Pc = Pa Pd 4. (A) For adiabatic process (from A to B) ΔW = −322J ΔU = −ΔW = 322J For another process (from A to B) ΔQ = ΔU + ΔW 100J = 322J + ΔW ⇒ ΔW = −222J 5. (B) C = Cv + RT V dV dT ; R = 3R 2 + RT V dV dT ; dV V + 1 2 dT T = 0 ⇒ VT 1/2 = constant 6. (C) δ = ΔV VΔT from PV = nRT ⇒ PΔV + VΔP = nRΔT Since ' P ' is constant ⇒ PΔV = nRΔT ⇒ δ = ( nR P ) ⋅ 1 T 7. (B) At constant pressure Q = nCpΔT; also Q = ΔU + w ⇒ nCpΔT = nCVΔT + w ⇒ n(Cp − CV)ΔT = w ⇒ nRΔT = 25 ΔU = n 6 2 RΔT = 75 J&Q = 25 + 75 = 100 J 8. (B) Q = ΔU + W ⇒ C = CV + RT V dV dT ⇒ C = CV − R αV (1 − αV) 9. (D) U = 2 × ( 5R 2 ) T + 4 ( 3R 2 ) T = 11RT 10. (A) Sina WA > WB and ΔU is same in both process ⇒ ΔQA > ΔQB 11. (B) PV γ = constant ⇒ P ργ = constant ⇒ P ′ p = (32) 7/5 12. (B) U = a + bPV = a + nbRT ⇒ dU dT = nbR But dU dT = nR γ−1 ⇒ nbR = nR γ−1 ⇒ γ = b+1 b 13. (C) Along a to b, Δu = 0 and along b to c, Δu = −w = −4j ⇒ a to c, Δu = 0 + (−4) = −4J (for any path) 14. (A) √ γRT M = √ 3RT M ⋅ (0.68) ⇒ γ = 1.3872, f = 2 γ−1 ≈ 5.16 15. (C) 2 P0V0 T0 = PV1 T1 + PV2 T0 and V1 + V2 = 2V0 V1 = x1 A, V2 = x2 A ⇒ x1 = 44 cm and x2 = 40 cm 16. (ABC) For adiabatic process : TiVi γ−1 = TfVf γ−1 For isothermal process : PiVi = PfVf 17. (ABCD) 18. (ABD) In case of cyclic process : η = 1 − QR Qg QR = Heat rejected ; Qq = Heat given 19. (ABD)from the graph we can conclude that Wgiven process > Wisothermal process For AB process P = −kV + C ⇒ T = − ( k nR) V 2 + CV
Which is the equation of parabola 20. (CD) For A, D T1Va γ−1 = T2Vd γ−1 ⇒ ( Va Vd ) γ−1 = T2 T1 For B, C T1Vb γ−1 = T2Vc γ−1 ⇒ ( Vb Vc ) γ−1 = T2 T1 ⇒ Vb Vc = ( T2 T1 ) 1 γ−1 So ( Va Vd ) = ( Vb Vc ) , i.e., i.e., choices C and D are correct. Hence choice A and B are wrong. 21. (CD) P = −mV + c ⇒ T = − ( m nR)V 2 + ( c nR)V ⇒ T3 > T1 = T2 ∴ 1 → 3 → expansion process accompanied by heating 3 → 2 → expansion process accompanied by cooling 22. (ABD)Degree of freedom of He = 3 Degree of freedom of H2 = 5 Average degree of freedom = 3×2+5×2 2+2 = 4 Now γ = 1 + 2 f = 3 2 For adiabatic process ΔQ = 0 and PV γ = cost or TV γ−1 = cost From first law ΔU = Δw [∵ ΔQ = 0] 23. (AD) This process is free expansion so ΔQ = ΔU = ΔW = 0 and P1V1 = P2V2 Only initial and final points are defined, process in between is not defined. 24. (AD) As, Wby = Q = 0, So U = constanct so 2 × 5 2 R × T0 + 3 × 3R × 2T0 = (2 × 5 2 R + 3 × 3R) T 23RT0 = 14RT ⇒ T = 23T0 14 By n1+n2 1−γmix = n1 1−γ1 + n2 1−γ2 ⇒ λmix = 19 14 25. (ABC) ΔQ = ΔU + ΔW For process A to B, ΔQ = ΔW For process B to C, ΔQ = ΔU For process C to D, ΔU = −ΔW For process D to A, ΔU = −ΔW 26. (BD) P 2 ρ = k ⇒ P 2RT PM = k ⇒ PT = ( kM R ) P 2 ρ = P ′2 ρ ⇒ P ′ = P √2 Hence from eq. (i) T ′ = T√2 PT = constant hence P − T curve is a hyperbola. 27. (AC) PV = 1RT (α − βV 2 )V = RT ; T = αV R − βV 2 R ; dT dV = α R − 3βV 2 R For maximum value of T; dT dV = 0 ⇒ V = √ α 3β 28. (AB) Work done in the process A to B = −nCVΔT nCV (TA − TB) = 2 × ( 3 2 ) × 8.314 × 150 = 3741 J For the process B to C : PB TB = PC TC ⇒ TC = PC PB TB = 425 K TC = PC PB TB = 425 K; ΔQ = nCVΔT = 2 × ( 3 2 ) × 8.314 × (425 − 850) = −10600 J 29. (ACD) (a) Process AB : PT = constant ⇒ nRT 2 V = constant ; W = ∫A B PdV = ∫A B Constant T dV ⇒ dV dT = 2nRT Constant ; W = ∫300 100 Constant T ⋅ 2nRT Constant dt ⇒ PA PB = TB TA ⇒ 1 3 = TB TA ⇒ TB = 300 3 = 100 ⇒ W = 2nR(100 − 300) ⇒ WAB = −400nR (b) Process CA : Isochoric P T constant : TA TC = PA PC ; TA TC = PA PB ⇒ TA TC = 1 3 ⇒ TC = 3TA TC = 900R ⇒ ΔU = nCVΔT = (1) 3 2 R × (TA − TC ) = 3 2 R × (300 − 900) ⇒ |ΔU| = (900R) (c) Process BC : Isobaric Q = nCPΔT
⇒ Q = (1) 5 2 R × (TC − TB) ; Q = 5 2 R × (900 − 100) ⇒ Q = 5 2 R × 800 ⇒ Q = 2000R 30. (AD) Point A and C are on the same line passing through origin ⇒ PA VA = PC VC Also TA = 200K = PAVA nR and also TC = 1800K = PCVC nR ⇒ PAVA PCVC = 1 9 From eq. (i) and (ii) VA VC = 1 3 31. (C) Vrms = √ 3RT M 32. (D) P = n V RT = 50 10−6×6.03×1023 × 8.314 × 20 = 1.4 × 10−14 N/m2 33. (B) Heat gain by left part = heat lost by right part ⇒ 3 2 nR(T − T0 ) = 3 2 nR(2T0 − T) ⇒ T = 3T0 2 Let final pressure = p ⇒ p T = P0 T0 ⇒ P = 3P0 2 34. (C) Heat flow = 3 2 nR ( 3T0 2 − T0) = 3 4 nRT0 = 3 4 P0 V0 2 = 3 8 P0V0 35. (C) P (final in left) = 3P0 2 = P (final in Right part). So when pin is removed, piston will not move. 36. (C) 37. (C) Q1 = nCΔT − n ( 11R 2 ) (√2T0 − T0) Q2 = nCpΔT = n ( 7R 2 ) (T0 − √2T0) Q3 = nRTln ( Vf Vi ) = −nRT0ln 2 2 η = Total work done by the system Positive energy supplied to the system For cyclic process, total work done (W) = total energy ( Q = nRT0 2 [4(√2 − 1) − ln 2] 1ln RT0 2 (√2 − 1) = 4(√2 − 1)ln 2 11(√2 − 1) = 4(0.40) − 0.7 11(0.40) = 1.6 − 0.7 4.4 = 9 44 × 100% = 20.5% 38. (A) 39. (D) p1V1 T1 = (p1+ kx A )(V1+Ax) T2 ; kx 2 + (p1A + kV1 A ) x + (p1V1 − p1V2T2 T1 ) = 0 2000x 2 + 4600x − 480 = 0 ; x = 0.1 m Wgas + Watmosphere + Wspring = 0 ; Wgas − PeAx − 1 2 kx 2 = 0 Wgas = peAx + 1 2 kx 2 = 310 J ; Q = ΔU + peAx + 1 2 kx 2 ; ΔU = f 2 NkΔT ΔU = f 2 p1V1 T1 ⋅ ΔT = 5 2 105Pa × 0.024 m3 300 K ⋅ 60 K = 1200.0 J 40. (D) 41. (C) 42. (D) (1 to 3) Since U = aV α ΔU = CvΔT = (aα)V (α−1)ΔV Where ΔV is the change in volume during the process or ΔU = α [ aV α V ] ΔV = αU V ΔV This gives ΔU U = α⋅ΔV V Work done during the process is ΔW = P ⋅ ΔV ; i.e., ΔW = P ⋅ [ V⋅ΔU Uα ] = PV α [ ΔU U ] = ΔU α [ PV U ] But PV = nRT = nCv(γ − 1)T ; = (γ − 1)[nCvT] = (γ − 1)U Hence ΔW = ( γ−1 α ΔU) Therefore ΔQ = ΔU + ΔW = ΔU [1 + γ−1 α ] Let C be the molar specific het in the process.
ΔQ = CΔT = ΔU [1 + γ − 1 α ] = CvΔT [1 + γ − 1 α ] ⇒ C = Cv + Cv(γ − 1) α = ( R γ − 1 ) + R α 43. [A − q, r, s;B − r; C − r; D − s] 44. [A − q;B − p; C − s; D − qr] Temperature at : J = 30 × 10 nR , K = 10 × 10 nR , L = 20 × 10 nR , M = 20 × 20 nR For JK: W = 0; ΔU < 0 ⇒ ΔQ < 0 For KL: W = 10 × 10; ΔU > 0 ⇒ ΔQ > 0 For LM: W = 0; ΔU > 0 ⇒ ΔQ > 0 For MJ: W < 0; ΔU < 0 ⇒ ΔQ < 0 45. [A − r;B − p; C − s; D − q] W = Area of triangle = 1 2 × 2P0 × V0 = P0V0 For CA: W = −P0V0 ⇒ ΔU = 3 2 R [ P0V R − 2P0V0 R ] = − 3 2 P0V0 & ΔQ = −P0V0 − 3 2 P0V0 = −5 2 P0V0 For BC: W = 1 2 [3P0 + P0 ]V0 = 2P0V0 ⇒ ΔU = 3 2 R [ 2P0V R − 3P0V0 R ] = − 3 2 P0V0 & ΔQ = P0V0 2 Maximum temperature will be for process BC. For BC: P = − 2P0 V0 V + 5P0 Using gas equation : T = − ( 2P0 V0R )V 2 + ( 5P0 R ) V By using maxima/minima : Tmax = 25 8R P0V0 46. (2) For gas in A, P1 = ( RT M ) mA V1 and P2 = ( RT M ) mA V2 ∴ ΔP = P1 − P2 = ( RT M ) mA ( 1 V1 − 1 V2 ) Putting V1 = V and V2 = 2V We get ΔP = RT M mA 2V Similarly for Gas in B, 1.5ΔP = ( RT M ) mB 2V From eq. (I) and (II) we get 2mB = 3mA 47. (7) For cylinder A. For cylinder B dQ = nCPdT dQ = nCVdT nCPdT = nCVdT ∴ dT = CP × 30 CV = 30 × 1.4 = 42 K 48. (1) Volume of the gas is constant V = constant ∴ P ∝ T ∴ P = 2P0 i.e. pressure will be doubled if temperature is doubled Now let F be the tension in the wire. Then equilibrium of any one piston gives F = (P − P0 )A = (2P0 − P0 )A = P0A 49. (5) Let T be the temperature of the mixture. Then U = U1 + U2 or f 2 (n1 + n2 )RT = f 2 (n1 )RT0 + f 2 (n2 )(R)(2T0 ) or (2 + 4)T = 2T0 + 8T0 (n1 = 2, n2 = 4) or T = 5 3 T0 50. (2) Process A to C Q = 210 J Work done WAC = area under AC = (10 × 4) + ( 10×4 2 ) = 60 J From Ist law of thermodynamics. ΔU = Q − WAC ⇒ UC − UA = 210 − 60 ∴ UC = UA + 150 = 30 + 150 = 180 J 51. (3) Path A to B UB = 60 J ∴ ΔU = Q − WAB UB −UA = Q − 0 ; 60− 30 = Q ∴ Q = 30J 52. (6) 5 2 nRΔT = 1 2 n × Mu 2 ⇒ ΔT = Mu 2 5R = 30×10−3×104 5×R = 60 R = 10 R × 6 ⇒ x = 6 53. (100) Process is polytropic C = Cv − R m−1 R 2 = 3 2 R − R m − 1 ⇒ m = 2 PV 2 = C 40 × V0 2 = P(2V0 ) 2 P = 10kPa PV T = P0V0 T0 ⇒ 10 × 2V0 T = 40 × V0 200 ⇒ T = 100 K