Tiêu đề 13. Thermodynamics hard Ans.pdf ✅

1. (c) P0V0 = nRT If temperature is changed to T without changing the pressure then Po V = nRT  P0(V -V0) = nRT(T-T0) =>P0 AV = n R AT using nR = Now thermal coefficient of volume expansion at constant pressure 2. (b) As the body is in thermal equilibrium with the environment then it must radiate at constant rate, that is equal to rate of absorption so as to keep its temperature constant. Thus E  t. 3. (c) If M0 is molecular mass of the gas then for initial condition PV= .RT ...(1) After 2M mass has been added P'. ...(2) By dividing (2) by (1) P' = 3P 4. (c) From freezing point to boiling point, celsius adn keivin scaies have 100 divisions whereas Fahrenheit scale has 180 divisions. Therefore lesser amount of heat is associated with 1°F then 1°C or 1K which have equal spacing. 5. (a) Vmp and T3 > T1 = T2>T4 =T5 (1 and 2 are on same isotherm also 4 and 5 are on same isotherm) 6. (b) Vrms 2 = = vrms = .v0. 7. (c) n radiation, body may be at lower temperature than surrounding but still transfer heat to the surrounding which is at higher temperature. 8. (b) Due to the pressure difference, there would be systematic motion within the gas according to the Newton's law. But as systematic motion is not possible, pressure must be uniform inside. To make the pressure uniform inside, distribution of the molecules is changed. 9. (b) According to KTG, for an ideal gas there exists no force of interaction between the molecules hence no potential energy can be defined associated with the interaction. As for (a) sum of kinetic energies of two colliding molecules is conserved. Also total sum of KE of individual molecules is internal energy of the gas which is not altered by collisions between the molecules. When a molecule collides with a wall it can exchange KE only if their temperatures are different. Collisions of the molecules with the wall are elastic if their temperatures are same. 10. (c) Object II remains in liquid state for maximum time, receiving energy at constant rate. It takes maximum energy to get vapourized. 11. (d) For cylinder A. For cylinder B dQ = nCPdT1 dQ = nCvdT2  nCPdT1 = nCvdT2 From (I) and (II) cvdT2=(cv +R)30  v v 2 c (c R)30 dT + = For diatomic gas R 2 5 cv =  dT2 = 42K . 12. (b) Crms = M 3RT or Crms  T ( ) ( ) 1 2 T1 rms T2 rms T T C C = = 100 400 = 2 (Crms) T2 = 2(Crms) T1 = 2  v = 2v 13. (a) V RT M rms = 3  2.0078x10 kg (1930) 3x8.31x300 V 3RT M 3 2 2 rms − = = = = 2.00 gm It is molecular weight of hydrogen (H ) 2 . 14. (d) From wine’s Law m T constant. Where T is temperature of black body and  m is wavelength corresponding to maximum energy of emission. Energy distribution of blackbody radiation is given below : 1. U1 and U3 are not zero because a blackbody emits nearly radiation’s of all wavelengths. 2. Since U1 corresponds lower wavelength and U3 corresponds higher wavelength and U2 corresponds medium wavelength. Hence U2  U1 .
15. (d) Po = 3 1 2 rms v v mn       P = 3 1 2 rms v v m n         where m = 2m, 2 v v rms  rms = putting the value  rms rms 2 rms 2 rms o 4 v v m 2m mv m v P P  =   = = 2 1 P = Po/2 16. (a) U = nCvT, Q = nCPT nC T nC T U Q v p   =   =  = 7/5 [for diatomic gas  = 7/5] 17. (d) dQ = n CPdT = n (Cv + R) dT dQ = n CvdT + n RdT n CPdT = n CvdT + n RdT n CvdT increases internal energy.  Fraction of energy that increases internal energy is P v C C = 7 5 Hence, D is correct Answer. 18. (c) Average degree of freedom fav = 5 11 2 3 2 3 1 5 = +  +  mix = 1 + av f 1 = 1 + 11 5 = 11 16 = 1.4 Hence, C is correct Answer. 19. (d) For cylinder A. For cylinder B dQ = nCPdT1 dQ = nCVdT2 = n(CV+R) dT1  nCVdT2 = n(CV+R) 30  v v 2 C (C R)30 dT + = For diatomic gas R 2 5 Cv =  dT2 = 42K . Hence (D) 20. (c) For gas in A, 1 A 1 V RT M m P       = 2 A 2 V RT M m P       =           −       = − = 1 2 2 1 A V 1 V 1 m M RT P P P Putting V1=Vand V2 =2V We get 2V m M RT P A        = Similarly for Gas in B, 2V m M RT 1.5 ΔP B       = From eq. (I) and (II) we get 2 mB = 3 mA 21. (d) KT 2 3 KE = , M 3RT Vrms = i.e. 2 T T KE KE 1 2 1 2 = =  21 2 1 KE 2KE 2x6.21x10− = = = 12.42  10-21J 1 2 rms,1 rms,2 T T V V = = 2  Vrms,2 = 2xVrms,1 = 684 m/s. Hence (D) is correct 22. (b) Rate of cooling  difference in temperature dT dt    dT dt = K in first case dT = 61− 59 = 2  = 60 − 30 = 30 dt = 4 minute  60 1 30x4 2 dt dT K = =  = For second case dT = 2  = 50 − 30 = 20  6min . x20 2 K dT dt 60 1 = =  = 23. (d) r 2 2 r P1V1 =P V ; r 2 1 2 1 V V P P         =
P(8) P(8) 32P 3 5 r = = = Hence (D) is correct. 24. (b) L1 = 1 [1 + 1 T] L2 = 2 [1 + 2 T] L1 – L2 = (1 - 2) + (T) (1 1 - 2 2) If L1 – L2 is to be independent of T Then 11 - 22 = 0 1 2   =     Hence (B) is correct. 25. (b) Work done = PV = 10 0 25 = 250 3 x . J Hence (B) is the correct answer. 26. (b) Slope of the curve dV dP = V P = − . As curve 2 is steeper, its  is greater. Also,  = 5/3 for monoatomic gas and  = 7/5 for diatomic gas. Hence (B) is the correct answer. 27. (b) T/R = 2(90`– T)/R, T=60 where R is thermal resistance of one rod. Hence (B) is the correct answer. 28. (b) When the pressure is first increased by 10% it becomes (110/100) P. when it is decreased by 10% from there, the pressure becomes 100 100 110 90   P = 100 99 P Thus the pressure decreases by 1% volume is increased by nearly 1% 29. (c) For a mole of an ideal gas, the equation of state is PV = RT or T = R PV which is proportional to the product pV At x , PV = (4 × 105 ) (1 × 10–4 ) = 40 Nm At y, pV = (1 × 105 ) (5 × 10–4 ) = 50 Nm At z, pV = (1 × 105 ) (1 × 10–4 ) = 10 Nm Thus, T is maximum at y since pV is the highest and T is minimum at z since pV is the smallest PV = RT 30. (b) dQ = – dU C = –CV = – 1 –R  = – 1 R  + + n P dT dV – 1 2R dT dV n P –  = T 5V = const. V = 5 T const. dT dV = – 5 6 T const PV = nRT P/n = RT/V + 5 T const. RT ×       6 T const – 5 = – 1 2R  2 5 = – 1 1    – 1 = 2/5  = 7/5 adiabatic compressibility  = P 1  = 7P 5 31. (c) Q = U + W W = 0 since volume is constant Q = U Vrms = 3RT / M Umix =U1 + U2 Umix = n1 v1 C T + n2 2 Cv T = (n1 + n2) (Cv)mix T (Uf) – (Ui) = nCv(T2 – T1) = (n1 + n2) (Cv)mix (T2–T1) Vrms = 2Vrms  T = 4T n1 = 4 8 = 2 ; n2 = 28 14 = 1/2 (Cv)mix = (n n ) n Cv n Cv 1 2 1 1 2 2 + + Uf – Ui = (n1 v1 C + n2 2 Cv ) (T2 – T1) =        +  2 5 2 1 2 3 2 R [1200 – 300] 4 17 × R × 100 = 3825 R 32. (b) 1 → 2 = isochoric 2 → 3 = isobaric 3 → 4 = isochoric
4 → 1 = isobaric 33. (a) U = nCv T = n R 2 5 (Tf – Ti) n = 1 1 1 RT P V U = 1 1 1 RT P V × 2 5 R (T2 – T1) = 390 300 10 0.08 3   × 2 5 (320 – 390) = 39 15 10 0.08 5 3    × (–70) = – 39 0.40 × 15 × 103 × 70 = – 39 40 ×150 × 70 = – 39 40 ×10500 U = –11000 Joule 34. (b) Work done is area under the curve W = – PoVo = –8 × (105 × 1.01) × 7 × 10–3 W = – 56 × 1.01 × 102 Joule W = –5656 Joule 35. (b) Heat is a path function. Heat transfer depends on process. Hence heat transfer is different for different paths between same initial & final status. 36. (c) QAB = nCPT = n 2 5 R(2T0 – T0) QBC = WBC = NR.2T0 ln 2  BC AB W Q = 4ln 2 5 37. (a) WAB = PV = (10) (2 – 1) = 10 J WBC = 0 From first law of thermodynamics Q = W + U U = 0 (process ABCA is cyclic)  Q = WAB + WBC + WCA  WCA = Q – WAB – WBC = 5 – 10 – 0 = – 5 J 38. (b) Internal energy change is U = n CV T = n 2 5 R (T – 0) = 2 5 nRT 39. (d) For adiabatic process : Slope : dT dP =           –1 T P  = dT dP 3 5 =       5 / 3−1 5 / 3 T P = 2 / 3 5 / 3 T P = 2.5 T P  = dT dP 5 7 =       2 / 5 7 / 5 T P = 3.5 T P  = dT dP 3 4 =       1/ 3 4 / 3 T P = 4 T P 40. (b) PA Kx Kx P = A 2Kx = 1 2100 × 2 1 = 100 N/m2 41. (b) U = U0V  nCVT = U0V  T  V isobaric process dT dV n P C = CV + dT dV = constant n P = V nRT = constantT nRT C = CV + constant R C = CV + R = 2 5 R + R = 2 7 R 42. (a) Isochoric process dV = 0 W = 0 Isobaric : W = PV = nRT adiabatic : W = 1 nR(T T ) i f  − −