Title 13. Thermodyanmaics Medium Ans.pdf ✅

1. (a): In thermal equilibrium the macroscopic variables like pressure, volume, temperature, mass and composition do not change with time 2. (d): The given statement is zero the law the thermodynamics it was formulated by R.H. Fowler in 1931. 3. (a): The internal energy of ideal gas depends only upon temperature of gas not on other factors. 4. (d): from the given initial state A to final state B. change in internal energy is same all the four cases, as it is dependent of the path from A to B 5. (b): According to first law of thermodynamics Q=U +W t W t U t Q   +   =    Here , 120W, t Q =   1 80 − =   Js t W 1 120 80 40 − = − =    Js t U 6. (b): Work done, W =PV J 4 5 3 2 10 2 10 (150 50) 10 =  =  −  − 7. (c): In adiabatic process or dU dW dQ dU dW =− =0 + =0 In the process of compression, work is done on the gas therefore dW is negative. Hence dU is positive i.e. internal energy of gas increase and therefore temperature of the gas also increase 8. (d): U =Q−W The internal energy is independent of path 9. (c): Work done on the system is always taken as negative 10. (b): since the gas is compressed adiabatically, then dQ=0 and dW =−150J From first law of thermodynamics dU dW J dQ dU dW  = − = −(−150) =150 = + 11. (c): As no work is done, therefore, W =0 According to first law of thermodynamics T R Q U W U nC T n V          −  = + = =  = 1 Here, 3 5 n =2,  = R R Q T T T K 100 300 1 3 5 2 2 1 (373 273) 100  =       −  =   = − = − = 12. (d): In adiabatic process Q = constant, then Q=0 and specific heat 0 0 =  =   = m T m T Q s 13. (a): 14. (b): here, CV R 2 3 = Since Cp −CV =R Cp CV R R R R 2 5 2 3  = + = + = 15. (c): Cp −CV =R 16. According to Mayer’s relation Cp −CV =R or p p V C R C C 1− = Or 1 R orC C 1 R or C C C 1 R 1 p p V p p  −  = =   −         =  =  − Specific heat capacity molecularweight molar heatcapacity = Specific heat capacity at constant pressure ( −1)  = M R 17. (c): molar specific heat of mixture at constant volume is R R R n n nC n C C V V V mixture 1.7 2 8 2 3 8 2 5 2 ( ) 1 2 1 1 2 2 = +  +  = + + =
18. (a): In a process x PV = constant, molar heat capacity is given by x R R C − +  − = 1 1 As the process is = V P constant i.e. = −1 PV constant, therefore, x= −1. for an ideal monatomic gas, 3 5  = R R R R R C 2 2 2 3 1 ( 1) 1 3 5 = + = − − + −  = 0 0 2 0 Q=nC(T)=1(2R)(2T −T )= RT 19. (c): An intensive property is that which does not depend on the quantity of matter or mass of they system, Refractive index is an intensive property. Volume , mass and weight are extensive properties. 20. (c): In the given P-V diagram pressure remains constant although volume increase hence the process is an isobaric process 21. (a): PV =cosnstant  22. (c): The given diagram shows that the curves move away from the origin at higher temperature. 23. (b): Useful work done per hour J 6 = 9.510 Heat absorbed per hour J 7 =6.210 Amount of heat wasted per hour = Heat absorbed per hours – useful work done per hour J J J J 7 7 6 6 5.25 10 6.2 10 9.5 10 10 (62 9.5) =  =  −  = − 24. (b): Total work done by the gas from D to E to F is equal to the area of DEF.  the area of DEF = DF  EF 2 1 Here, DF change in pressure =600− 200 1 400 − = Nm Also, EF = change in volume 3 3 3 =7m −3m =4m Area of DEF 400 4 800J 2 1  =   = Thus, the total work done by the gas form D to E to F is 800 J. 25. (c): During isothermal process internal energy of system always remains constant hence change in it will be zero everywhere during the process. 26. (a): For isothermal process, PV= constant Differentiating both side V p dV dP pdV Vdp or − + = 0 = ...... (i) Again for adiabatic process PV =Constant  Again differentiating both side 0 1 +  =  − dpV V dV p Or =−   V P dV dP  slope of adiabatic curve =   slope isothermal curve 27. (d): in a cyclic process, the system returns to its initial state. Since internal energy is a state variable U =0 for a cyclic process. 28. (a): Let an ideal gas goes isothermally from its initial state ( ) P1V1 to the final state ( ) P2V2 then work done  = 2 1 V V W PdV Taking V nRT PV =nRT or p= Then,   = = = 2 1 2 1 1 2 ln V V V V V V nRT V dV dV nRT V nRT W 29. (d): 30. (b): here V AT BT P 2 − = Or 2 PV = AT − BT Since p is a constant PdV =(A− 2BT)dT ..........(i) Work done = pdV ( 2 )   ( sin ( )) 2 1 2 1 2 W A BT dT AT BT u g i T T T T   = − = − ( ) ( ) 2 1 2 =A T2 −T1 − B T2 −T 31. (c): For adiabatic expansion 1 2 2 1 1 1 − − TV =TV 206 2 1 273 1 1.4 1 2 1 2 1  =      =         = − − V V T T J T T R dU dW 1373 (273 206.89) (1.4 1) 1 8.31 ( ) 1 1 2 = − −  − =  − = = 32. (a): for an adiabatic process,
 PV = constant   P1V1 =P2V2 Where subscripts 1 and 2 represent the initial and final states respectively. Or              =           =         = 1 1 1 2 2 1 1 2 n V V P P Or ( ) P2 =P1n =n P P1 =P    33. (d) 150 J of heat has been added to the gas. 34. (b): In the adiabatic process  PV = constant (K) If an ideal gas is changed from state ( ) P1V1T1 to state ( ) P2V2T2 adiabatically, then work done.    = = 2 1 2 1 V V V V V dV W PdV K       − −  =       −  + = − − −+ 1 1 1 2 1 1 1 1 1 2 1 V V V K K V V [ ] 1 1 1 1 1 2 2 1 1 1 1 1 1 2 2 2 PV PV V PV V PV W − −  =      − −  = −  −  [ ] ( 1) ( ) ( ) 1 1 1 1 2 2 2 1 2 2 1 PV nRT and PV nRT nR T T T T nR = =  − − − = −  =  35. (c): In isometric process volume is constant P T 36. (c): Here, dV ( PV RT ).....(i) V RT dW =PdV =  = Given T dT 3 2 V KT dV K 2 / 3 −1/ 3 =  = ..... ( ) 3 3 2 2 2 / 3 1/ 3 ii T dT KT K T dT V dV  = = − Form (i)  = 2 1 T T V dV W RT  = 2 1 ( sin ( )) 3 2 T T u g ii T dT RT W R T T R 60 40R 3 2 ( ) 3 2  = 2 − 1 =  = 37. (a): Type of processes Feature Isothermal Temperature constant Isobaric Pressure constant Isochoric Volume constant Adiabatic Q=0 38. (a): At constant pressure W P V P (V V )  =  =  f − i ..... (i) For an idea gas PV =nRT Or PVf =nRT and PVi =nRTi Form (i) W nR(T T ) ..... (ii)  = f − i For constant temperature, PV = constant         =         = f i i f f f i i P P V V P V PV or f i P P W =nRT ln ...... (iii) So work done for path AB, BC, CD and DA respectively will be WAB =nR(Tf −Ti ) =2 R(400− 300) =200R (using (ii)) ln 2 R 400ln 2 800Rln 2 P P W nRT f i BC =   =          = (using (iii)) WCD =nR(Tf −Ti ) =2 R[300− 400] = −200R (using (ii)) 600 ln 2 2 1 ln 2 R 300ln R P P W nRT f i DA = −      =            = (using (iii)) Hence, the work done in the complete cycle W =WAB +WBC +WCD +WDA =200R+800Rln2−200R−600Rln2=200Rln2 39. (b): since in an adiabatic process 1 2 2 1 1 1 − − TV =TV       − −       =         = 1 3 5 1 2 1 2 1 4 2 300 V V V V T T [ 2 4 5/ 3] V1 = V andV2 = V formonoatomicgas  = 188.98 189K 2 300 2 1 300 2 / 3 2 / 3  = =       = 40. (a): as slope of curve B is more than slope of A therefore curve A is showing isothermal process and curve B is showing adiabatic process.
41. (d): According to an ideal gas equation PV =nRT Or P nRT V = (Given) T a P= ..... (i) a nRT V 2  = .....( ) 2 dT ii a nRT dV = Work done by the gas, dW = PdV Or  = T T dT u g i and ii a nRT T a W 4 ( sin ( ) ( )) 2  nRT nRT T 2 T 6 4 = = 42. (b): in an adiabatic operation   P1V1 =P2V2            =         =           = 1 2 2 2 1 1 1 2 d d m d d m V V P P (32) (2 ) 2 128 7 / 5 5 7 / 5 7 1 2  = = = = P P 128 1 2  = P P 43. (d): since gas is suddenly expanded it means the process is adiabatic process, then 1 2 2 1 1 − − TiV =T V Putting 1 2 8 1 T =273+ 20 = 293K,V = V 1 2 1 1 1 (293)( ) (8 ) − − V =T V 1 293 28 − =T       = =  = − − 3 5 8 293 8 293 1 3 T2 1 5  T 73.25K 4 293 (2 ) 293 8 293 2 2 / 3 3 2 / 3 = = = = 44. (b): since DA involves compressing of gas from VD to VA at constant pressure PA . The work will be done on gas. ( ) ( )( ) 2 3 ( ) ( ) 2 3 A A D A A D A D A A D D A A D DA DA DA P V V P V V P P P V P V P V V Q U W = − + − = = − + −  = +  ( ) 2 5 ( ) 2 5 =− PA VD −VA = PA VA −VD 45. (b): In an adiabatic expansion. PV= constant V RT P= (for 1 mole of gas)  =        V V RT constant − − =  1 1 TV constant of T V As 1.5 2 3 2 1 1 2 1 1 1  = + = =   −  =− V T 46. (a): 47. (b): The efficiency  =1 is maximum efficiency, it shows that the heat engine is 100% efficient which is practically not possible. 48. (d): Here, 1 2 1 1 T T  = − Or 1 2 1 2 T T 1 4 1 T T 0.25=1−  = − 4 3 4 1 1 T T 1 2 = − = According to question 2 1  = 2 and T2 = T2 − 58oC 1 2 1 2 T T 58oC 2 1 1 T (T 58oC) 1 4 1 2 −  − = −   = − T 232oC T 58 2 1 4 3 T 58o T T 2 1 1 1 1 1 1 = = −  − =  = 49. (b) Here, T 750K;T 350K; 1 = 2 = W =1.25kJ =1250J Now efficiency 15 8 750 400 750 350 1 T T 1 1 2  = − = − = = As , Q W 1  = 15 2343.75J 2.34kJ 8 W 1250 Q1 =  = =   = 50. (a) 1 1 2 1 2 T T T T T 1 −  = − =