1.In a constant volume process (A)Pressure does not depend on mass

1. 1.In a constant volume process (A)Pressure does not depend on mass (B)Pressure directly proportional to the Temp. (C)Pressure inversely proportional to the Temp. (D)None of these Ans.(B)Pressure directly proportional to the Temp.

2. 2.In the polytropic process PVn=C,if n=1 process is (A)Adiabatic (B)Isothermal (C)constant pressure (D)constant volume Ans. (B)Isothermal

3. 3. The ideal gas equation is (A)PV=RT (B) PV=MRT (C) PVn=RT (D) PVn=MRT Ans. (B) PV=MRT

4. 4. For isobaric process, n would be • 0 • 1 • 7 • infinity  Ans: (A)

5. 5. Work done in free expansion process is • 0 (B )maximum (C)Negative (D)minimum Ans. (A) 0

6. 6. In reversible polytropic process (A)true heat transfer occurs (B)the entropy remains constant (C) the enthalpy remains constant (D) the internal energy remains constant Ans. (A)true heat transfer occurs

7. 7. An isentropic process is always (A)irreversible and adiabatic (B) reversible and isothermal (C) frictionless and irreversible (D) reversible and adiabatic  Ans. (D) reversible and adiabatic 

8. 8. The net work done per kg of gas in a polytropic process is equal to • p1v1 loge v2/v1 (B)p1 (v1 – v2) (C) p1v1-pp2v2/n-1 (D)p2v1-pp2v2/n-1  Ans.(C) p1v1-p2v2/n-1

9. 9. A reversible process requires that • there be no heat transfer (B) temperature of system and surroundings be equal (C) there be no viscous or coulomb friction in the system (D) Heat transfer occurs from surroundings to system only.  Ans. (C) there be no viscous or coulomb friction in the system

10. 10. A process which is locus of all equilibrium point is called (A)reversible process (B)irreversible process (C)quasi-static process (D)poly-tropic process Ans. (D)poly-tropic process

11. 11.A process which is locus of all equilibrium point is called (A)reversible process (B)irreversible process (C)quasi-static process (D)poly-tropic process Ans.(D)poly-tropic process

12. 12. The processes or systems that do not involve heat are called • isothermal processes (B) equilibrium processes (C) thermal processes (D) adiabatic processes. Ans.(D) adiabatic processes

13. 13. In a reversible adiabatic process the ratio (T1/T2) is equal to (A)( p1/ p2)Ƴ-1/Ƴ (B)( p2/ p1)Ƴ-1/Ƴ (C)( v1/ v2) Ƴ (D) )( v1/ v2) Ƴ-1 Ans.(A)( p1/ p2)Ƴ-1/Ƴ

14. 14. In isothermal process • temperature increases gradually (B) volume remains constant (C) enthalpy change is maximum (D) change in internal energy is zero.  Ans. (D) change in internal energy is zero. 

15. 15. During throttling process • internal energy does not change (B) pressure does not change (C) entropy does not change (D) enthalpy does not change  Ans. (D) enthalpy does not change

16. 16. When a gas is to be stored, the type of compression that would be ideal is (A) isothermal (B) Adiabatic (C) Polytropic (D) constant volume   Ans. (A) isothermal

17. 17. For a reversible adiabatic process, the change in entropy is (A)Zero (B) Minimum (C) Maximum (D) infinite. Ans. (A)Zero

18. 18. For any irreversible process the net entropy change is (A)Zero (B) Positive (C) Negative (D) infinite.  Ans. (B) Positive

19. 19. In a reversible cycle, the entropy of the system • Increases (B) does not change (C) first increases and then decreases (D) depends on the properties of working substance Ans.(C) first increases and then decreases

20. 20. Which of the following is defined by second law of thermodynamics? • Internal energy (B)Enthalpy (C)Entropy (D)Heat  Ans.(C) Entropy

21. 21. The change of entropy, when heat is absorbed by the gas is (A)Positive (B) Negative (C) positive or negative. Ans.(A)Positive

22. 22. Which of the following statements is correct ? (A) The increase in entropy is obtained from a given quantity of heat at a low temperature (B) The change in entropy may be regarded as a measure of the rate of the availability of heat for transformation into work (C) The entropy represents the maximum amount of work obtainable per degree drop in temperature (D) All of the above.. Ans.(D) All of the above.

23. 23. The condition for the reversibility of a cycle is • the pressure and temperature of working substance must not differ, appreciably from those of the surroundings at any stage in the process (B) all the processes taking place in the cycle of operation, must be extremely slow (C) the working parts of the engine must be friction free (D) all of the above.  Ans. (D) all of the above

24. 24. The main cause for the irreversibility is • mechanical and fluid friction • unrestricted expansion (C) heat transfer with a finite temperature difference (D) all of the above.   Ans. (D) all of the above.

25. 25. Which of the following is defined by second law of thermodynamics? (A)Internal energy (B)Enthalpy (C)Entropy (D)Heat Ans.(C)Entropy

26. 26. A mono-atomic ideal gas (γ = 1.67,molecular weight = 40) is compressed adiabatically from 0.1MPa, 300 K to 0.2MPa. The universal gas constant is 8.314 kJkg−1mol−1K−1. The work of compression of the gas (in kJkg−1) is (A) 29.7 (B) 19.9 (C) 13.3 (D) 0 Ans. (A) 29.7 kJ/kg Gas constant = Universal Gas constant/molecular weight =8.314/40= 0.20785 kJ/kg K For adiabatic process, (T2/T1)=(P2/P1)γ-1/γ= (T2/300)=(0.2/0.1)1.67-1/1.67= 396 K Work done in adiabatic process is given by, =R(T1 - T2 ) /γ-1=0 20785(300-396)/(1.67-1) =− 29.7 kJ/kg

27. 27. Consider the following two processes ; (a) A heat source at 1200 K loses 2500 kJ of heat to a sink at 800 K (b) A heat source at 800 K loses 2000 kJ of heat to a sink at 500 K Which of the following statements is true ? (A)Process I is more irreversible than Process II (B) Process II is more irreversible than Process I (C) Irreversibility associated in both the processes are equal (D) Both the processes are reversible Ans. (B) Process II is more irreversible than Process I We know from the clausius Inequality, If T ɠdQ/T = 0, the cycle is reversible ɠdQ/T < 0, the cycle is irreversible and possible For case (a), ɠdQ/T=2500/1200-2500/800=- 1.041 kJ/kg For case (b), ɠdQ/T=2000/1200-2000/800=-1.5 So, process (b) is more irreversible than process (a)

28. 28. A frictionless piston-cylinder device contains a gas initially at 0.8MPa and 0.015m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ) during this process will be (A)8.32 (B) 12.00 (C) 554.67 (D) 8320.00  Ans. (A)8.32 P1=0.8 MPa, v1=0.015 m3,v2=0.030 m3,T= Constant We know work done in a constant temperature (isothermal) process W= P1 v1ln (v2/ v1)=(0.8*106)(0.015)ln(0.030/0.015) = 8.32 kJ

29. 29. 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by (A) −Rln2 (B) 0 (C) Rln2 (D) Rln4 Ans.(B) 0 T1= T2, p1= p2 Universal Gas constant = R Here given oxygen are mixed adiabatically So, dQ = 0 We know,ds=dQ/T,0/T=0

30. 30. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01m3. The maximum amount of work that could be utilized from the above process is (A) 0 kJ (B) 1 kJ (C) 2 kJ (D) 3 kJ   Ans.Pa=100kPa,Ps=300kPa, Δν=0.01m3 Net pressure work on the system, P= Ps - Pa= 300 − 100 = 200 kPa For constant pressure process work done is given by W = pΔν = 200*0.01 = 2 kJ

31. 31. A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to be heated. The changes in internal energy of the gas during the process is (A) −7000 kJ (B) −3000 kJ (C) +3000 kJ (D) +7000 kJ   Ans.(C) 3000 kJ W =− 5000 kJ (Negative sign shows that work is done on the system) Q =− 2000 kJ (Negative sign shows that heat rejected by the system) From the first law of thermodynamics, ΔQ = ΔW+ΔU So, ΔU = ΔQ −ΔW =− 2000 − (− 5000) = 3000 kJ

32. 32. A system at 500 K receives 7200 kJ/min from a source at 1000 K. The temperature of atmosphere is 300 K. Assuming that the temperatures of system and source remain constant during heat transfer find out The entropy produced during heat transfer (A) 7.2kJ/min-K. (B)5.3 kJ/min-K.(C) 6.2kJ/min-K. (D) 8.1 kJ/min-K. Ans.(A) 7.2kJ/min-K.  Net change of entropy : Change in entropy of the source during heat transfer = -Q/T1=-7200/1000=-7.2 kJ/min-K Change in entropy of the system during heat transfer =-Q/T2= 7200/500=14.4 kJ/min-K The net change of entropy, ΔS = – 7.2 + 14.4 = 7.2 kJ/min-K

33. 33. A closed system of constant volume experiences a temperature rise of 25°C when a certain process occurs. The heat transferred in the process is 30 kJ. The specific heat at constant volume for the pure substance comprising the system is 1.2 kJ/kg°C, and the system contains 2.5 kg of this substance. Determine The change in internal energy (A) 70kJ (B)65 kJ (C) 75 kJ (D)55 kJ Ans.(C)75 kJ Δ U = mʃcv dT=2.5*1.2*(25) =75 kJ

34. 34. Air at 1.02 bar, 22°C, initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate :The final temperature (A)234.24 ̊ C(B)245.39 ̊ C (C)285.11̊ C (D)239 ̊ C Ans.(A) 234.24°C. (T2/T1)= (P2/P1)γ-1/γ = (T2/295) = (6.8/1.02)1.4-1/1.4 =507.24 ̊ K Final temperature = 507.24 – 273 = 234.24°C.