Thermodynamics Questions and Answers – Maximum Work in a Reversible Process

«
»

This set of Thermodynamics Multiple Choice Questions & Answers focuses on “Maximum Work in a Reversible Process”.

1. For a process from state 1 to state 2, heat transfer in a reversible process is given by
a) Q for reversible=(To)*(S1-S2)
b) Q for reversible=(To)*(S2-S1)
c) Q for reversible=(To)/(S1-S2)
d) Q for reversible=(To)/(S2-S1)
View Answer

Answer: b
Explanation: To is the temperature of the surroundings and S1,S2 are the entropies at state 1 and 2 respectively and ΔS(universe)=0.
advertisement

2. For a process from state 1 to state 2, heat transfer in an irreversible process is given by
a) Q for irreversible=(To)*(S1-S2)
b) Q for irreversible>(To)*(S1-S2)
c) Q for irreversible<(To)*(S1-S2)
d) none of the mentioned
View Answer

Answer: c
Explanation: To is the temperature of the surroundings and S1,S2 are the entropies at state 1 and 2 respectively ans ΔS(universe)>0.

3. Which of the following is true?
a) Q for reversible > Q for irreversible and work for reversible < work for irreversible
b) Q for reversible < Q for irreversible and work for reversible > work for irreversible
c) Q for reversible < Q for irreversible and work for reversible < work for irreversible
d) Q for reversible > Q for irreversible and work for reversible > work for irreversible
View Answer

Answer: d
Explanation: This is because, Q for reversible=(To)*(S2-S1) and Q for irreversible<(To)*(S1-S2).
advertisement
advertisement

4. Work done in all reversible processes is equal.
a) true
b) false
View Answer

Answer: a
Explanation: Reversible processes between the same end states must coincide and and produce equal amounts of work.

5. In an open system, for maximum work, the process must be entirely
a) irreversible
b) reversible
c) adiabatic
d) none of the mentioned
View Answer

Answer: b
Explanation: A reversible process gives the maximum work.
advertisement

6. Which of the following is true for a steady flow system?
a) mass entering = mass leaving
b) mass does not enter or leave the system
c) mass entering can be more or less than the mass leaving
d) none of the mentioned
View Answer

Answer: a
Explanation: For a steady flow process, mass entering the system is equal to the mass leaving the system.

7. Which of the following is true for a closed system?
a) mass entering = mass leaving
b) mass does not enter or leave the system
c) mass entering can be more or less than the mass leaving
d) none of the mentioned
View Answer

Answer: b
Explanation: For a closed system mass does not change.
advertisement

8. Which of the following is mostly neglected while doing calculations for finding maximum work?
a) KE
b) PE
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: c
Explanation: The changes in KE and PE are very small, hence they are neglected.

9. The work done by a closed system in a reversible process is always ___ that done in an irreversible process.
a) less than or more than
b) equal to
c) less than
d) more than
View Answer

Answer: d
Explanation: A reversible process always produces maximum work.
advertisement

10. The proof that work done in all reversible processes is same can be done by violating Kelvin-Planck statement.
a) true
b) false
View Answer

Answer: a
Explanation: During the proof, we end up violating the Kelvin-Planck statement.

11. A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01m^3. A constant pressure process gives 54 kJ of work out. Find the final volume of the air.
a) 0.05 m^3
b) 0.01 m^3
c) 0.10 m^3
d) 0.15 m^3
View Answer

Answer: c
Explanation: W = ∫ P dV = PΔV
ΔV = W/P = 54/600 = 0.09 m^3
V2 = V1 + ΔV = 0.01 + 0.09 = 0.1 m^3.

12. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine the total heat transferred to the air.
a) 747 kJ
b) 757 kJ
c) 767 kJ
d) 777 kJ
View Answer

Answer: c
Explanation: Qin – Wout = ΔU = m(u3 –u1)
m = P1V1/RT1 = 0.697 kg
u1 = [email protected] 300 K = 214.36 kJ/kg
u3 = [email protected] 1400 K = 1113.43 kJ/kg
Therefore Qin = 767 kJ.

13. A piston cylinder contains 0.5 kg of air at 500 kPa and 500 K. The air expands in a process so pressure is linearly decreasing with volume to a final state of 100 kPa and 300 K. Find the work in the process.
a) 56.1 kJ
b) 66.1 kJ
c) 76.1 kJ
d) 86.1 kJ
View Answer

Answer: d
Explanation: Work = ⌠ PdV = (1/2)(P1 + P2)(V2 – V1)
V1 = mR T1/ P1 = 0.5 × 0.287 × (500/500) = 0.1435 m^3
V2 = mR T2/ P2 = 0.5 × 0.287 × (300/100) = 0.4305 m^3
W = (1/2)(500 + 100)(0.4305 – 0.1435) = 86.1 kJ.

Sanfoundry Global Education & Learning Series – Thermodynamics.
To practice all areas of Thermodynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter