# Thermodynamics Questions and Answers – Maximum Work in a Reversible Process

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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)

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

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

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

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

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

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
d) none of the mentioned

Explanation: A reversible process gives the maximum work.

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

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

Explanation: For a closed system mass does not change.

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

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

Explanation: A reversible process always produces maximum work.

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

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

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

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

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.

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