A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :
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- 10 J of the work will be done by the gas.
- 6 J of the work will be done by the gas.
- 10 J of the work will be done by the surrounding on gas.
- 6 J of the work will be done by the surrounding on gas.
The Correct Option is D
Solution and Explanation
\(\Delta U_{A\to B} = q + w = 5 - 8 = -3\) J.
Internal energy is a state function, so \(\Delta U_{B\to A} = +3\) J.
For B to A, heat is evolved: \(q = -3\) J.
\(\Delta U = q + w \Rightarrow 3 = -3 + w \Rightarrow w = +6\) J. Positive work means 6 J of work is done on the gas by the surroundings.
Answer: (4) 6 J of the work will be done by the surrounding on gas.
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