Question
Gas 'A' undergoes change from state ' X ' to state ' Y '. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state ' X ' by another process during which 6 J of heat is evolved. In the reverse process of ' Y ' to ' X ',
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Solution
Sol. [Key Concept: First law $\Delta\ U = q + w$; use $\Delta\ U_{Y \rightarrow \ X} = - \Delta\ U_{X \rightarrow \ Y}$ to find the reverse-process work.]
For process $X \rightarrow Y$
$${w = - 18\text{ }J }{q = + 10\text{ }J }{\Delta U = q + w }{\Delta U_{X \rightarrow Y} = 10 + ( - 18) = - 8\text{ }J }$$For reverse process:
$${\Delta U_{Y \rightarrow X} = 8\text{ }J }{q = - 6\text{ }J }{\Delta U = q + w }{8 = - 6 + w }{w = 14\text{ }J }$$
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