Q132 / 180 · KCET 2025 · Physics

A gas is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these two paths are $\mathrm{Q}_1$ and $\mathrm{Q}_2$ and $\mathrm{W}_1$ and $\mathrm{W}_2$ respectively. Then

A
$\mathrm{W}_1=\mathrm{W}_2$
Correct answer
B
$\mathrm{Q}_1-\mathrm{W}_1=\mathrm{Q}_2-\mathrm{W}_2$
C
$\mathrm{Q}_1+\mathrm{W}_1=\mathrm{Q}_2+\mathrm{W}_2$
D
$Q_1=Q_2$

Solution

$\begin{aligned} & \Delta \mathrm{U}=\mathrm{Q}-\mathrm{W} \text { (By Ist law of thermodynamics) } \\ & \Delta \mathrm{U}_1=\Delta \mathrm{U}_2 \\ & \mathrm{Q}_1-\mathrm{W}_1=\mathrm{Q}_2-\mathrm{W}_2\end{aligned}$

Hence, the correct answer is option: 'B'.

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