KCET · Physics · Thermodynamics
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 \(\mathrm{W}_1=\mathrm{W}_2\)
- 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 \(Q_1=Q_2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{Q}_1-\mathrm{W}_1=\mathrm{Q}_2-\mathrm{W}_2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{U}=\mathrm{Q}-\mathrm{W} \text { (By Ist law of thermodynamics) } \\ & \Delta \mathrm{U}_1=\Delta \mathrm{U}_2 \\ & \mathrm{Q}_1-\mathrm{W}_1=\mathrm{Q}_2-\mathrm{W}_2\end{aligned}\)
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