COMEDK · Physics · 13. Thermodynamics
An ideal gas goes from state \(A\) to state \(B\) via three different processes as indicated in the \(p-V\) diagram.
If \(Q_1, Q_2\) and \(Q_3\) indicate the heat absorbed by the three processes and \(\Delta U_1, \Delta U_2\) and \(\Delta U_3\) indicate the change in internal energy along the three processes respectively, then
- A \(Q_1>Q_2>Q_3\) and \(\Delta U_1=\Delta U_2=\Delta U_3\)
- B \(Q_3>Q_2>Q_1\) and \(\Delta U_1=\Delta U_2=\Delta U_3\)
- C \(Q_1=Q_2=Q_3\) and \(\Delta U_1>\Delta U_2>\Delta U_3\)
- D \(Q_3>Q_2>Q_1\) and \(\Delta U_1>\Delta U_2>\Delta U_3\)
Answer & Solution
Correct Answer
(A) \(Q_1>Q_2>Q_3\) and \(\Delta U_1=\Delta U_2=\Delta U_3\)
Step-by-step Solution
Detailed explanation
For all processes 1,2 and 3 Change in internal energy, i.e. \(\begin{gathered} \Delta U=U_B-U_A \\ \therefore \Delta U_1=\Delta U_2=\Delta U_3 \Rightarrow Q=\Delta U+\Delta W \end{gathered}\) Now, \(\Delta W=\) work done by the gas, i.e. area under \(p-V\) curve. As, area…
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