JEE Advanced · Physics · 13. Thermodynamics
One mole of an ideal gas in initial state A undergoes a cyclic process \(A B C A\), as shown in the figure. Its pressure at \(A\) is \(p_0\). Choose the correct option(s) from the following

- A Internal energies at \(A\) and \(B\) are the same
- B Work done by the gas in process \(A B\) is \(p_0 V_0 \ln 4\)
- C Pressure at \(C\) is \(\frac{p_0}{4}\)
- D Temperature at \(C\) is \(\frac{T_0}{4}\)
Answer & Solution
Correct Answer
(B) Work done by the gas in process \(A B\) is \(p_0 V_0 \ln 4\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& T_A=T_B \quad \therefore U_A=U_B \\
& W_{A B}=(1)(R) T_0 \ln \left(\frac{V_f}{V_i}\right) \\
& =R T_0 \ln \left(\frac{4 V_0}{V_0}\right) \\
& =p_0 V_0 \ln (4)
\end{aligned}
\)
Information regarding \(p\) and \(T\) at \(C\) can not be obtained from the given graph. Unless it is mentioned that line \(B C\) passes through origin or not.
Hence, the correct options are (a) and (b).
\begin{aligned}
& T_A=T_B \quad \therefore U_A=U_B \\
& W_{A B}=(1)(R) T_0 \ln \left(\frac{V_f}{V_i}\right) \\
& =R T_0 \ln \left(\frac{4 V_0}{V_0}\right) \\
& =p_0 V_0 \ln (4)
\end{aligned}
\)
Information regarding \(p\) and \(T\) at \(C\) can not be obtained from the given graph. Unless it is mentioned that line \(B C\) passes through origin or not.
Hence, the correct options are (a) and (b).
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