One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is $P_{0}$ . Choose the correct option(s) from the following.

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Internal energies at A and B are the same

Work done by the gas in process AB is $P_{0} V_{0} \log_{e} 4$

Pressure at C is $\frac{P_{0}}{4}$

Temperature at C is $\frac{T_{0}}{4}$

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} (a) Δ U = n C_{V} d T = 0 \\ U = c o n s \tan t \end{array}$

$U_{A} = U_{B}$

$\begin{array}{l} (b) W_{A B} = n R T_{0} \log_{e} (\frac{V_{2}}{V_{1}}) = n R T_{0} \log_{e} (\frac{4 V_{0}}{V_{0}}) \\ W_{A B} = n R T_{0} \log_{e} (4) = P_{0} V_{0} \log_{e} (4) \end{array}$

$\begin{array}{l} (c) A l o n g A B T_{0} i s c o n s \tan t \\ P_{1} V_{1} = P_{2} V_{2} \\ P_{0} V_{0} = P_{B} \times 4 V_{0} \\ P_{B} = \frac{P_{0}}{4} \\ (d) A l o n g B C, V α T ∵ P i s c o n s \tan t \\ P_{C} = \frac{P_{0}}{4} \\ \frac{V_{B}}{T_{B}} = \frac{V_{C}}{T_{C}} \Rightarrow \frac{4 V_{0}}{T_{0}} = \frac{V_{0}}{T_{C}} \Rightarrow T_{C} = \frac{T_{0}}{4} \end{array}$