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

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

work done by the gas in process $A B$ is $P_{\circ} V_{\circ} l n 4$

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

internal energies at $A$ and $B$ are the same

answer is A, B, C, D.

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Detailed Solution

Internal energy of an ideal gas depends on temperature . Since $T_{A} = T_{B} = T_{0}, w e c a n s a y U_{A} = U_{B} .$

$W_{A B} = n R T l n \frac{V_{2}}{V_{1}} = (1) (R) \frac{P_{0} V_{0}}{R} l n \frac{4 V_{0}}{V_{0}}$ = $P_{0} V_{0} \ln 4$

Since slope of the line BC is constant and it passes through the origin , process BC is isobaric process .So $P_{C} = P_{B} = \frac{P_{0} V_{0}}{4 V_{0}} = \frac{P_{0}}{4}$ .Also $\frac{\frac{P_{0}}{4} V_{0}}{T_{C}} = \frac{P_{0} V_{0}}{T_{0}} \Rightarrow T_{C} = \frac{T_{0}}{4}$