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(B) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by $p V^\gamma = 	ext{constant}$.For the adiabatic process $A

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$\frac{p_0 V_0(2^{1/\gamma}-2)}{1-\gamma}$

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(B) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by $p V^\gamma = 	ext{constant}$.For the adiabatic process $A ightarrow B$,the states are $(2p_0, V_0)$ and $(p_0, V_1)$.Thus,$(2p_0) V_0^\gamma = p_0 V_1^\gamma$.Dividing by $p_0$,we get $2 V_0^\gamma = V_1^\gamma$,which implies $V_1 = 2^{1/\gamma} V_0$.The work done $W$ in an adiabatic process is given by $W = \frac{p_i V_i - p_f V_f}{\gamma - 1}$.Substituting the values for process $A ightarrow B$:$W = \frac{(2p_0)(V_0) - (p_0)(V_1)}{\gamma - 1} = \frac{2p_0 V_0 - p_0 (2^{1/\gamma} V_0)}{\gamma - 1}$.Rearranging the terms,we get $W = \frac{p_0 V_0 (2 - 2^{1/\gamma})}{\gamma - 1} = \frac{p_0 V_0 (2^{1/\gamma} - 2)}{1 - \gamma}$.

An ideal gas undergoes a process $A \rightarrow B \rightarrow C \rightarrow A$ cycle. The process $A \rightarrow B$ is adiabatic. Calculate the work done in the process $A \rightarrow B$.

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