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(B) Step $1$: In the isothermal process,the temperature remains constant,so $P_1V_1 = P_2V_2$.Given $P_1 = P$,$V_1 = V$,and $V_2 = 4V$.$P 	imes V = P

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$2P$

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(B) Step $1$: In the isothermal process,the temperature remains constant,so $P_1V_1 = P_2V_2$.Given $P_1 = P$,$V_1 = V$,and $V_2 = 4V$.$P 	imes V = P_2 	imes 4V \implies P_2 = \frac{P}{4}$.Step $2$: In the adiabatic process,$P_2V_2^\gamma = P_3V_3^\gamma$,where $\gamma = 1.5$ (assuming a monatomic gas or given value).$P_2 = \frac{P}{4}$,$V_2 = 4V$,$V_3 = V$.$\frac{P}{4} 	imes (4V)^{1.5} = P_3 	imes V^{1.5}$.$P_3 = \frac{P}{4} 	imes 4^{1.5} = \frac{P}{4} 	imes (2^2)^{1.5} = \frac{P}{4} 	imes 2^3 = \frac{P}{4} 	imes 8 = 2P$.Thus,the final pressure is $2P$.

Initial pressure and volume of a gas are $P$ and $V$ respectively. First,it is expanded isothermally to volume $4V$ and then compressed adiabatically to volume $V$. The final pressure of the gas will be

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