n moles of an ideal gas undergo a process A t | vedclass

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(D) For the given graph,the equation of the $P-V$ line passing through $(V_0, 2P_0)$ and $(2V_0, P_0)$ is:$P - 2P_0 = \frac{P_0 - 2P_0}{2V_0 - V_0} (V

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$\frac{9 P_0 V_0}{4nR}$

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(D) For the given graph,the equation of the $P-V$ line passing through $(V_0, 2P_0)$ and $(2V_0, P_0)$ is:$P - 2P_0 = \frac{P_0 - 2P_0}{2V_0 - V_0} (V - V_0)$$P - 2P_0 = -\frac{P_0}{V_0} (V - V_0)$$P = 3P_0 - \frac{P_0}{V_0} V$Using the ideal gas equation $PV = nRT$,we have $T = \frac{PV}{nR}$.Substituting $P$ in terms of $V$:$T = \frac{1}{nR} (3P_0 - \frac{P_0}{V_0} V) V = \frac{1}{nR} (3P_0 V - \frac{P_0}{V_0} V^2)$For maximum temperature,$\frac{dT}{dV} = 0$:$\frac{d}{dV} (3P_0 V - \frac{P_0}{V_0} V^2) = 0$$3P_0 - \frac{2P_0}{V_0} V = 0$$V = \frac{3}{2} V_0$Substituting $V = \frac{3}{2} V_0$ back into the pressure equation:$P = 3P_0 - \frac{P_0}{V_0} (\frac{3}{2} V_0) = 3P_0 - \frac{3}{2} P_0 = \frac{3}{2} P_0$Now,calculate the maximum temperature:$T_{max} = \frac{P V}{nR} = \frac{(\frac{3}{2} P_0) (\frac{3}{2} V_0)}{nR} = \frac{9 P_0 V_0}{4nR}$

$n$ moles of an ideal gas undergo a process $A \to B$ as shown in the figure. The maximum temperature of the gas during the process is

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