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

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(C) The equation for the line passing through points $(V_0, 2P_0)$ and $(2V_0, P_0)$ is:Slope $m = \frac{P_0 - 2P_0}{2V_0 - V_0} = \frac{-P_0}{V_0}$.U

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$\frac{9P_0V_0}{4nR}$

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(C) The equation for the line passing through points $(V_0, 2P_0)$ and $(2V_0, P_0)$ is:Slope $m = \frac{P_0 - 2P_0}{2V_0 - V_0} = \frac{-P_0}{V_0}$.Using the point-slope form $P - P_0 = m(V - V_0)$:$P - 2P_0 = \frac{-P_0}{V_0}(V - V_0)$$P = \frac{-P_0}{V_0}V + P_0 + 2P_0 = \frac{-P_0}{V_0}V + 3P_0$.From the ideal gas law,$PV = nRT$,so $T = \frac{PV}{nR}$.Substituting $P$:$T = \frac{1}{nR} \left( \frac{-P_0}{V_0}V + 3P_0 ight)V = \frac{1}{nR} \left( \frac{-P_0}{V_0}V^2 + 3P_0V ight)$.For maximum temperature,$\frac{dT}{dV} = 0$:$\frac{dT}{dV} = \frac{1}{nR} \left( \frac{-2P_0}{V_0}V + 3P_0 ight) = 0$.$\frac{2P_0}{V_0}V = 3P_0 \implies V = \frac{3}{2}V_0$.Substituting $V = \frac{3}{2}V_0$ back into the equation for $T$:$T_{	ext{max}} = \frac{1}{nR} \left( \frac{-P_0}{V_0} \left( \frac{3}{2}V_0 ight)^2 + 3P_0 \left( \frac{3}{2}V_0 ight) ight)$$T_{	ext{max}} = \frac{1}{nR} \left( \frac{-9}{4}P_0V_0 + \frac{9}{2}P_0V_0 ight) = \frac{1}{nR} \left( \frac{9}{4}P_0V_0 ight) = \frac{9P_0V_0}{4nR}$.

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

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