One mole of a gas expands obeying the relatio | vedclass

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(D) The graph is a straight line passing through points $(V, 2P)$ and $(2V, P)$.The equation of the line is $P - P_1 = m(V - V_1)$,where $m = \frac{P

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(D) The graph is a straight line passing through points $(V, 2P)$ and $(2V, P)$.The equation of the line is $P - P_1 = m(V - V_1)$,where $m = \frac{P - 2P}{2V - V} = \frac{-P}{V}$.So,$P - 2P = -\frac{P}{V}(V - V_1) \Rightarrow P = -\frac{P}{V}V + 3P$.From the ideal gas equation for $1$ mole,$PV = RT \Rightarrow P = \frac{RT}{V}$.Substituting $P$ in the line equation: $\frac{RT}{V} = -\frac{P}{V}V + 3P$.$T = \frac{1}{R}(-PV + 3PV) = \frac{1}{R}(-PV + 3PV)$. Wait,let's re-evaluate: $T = \frac{PV}{R} = \frac{V}{R}(-\frac{P}{V}V + 3P) = \frac{1}{R}(-PV + 3PV)$.Actually,$P = -\frac{P}{V}V + 3P$. Multiplying by $V$: $PV = -\frac{P}{V}V^2 + 3PV$.$RT = -\frac{P}{V}V^2 + 3PV$.To find maximum $T$,differentiate with respect to $V$: $\frac{dT}{dV} = \frac{1}{R}(-\frac{2PV}{V} + 3P) = 0$.$2V = 3V \Rightarrow V = 1.5V$.At $V = 1.5V$,$P = -\frac{P}{V}(1.5V) + 3P = 1.5P$.$T_{max} = \frac{PV}{R} = \frac{(1.5P)(1.5V)}{R} = \frac{2.25PV}{R} = \frac{9PV}{4R}$.

One mole of a gas expands obeying the relation as shown in the $P-V$ diagram. The maximum temperature in this process is equal to

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