The process AB is shown in the diagram. As th | vedclass

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(A) The equation of the straight line $AB$ passing through $(V, 2P)$ and $(2V, P)$ is given by:$\frac{P' - 2P}{V' - V} = \frac{P - 2P}{2V - V} = \frac

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initially increases then decreases

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(A) The equation of the straight line $AB$ passing through $(V, 2P)$ and $(2V, P)$ is given by:$\frac{P' - 2P}{V' - V} = \frac{P - 2P}{2V - V} = \frac{-P}{V}$$P' - 2P = -\frac{P}{V}(V' - V)$$P' = -\frac{P}{V}V' + 3P$From the ideal gas equation,$PV = nRT$,so $T \propto PV$.Let $f(V') = P'V' = V'(-\frac{P}{V}V' + 3P) = -\frac{P}{V}(V')^2 + 3PV'$.This is a downward-opening parabola with respect to $V'$. The maximum value occurs at $V' = -\frac{b}{2a} = -\frac{3P}{2(-P/V)} = 1.5V$.Since the volume $V'$ increases from $V$ to $2V$,the product $PV$ (and thus temperature $T$) first increases until $V' = 1.5V$ and then decreases as $V'$ approaches $2V$.

The process $AB$ is shown in the diagram. As the gas is taken from $A$ to $B$,its temperature

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