One mole of an ideal monoatomic gas is taken | vedclass

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(A) The path $BC$ is a straight line passing through points $(V_0, 3P_0)$ and $(2V_0, P_0)$.The slope of the line $BC$ is $m = \frac{P_0 - 3P_0}{2V_0

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$\frac{25}{8} \frac{P_0 V_0}{R}$

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(A) The path $BC$ is a straight line passing through points $(V_0, 3P_0)$ and $(2V_0, P_0)$.The slope of the line $BC$ is $m = \frac{P_0 - 3P_0}{2V_0 - V_0} = \frac{-2P_0}{V_0}$.The equation of the line $BC$ is $P - 3P_0 = \frac{-2P_0}{V_0}(V - V_0)$,which simplifies to $P = 3P_0 - \frac{2P_0}{V_0}(V - V_0) = P_0(5 - \frac{2V}{V_0})$.Using the ideal gas law $PV = nRT$ with $n = 1$,we have $T = \frac{PV}{R} = \frac{P_0}{R}(5V - \frac{2V^2}{V_0})$.To find the maximum temperature,we set $\frac{dT}{dV} = 0$:$\frac{dT}{dV} = \frac{P_0}{R}(5 - \frac{4V}{V_0}) = 0 \implies V = \frac{5}{4}V_0$.Substituting $V = \frac{5}{4}V_0$ back into the expression for $T$:$T_{max} = \frac{P_0}{R}(5(\frac{5}{4}V_0) - \frac{2}{V_0}(\frac{25}{16}V_0^2)) = \frac{P_0}{R}(\frac{25}{4}V_0 - \frac{25}{8}V_0) = \frac{25}{8} \frac{P_0 V_0}{R}$.

One mole of an ideal monoatomic gas is taken along the path $ABCA$ as shown in the $PV$ diagram. The maximum temperature attained by the gas along the path $BC$ is given by

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