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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$\frac{25}{8} \frac{P_{0} V_{0}}{R}$

$\frac{25}{4} \frac{P_{0} V_{0}}{R}$

$\frac{25}{16} \frac{P_{0} V_{0}}{R}$

$\frac{5}{8} \frac{P_{0} V_{0}}{R}$

answer is A.

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Detailed Solution

Equation of the BC is given by two point straight line equation
$\Rightarrow \frac{P - P_{0}}{V - 2 V_{0}} = \frac{3 P_{0} - P_{0}}{V_{0} - 2 V_{0}} \Rightarrow P = P_{0} - \frac{2 P_{0}}{V_{0}} (V - 2 V_{0})$
using PV = nRT
Temperature, $T = \frac{P_{0} V - \frac{2 p_{0} V^{2}}{V_{0}} + 4 P_{0} V}{1 \times R}$ $(∴ n = 1 mole given)$ $T = \frac{P_{0}}{R} [5 V - \frac{2 V^{2}}{V_{0}}], \frac{d T}{d V} = 0 \Rightarrow 5 - \frac{4 V}{V_{0}} = 0 \Rightarrow V = \frac{5}{4} V_{0}$ $T = \frac{P_{0}}{R} [5 \times \frac{5 V_{0}}{4} - \frac{2}{V_{0}} \times \frac{25}{16} V_{0}^{2}] = \frac{25}{8} \frac{P_{0} V_{0}}{R}$