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An ideal diatomic gas undergoes a thermodynamic process where pressure (P) as a function of volume (V) is given by $P = P_{o} e^{\frac{3 V_{o}}{V}}$ The average rotational kinetic energy of the gas molecules at volume $V = \frac{V_{0}}{2} is$

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$\frac{P_{0} V_{0}}{2} e^{6}$

$P_{0} V_{0}$

$\frac{3 P_{0} V_{0}}{4} e^{6}$

$\frac{5}{4} P_{0} V_{0} e^{6}$

answer is A.

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Detailed Solution

${K E}_{R} = \frac{2}{2} n R T = P V = P_{0} e^{6} (\frac{V_{0}}{2}) = \frac{P_{0} V_{0}}{2} e^{6}$

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