One mole of an ideal gas undergoes a process $P = P_{0} {[1 + {(\frac{2 V_{0}}{V})}^{2}]}^{- 1}$ , $where P_{0}, V_{0} are$ constants. Change in temperature of the gas when volume is changed from V=V₀ to V₀ = 2V is:

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$\frac{3}{4} \frac{P_{0} V_{0}}{n R}$

$\frac{7}{9} \frac{P_{0} V_{0}}{n R}$

$\frac{2}{3} \frac{P_{0} V_{0}}{n R}$

$\frac{4}{5} \frac{P_{0} V_{0}}{n R}$

answer is A.

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Detailed Solution

$\begin{array}{l} P = P_{0} {[1 + {(2 V_{0} / V)}^{2}]}^{- 1} \\ At V = V_{0}, P = P_{0} / 5 \\ T_{i} = \frac{P V}{n R} = \frac{(P_{0} / 5) V_{0}}{n R} = \frac{P_{0} V_{0}}{5 n R} \\ At V = 2 V_{0}, P = P_{0} / 5 \\ T_{i} = \frac{P V}{n R} = \frac{(P_{0} / 5) V_{0}}{n R} = \frac{P_{0} V_{0}}{5 n R} \\ At V = 2 V_{0}, P = P_{0} / 2 \\ T_{f} = \frac{P V}{n R} = \frac{(P_{0} / 2) (2 V_{0})}{n R} = \frac{P_{0} V_{0}}{n R} \\ Δ T = T_{f} - T_{i} = \frac{4 P_{0} V_{0}}{5 n R} \end{array}$