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Gas laws from KTG

One mole of a...

Question

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

- \frac{2 P_{0} V_{0}}{5 R}

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\frac{11 P_{0} V_{0}}{10 R}

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- \frac{5 P_{0} V_{0}}{4 R}

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P_{0} V_{0}

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Solution

The correct option is B $\frac{11 P_{0} V_{0}}{10 R}$
At $V = V_{0}, P = \frac{P_{0}}{2}$
$∴$ $T_{i} = \frac{P V}{n R} = \frac{(\frac{P_{0}}{2}) (V_{0})}{R} = \frac{P_{0} V_{0}}{2 R}$ $(n = 1)$
and at $V = 2 V_{0}, P = \frac{4 P_{0}}{5}$
$∴ T_{f} = \frac{P V}{n R} = \frac{(2 V_{0}) (\frac{4 P_{0}}{5})}{R} = \frac{8 P_{0} V_{0}}{5 R}$
$∴ Δ T = T_{f} - T_{i} = (\frac{8}{5} - \frac{1}{2}) \frac{P_{0} V_{0}}{R} = \frac{11 P_{0} V_{0}}{10 R}$

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