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Gas laws from KTG

One mole of a...

Question

One mole of an ideal gas undergoes a process :
$P = \frac{P_{0}}{1 + (V_{0} / V)^{2}}$
Here $P_{0}$ and $V_{0}$ are constants. Change in temparature 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 $V = V_{0} : P = \frac{P_{0}}{2}$
$∴ T_{1} = \frac{P V}{n R} = \frac{(\frac{P_{0}}{2}) V_{0}}{R} = \frac{P_{0} V_{0}}{2 R} (n = 1)$
and at $V = 2 V_{0} : P = \frac{4 P_{0}}{5}$

$∴ T_{f} = \frac{P V}{n R} = \frac{(2 V_{0}) (\frac{4 P_{0}}{5})}{R} = \frac{8 P_{0} V_{0}}{5 R}$

$∴ Δ T = T_{f} - T_{i} = (\frac{8}{5} - \frac{1}{2}) \frac{P_{0} V_{0}}{R} = \frac{11 P_{0} V_{0}}{10 R}$
Hence, the correct answer is option (b).

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