A gas has found to obey $P^{2} V$ = constant. The initial temperature and volume of gas are $T_{0}$ & $V_{0}$ . If the gas expands to volume $3 V_{0}$ . Find the final temperature of gas

T_{0}

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\sqrt{3} T_{0}

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2 T_{0}

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\frac{T_{0}}{2}

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Solution

The correct option is B $\sqrt{3} T_{0}$
Given: Initial temperature and volume of gas is $T_{o} a n d V_{o}$ . The gas follows $P^{2} V = C o n s t a n t$
Solution:
$P^{2} V = K$
Using ideal gas law we can write,
$(\frac{n R T}{V})^{2} V = K$
As n, R are constant so.
$\frac{T^{2}}{V} = K$
$\Rightarrow T^{2} = K V$
$\Rightarrow T \propto \sqrt{V}$
$\Rightarrow \frac{T_{f}}{T_{i}} = \sqrt{(} \frac{V_{f}}{V_{i}})$
$\Rightarrow \frac{T_{f}}{T_{o}} = \sqrt{(} \frac{3 V_{o}}{V_{o}})$
$\Rightarrow T_{f} = \sqrt{3} T_{o}$
Hence correct option is B.

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