During an experiment an ideal gas is found to obey an additional law $V p^{2} =$ constant. The gas is initially at temperature $T$ and volume $V$ , when it expands to volume $2 V$ , the resulting temperature is $T_{2}$ :

\frac{T}{2}

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2 T

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\sqrt{2} T

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\frac{T}{\sqrt{2}}

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Solution

The correct option is C $\sqrt{2} T$
Ideal gas: $V P^{2} =$ constant
Again $P V = n R T$ from equation of state,
Hence, $V P \times P =$ constant i.e. $n R T \times P =$ constant
Again, $P = \frac{n R T}{V}$
$∴$ $\frac{(n R T)^{2}}{V} =$ constant
$\frac{T^{2}}{V} =$ constant
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$T_{2} = \sqrt{\frac{2 v}{v}} = \sqrt{2} T_{1}$

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