During an experiment an ideal gas is found to obey an additional law ${VP}^{2} =$ constant. The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2V?

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$(\sqrt{3}) T$

$(\sqrt{2}) T$

$\frac{3 T}{2}$

answer is A.

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Detailed Solution

According to the given problem ${VP}^{2} = constant$ . So the gas equation PV = nRT in the light of above (eliminating P) yields, $(\frac{K}{\sqrt{V}}) V = nRT$

i.e., $\sqrt{V} = \frac{nR}{K} T$

$\sqrt{\frac{V_{1}}{V_{2}}} = (\frac{T_{1}}{T_{2}}), i . e ., \sqrt{\frac{V}{2 V}} = \frac{T}{T^{'}} or T^{'} = (\sqrt{2}) T$

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