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An open flask has Helium gas at 2atm and $327^{0} C$ . The flask is heated to $527^{0} C$ at the same pressure. The fraction of original gas remaining in the flask is

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$\frac{3}{4}$

$\frac{2}{5}$

$\frac{1}{4}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$T_{1} W_{1} = T_{2} W_{2}$

$600 (1) = 800 W_{2}$

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