An open flask has helium gas at 2atm and 327^oC. The flask is heated to 527^oC at the same pressure. The fraction of original gas remaining in the flask is

Given data:

Initial pressure, ${\mathrm{P}}_1$=1 atm

Initial temperature, ${\mathrm{T}}_1$ =327° C=600 K

Final pressure, ${\mathrm{P}}_2$ =1 atm

Final temperature, ${\mathrm{T}}_2$ =527° C=800 K

According to ideal gas equation PV = nRT

$\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

${\mathrm{V}}_2=\frac{1\times {\mathrm{V}}_1}{600}$

${\mathrm{V}}_2=\frac{4}{3}{\mathrm{V}}_1$

But given the hold of volume remaining the flask 

${\mathrm{V}}_2=\frac{4}{3}{\mathrm{V}}_1-{\mathrm{V}}_1$

${\mathrm{V}}_2=\frac{1}{3}{\mathrm{V}}_1$

Air Expelled out of fraction.

$\frac{{\mathrm{V}}_1/3}{4{\mathrm{V}}_1/3}=\frac{1}{4}$

Remaining air in the flask

$1-\frac{1}{4}=\frac{3}{4}$

Hence the required answer is A