Answer
V= 3.0 L
The volume would not be different if the gas was Ar.
Work Step by Step
According to the ideal gas law,
$pV= nRT$
or $V= \frac{nRT}{p}$
n= 0.118 mol
p= 0.97 atm
T= 305 K
and R= $0.082057\,L\,atmK^{-1}mol^{-1}$
Therefore, V=$\frac{ 0.118 mol\times0.082057\,L\,atmK^{-1}mol^{-1}\times305 K}{0.97 atm}=3.0\,L$
As the nature of the gas particles doesn't matter, the volume wouldn't be different if the gas was argon.
