Answer
V = 7.88 L
The volume would be different if the sample were He.
Work Step by Step
According to the ideal gas law,
$pV = nRT$
or $V = \frac{nRT}{p}$
n = $\frac{Mass\,in\,grams}{Molar\,mass}=\frac{12.5g}{39.948\,g\,mol^{-1}}=0.313\,mol$
p = 1.05 atm
T = 322 K
and R = $0.082057\,L\,atmK^{-1}mol^{-1}$
Therefore, V =$\frac{ 0.313 mol\times0.082057\,L\,atmK^{-1}mol^{-1}\times322 K}{1.05 \,atm}=7.88\,L$
As the number of gas particles changes, the volume would be different if the sample were 12.5 g He.
