#### Answer

V = 7.88 L
If the sample were the same mass of Helium, the volume would be completely different. Since the molar mass would be different, the amount of moles of gas would also be different, affecting the Ideal Gas Law.

#### Work Step by Step

$ Ar $ : 39.95 g/mol
1. Using the molar mass as a conversion factor, find the amount in moles:
$$ 12.5 \space g \times \frac{1 \space mole}{ 39.95 \space g} = 0.313 \space mole$$
2. According to the Ideal Gas Law:
$$V = \frac{nRT}{P} = \frac{( 0.313 \space mol)( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 322 \space K)}{ 1.05 \space atm }$$
$$V = 7.88 \space L $$
3. If the sample were 12.5 g of He:
$ He $ : 4.003 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 12.5 \space g \times \frac{1 \space mole}{ 4.003 \space g} = 3.12 \space moles$$
4. According to the Ideal Gas Law:
$$V = \frac{nRT}{P} = \frac{( 3.12 \space mol)( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 322 \space K)}{ 1.05 \space atm }$$
$$V = 78.5 \space L $$