# What is the volume of 2.10 moles of chlorine gas (Cl_2) at 273 K and 1.00 atm?

$47.0 L$
$P V = n R T$, where $R = 0.082057 \text{ L atm } m o {l}^{-} 1 {K}^{-} 1$
$\therefore V = \frac{n R T}{P}$
$V = \frac{2.10 m o l \times R \times 273 K}{1.00 a t m} = 47.0 L$