# What volume would 5 moles of Helium occupy at 278°C and 765 mmHg?

Approx. $220 \cdot L$.
We use the ideal gas equation with the appropriate constant. We know that $1 \cdot a t m$ $\equiv$ $760 \cdot m m \cdot H g$ (or more correctly, $1 \cdot a t m$ will support a column of mercury $760 \cdot m m$ high).
$V = \frac{n R T}{P}$ $=$ $\left(5 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 546 \cdot K\right) \times {\left(\frac{765 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}\right)}^{-} 1$
$=$ ??L