# If 1 mole of H2 gas was collected under 294.2K and 746.7 mmHg, what volume will it occupy?

Apr 28, 2017

Well, $V = \frac{n R T}{P}$, and of course, we must choose the appropriate gas constant; $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$ is used here.
The key to solving these problems is to recall that $1 \cdot a t m$ of pressure will support a column of mercury that is $760 \cdot m m$ high. A column of mercury may thus be used to measure pressures up to $1 \cdot a t m$.
And so V=(1*molxx0.0821*(L*atm)/(K*mol)xx294.2*K)/((746.7*mm*Hg)/(760*mm*Hg*atm^-1)
=??L. It should be about $25 \cdot L$.