# If 37.5 mol of an ideal gas occupies 38.5 Lat 85.00°C, what is the pressure of the gas?

Oct 31, 2016

$P \cong 29 \cdot a t m$
$P = \frac{n R T}{V}$ $=$ $\frac{37.5 \cdot \cancel{m o l} \times 0.0821 \cdot \cancel{L} \cdot a t m \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 358.15 \cdot \cancel{K}}{38.5 \cdot \cancel{L}}$
Note that the equation is consistent dimensionally. We wanted an answer in $\text{atmospheres}$ and we got one.
When we do equations like this, the gas constant $R$ can have many different values, each with different units. For chemists, probably the most useful one is $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$. These need not be learned, because they will appear as supplementary material on the exam paper. They do have to be used correctly.