# Question #d9052

Apr 26, 2016

Under $3$ $\text{atmospheres.}$

#### Explanation:

$P = \frac{n R T}{V}$ $=$ $\frac{\left(\frac{0.390 \cdot \cancel{g}}{4.00 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}\right) \times 0.0821 \cdot \cancel{L} a t m \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 301 \cancel{K}}{0.836 \cdot \cancel{L}}$

The only real problem in answering this question is the choice of $R$, the universal gas constant. Chemists would tend to use $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$ because it gives an answer in atmospheres, whose magnitude can be readily appreciated.