# How do you calculate the number of mol of a gas that is present in a 5.67 L container at 44.7 C, if the gas exerts a pressure of 614.0 mmHg.?

Aug 19, 2016

$n = \frac{P V}{R T}$ $\cong 0.17 \cdot m o l$

#### Explanation:

$n = \frac{P V}{R T}$ $=$ $\frac{\left(\frac{614 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 5.67 \cdot L\right)}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 317.9 \cdot K}$

$\cong 0.17 \cdot m o l$

This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant, $R$. The most convenient unit of pressure is still the $\text{atmosphere}$, and this is a unit that is very intuitive. Given this, most chemists would use $m m \cdot H g$, knowing that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$, or rather that $1$ $a t m$ of pressure will support a column of mercury that is $760 \cdot m m$ high. And thus conversion of $m m \cdot H g$ to $\text{atmospheres}$ makes the choice of gas constant easy.