# What is the molar quantity of a gas at 742*mm*Hg pressure, at 298.0*K temperature, confined to a volume of 6.0*L?

Mar 30, 2017

$n = 0.240 \cdot m o l$

#### Explanation:

We use the old ideal gas equation, i.e. $P V = n R T$

$n = \frac{P V}{R T} = \frac{\frac{742 \cdot \cancel{m m \cdot H g}}{760 \cdot \cancel{m m \cdot H g \cdot a t {m}^{-} 1}} \times 6.0 \cdot \cancel{L}}{0.0821 \cdot \cancel{L \cdot a t m \cdot {K}^{-} 1} \cdot m o {l}^{-} 1 \times 298.0 \cdot \cancel{K}}$

$= 0.240 \cdot m o l$

What mass of $H C \equiv N$ is present? Would you expect $H C \equiv N$ to behave ideally? Why or why not? What hazards would be associated with this experiment?

Note the dimensional consistency of our calculation. We sought an answer in $\text{moles}$; the calculation gave us an answer in $\text{moles}$ or at least $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$, so perhaps we are doing something right (for once!).

The key knowledge is the idea that $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high. And thus we may use a unit of length, the height of a mercury column, to give a useful and reproducible measurement of pressure.

Note that these days, mercury columns are a bit of a rarity. Has your teacher shown you a mercury column?