# Question #caa3e

Apr 27, 2017

All of these problems depend on the choice of gas constant $R$.....

#### Explanation:

And typically, chemists, measure volume in $\text{litres}$, and pressure in $m m \cdot H g$, where $\text{760 mm Hg} \equiv 1 \cdot a t m$. (NB, you do not use a mercury column to measure pressures > $1 \cdot a t m$.

And thus (at least in my opinion) the gas constant that is generally most useful is $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$.

The old standby, $P = \frac{n R T}{V}$. Which equation is this?

$= \frac{\frac{25 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K}{0.500 \cdot L} = 27.8 \cdot a t m$ (which is rather high pressure!).

We might have expected the high pressure, because the molar volume at $1 \cdot a t m$ is approx. $25 \cdot L$, and here we have compressed the gas into a much smaller volume.