# How do you calculate the volume occupied by 64.0 grams of CH_4 at 127°C under a pressure of 1535 torr?

Mar 2, 2017

#### Answer:

For a start let's specify the pressure as $\frac{1535 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 2.02 \cdot a t m$.

#### Explanation:

$m m \cdot H g$ are used for low pressures, or for pressures round about $1 \cdot a t m$. You do NOT use a mercury column for pressures over $1 \cdot a t m$. Why not? Because you risk contaminating your laboratory with elemental mercury, where it will inhabit every crack and every hole - this is major clean-up job which contract cleaners won't touch.

So now we can use the Ideal Gas Equation, $P V = n R T$, and thus, $V = \frac{n R T}{P} = \frac{\frac{64.0 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 400 \cdot K}{2.02 \cdot a t m}$

$\cong 65 - 70 \cdot L$