Question

Find the density of methane gas at 20$^{\circ}$ C and 5 atm pressure.

Answer 1

We will require an ideal gas equation to obtain the required density. In the ideal gas equation, n is the amount of substance or number of moles for the concerned gas. If we determine the molecular mass of methane, it will be the mass of methane for 1 mole amount of molecules.
Formula used:
Ideal gas equation:
PV = nRT
where P is the pressure of the gas, V is the volume, T is the temperature, R is universal gas constant and n is the number of moles.

Complete answer:
First we determine the molecular mass (or weight) for methane,
m = (12 + 4) g = 16 g;
where we know the atomic mass of carbon is 12 g and for hydrogen is 1 g. The molecule of methane has 1 carbon atom and 4 hydrogen atoms as the molecular formula is $CH_4$.
If we consider a volume V, then the total mass of methane in that volume would be:
M = nm,
where n is the total number of moles in the concerned volume and m is the molecular mass per mole (same as we derived previously).
The density will be then,
$d = \dfrac{nm}{V}$.
Now, we make substitution of n/V with the help of ideal gas equation as,
$\dfrac{n}{V} = \dfrac{P}{RT}$.
So, the density now can be written as,
$d = \dfrac{Pm}{RT}$.
We are given T = 20$^{\circ}$ C = 293 K, P = 5 atm. We keep R = 8.314 m$^3$ Pa K$^{-1}$ mol$^{-1}$, which gives us:
$d = \dfrac{5 \times 16}{8.314 \times 293} = 0.0328 g m^{-3}$.
Therefore, the required density of methane in standard units is $3.28 \times 10^{-5} kg m^{-3}$.

Note:
The derivation might get a little confusing if one does not know what n stands for. One might otherwise, directly keep the mass to be 16 grams and not bother about the number of moles. This might produce wrong answers so to avoid such mistakes one should clear up the concepts first regarding molecular mass and number of moles.