# Calculating partial pressures from mixture of gases

A gas is enclosed in a container at $$35~\mathrm{m^3\,mol^{-1}}$$ and has the following composition at $$49\,\mathrm{^\circ C}$$

$$\ce{N_2} \ 2\,\%$$

$$\ce{CH_4} \ 79 \,\%$$

$$\ce{C_2H_6} \ 19 \,\%$$

Assuming an ideal gas, calculate the partial pressure of each component (in kPa).

My working so far: I chose $$100 \,\mathrm{mol}$$ because I've been given percentages and it's a nice number to work with, and I multiplied this with $$35 \,\mathrm{m^3\, mol^{-1}}$$ to get $$3500 \,\mathrm{m^3}$$

I used $$P=\frac{nRT}{V}$$ to calculate an overall pressure but got $$76\, \mathrm{Pa}$$ which seems way too low. Have I done anything wrong?

• $35\,\mathrm{\frac{m^3}{mol}}$ is, in fact, in the question?
– Jan
Oct 4 '15 at 16:29
• $35~\mathrm{m^3\,mol^{-1}}$ means that 1 mole of gas occupies 35 $\mathrm{m^3}$ of volume. What is the volume of 1 mole of gas under normal pressure and how it compares to that? Oct 4 '15 at 16:30

First calculate the total pressure from the ideal gas law, explicitly keeping track of units to avoid confusion and mistakes:

$$\text{Ideal Gas Law:}~~~~{P=\frac{n}{v}*R*T}$$

where:
$\frac{n}{v}=\pu{\frac{1}{35}\frac{mol}{m^3}}$
$R=\pu{8.31\frac{m^3~Pa}{mol~K}}$
$T=\pu{322K}$

Then plug the values in to the ideal gas equation to get the total pressure:

$$P_{total}=\pu{\frac{1}{35}\frac{mol}{m^3}*8.31\frac{m^3~Pa}{mol~K}*322~K}$$

$${P_{total} =\pu{76.5Pa}}$$

Then just multiply the total pressure by the fraction of each gas as stated in the problem:

${P_{\ce{N2}} = \pu{76.5Pa * 0.02 = 1.5 Pa}}$
${P_{\ce{CH4}} = \pu{76.5Pa * 0.79 = 60 Pa}}$
${P_{\ce{C2H6}} = \pu{76.5Pa * 0.19 = 15 Pa}}$