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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?

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  • $\begingroup$ $35\,\mathrm{\frac{m^3}{mol}}$ is, in fact, in the question? $\endgroup$ – Jan Oct 4 '15 at 16:29
  • $\begingroup$ $35~\mathrm{m^3\,mol^{-1}}$ 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? $\endgroup$ – Ivan Neretin Oct 4 '15 at 16:30
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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:

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

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