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?
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:  

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