I have to decide whether a $\pu{20 L}$ container with $\pu{0.8 kg}$ ethane can withstand the pressure at $\pu{373.15 K}$. The following information is given: \begin{align} b &= \pu{6.448x10^-2 dm3/mol}\\ a &= \pu{5.572 dm6 bar/mol2}\\ p_\mathrm{c} &= \pu{49.64 bar}\\ T_\mathrm{c} &= \pu{308 K}\\ \end{align}

The container can withstand any pressure up to $\pu{3000 kPa}$.

I am not sure how to solve this, so far I've tried inserting my values in the van der Waals equation of state, but I could not arrive at the right result with this approach. The correct answer should be $\pu{3.528E6 Pa}$ and I got $\pu{3.14E6 Pa}$.

I inserted the following but the result is wrong:

\begin{align} p &=\frac{\pu{26.60 mol} \times 8.3145 \times \pu{373.15 K}} {\pu{0.02 m3} - \pu{26.6 mol} \times 6.44 \times 10^{-8}} - (\pu{0.557 m6//mol2} ) \times \frac{\pu{26.6 mol2}}{(\pu{0.02 m3})^2}\\ &= \pu{3.14E6 Pa} \end{align}

  • $\begingroup$ I plugged the numbers and ended up within a rounding error from the correct answer. Not sure what else can be done. $\endgroup$ – Ivan Neretin Feb 16 '18 at 15:34
  • $\begingroup$ did you type in the following: p=(26,60 mol*8,3145*373,15K)/(0,02m^3-26,6 mol*6,44*(〖10〗^(-8) m^3)/mol)-(0,557m^6)/(mol^2 )*(〖26,6mol〗^2/(0,02〖m^3〗^2 ))=3,14*〖10〗^6 Pa $\endgroup$ – Jacob Andersen Feb 16 '18 at 16:53

Having edited the expression that caused all the problems I can now see that the units for $b$ were converted incorrectly: $$6.448\times10^{-2}\frac{\text{dm}^3}{\text{mol}}\left(\frac{1\,\text{m}}{10\,\text{dm}}\right)^3=6.448\times10^{-5}\frac{\text{m}^3}{\text{mol}}$$ I got $$v=\frac{\left(20\,\text{L}\right)\left(\frac{1\,\text{dm}^3}{1\,\text{L}}\right)}{\left(0.8\,\text{kg}\right)\left(\frac{1000\,\text{g}}{1\,\text{kg}}\right)\left(\frac{1\,\text{mol}}{30.070\,\text{g}}\right)}=0.75175\frac{\text{dm}^3}{\text{mol}}$$ $$P=\frac{\left(8.314\frac{\text{J}}{\text{K}\,\text{mol}}\right)\left(\frac{1\,\text{m}^3\,\text{Pa}}{1\,\text{J}}\right)\left(\frac{10\,\text{dm}}{1\,\text{m}}\right)^3\left(\frac{1\,\text{bar}}{10^5\,\text{Pa}}\right)\left(373.15\,\text{K}\right)}{0.75175\frac{\text{dm}^3}{\text{mol}}-6.448\times10^{-2}\frac{\text{dm}^3}{\text{mol}}}-\frac{5.572\frac{\text{dm}^6\text{bar}}{\text{mol}^2}}{\left(0.75175\frac{\text{dm}^3}{\text{mol}}\right)^2}=35.281\,\text{bar}$$ So convert units correctly and all should be well.

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