A gas in a vessel is under pressure of $\pu{1800KPa}$. The design temperature of the tank is $\pu{423K}$. The gas consists of (by volume) $20\%$ of $\ce{CH4}$ and $80\%~\ce{N2}$. Estimate the density ($\pu{kg/m^3}$) of the gas.

I can't just use the ideal gas law to find out the density of the mixture as there is 2 components, right?

density = $\frac{P \times M_r}{RT} $

From the ideal gas law, I can also predict that vol% equals mol%. But with this, am I supposed to find the average value of $M_r$ and then substitute it into the equation above?

  • $\begingroup$ Hi, welcome to Chem.SE! Is the last line actually "...of the gas mixture" in your question? $\endgroup$ – Gaurang Tandon Feb 22 '18 at 9:17
  • $\begingroup$ @GaurangTandon nope. $\endgroup$ – user185692 Feb 22 '18 at 9:27
  • $\begingroup$ Yes. What is the molar average molecular weight? (Hint: when we say 20% by volume, we mean that the mole fraction is 0.2) $\endgroup$ – Chet Miller Feb 22 '18 at 13:27

You originally said that:

The gas consists of 20vol%,CH4 and 80% N2...

This is a fault in the question. A gas cannot consist of two other gases and still remain a gas. Instead, it is now a gas mixture. (hence I had added the term "mixture" into your question, which I've now removed to retain the original meaning)

Once this is out of the way, you can think of density as: $$\begin{align} \text{density}&=\frac{\text{total mass}}{\text{total volume}}\\\\\ &=\frac{n_1M_1+n_2M_2}{(n_1+n_2)RT/P}\\\\ &=\frac{P(n_1M_1+n_2M_2)}{(n_1+n_2)RT}\\\\ &=\frac{PM_{\text{avg}}}{RT} \end{align}$$

and recall that volume percent is equivalent to mole percent, as you correctly did already. Substitute the correct values into this equation and you're good to go.

PS: I do not understand what you're denoting by $M_\mathrm r$. But, the formula you were using is identical to mine if you swap $M_\mathrm r$ with $M_\text{avg}$.

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