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Typically all the problems which I've seen regarding rate of diffusion involve relating the rate of one gas to the other and from there finding either speed of effusion or diffusion or getting the formula weight. However what sort of procedure should be used when it is needed to find the composition of a mixture?

The problem is as follows:

A certain volume of nitrogen gas $(\ce{N2})$ pass through a pinhole in $\pu{20 s}$. Find the composition of a mixture between $\ce{O2}$ and $\ce{CO2}$ if this mixture is let pass through the same slit and the measured time for this effusion is $\pu{24 s}$ and of Avogadro's conditions.

  1. 45.5% $\ce{O2}$
  2. 69.3% $\ce{CO2}$
  3. 12.4% $\ce{O2}$
  4. 69.3% $\ce{O2}$
  5. 45.5% $\ce{CO2}$

I'm not very sure if it is possible to find what it is being asked here. It's obvious that the approach here is to use Graham's law of gas diffusion. But the problem arises on what volume should be used to establish an equation from where to find the composition of the mixture.

What I thought was: $$\frac{v_{\ce{N2}}}{v_\text{mixture}} = \sqrt{\frac{FW_\text{mixture}}{FW_{\ce{N2}}}}$$

The thing here is that the average molecular weight can be used to find the composition of the mixture.

$$\frac{v_{\ce{N2}}}{v_\text{mixture}}=\sqrt{\frac{\chi_{\ce{O2}} \cdot FW_{\ce{O2}} + \chi_{\ce{CO2}} \cdot FW_{\ce{CO2}}}{FW_{\ce{N2}}}}$$

But the thing here is I am still stuck on where to go from here. What should be the right way to go to get the composition of the mixture. Could it be that the volume of the nitrogen is needed?. So far I'm going in circles with this. Regarding the source of this problem it doesn't indicate as far I'm aware since it is a collection of problems from different topics in gases.

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Go on! You are nearly at the end. If you square your equation, and replace $y$ by $1-x$, you obtain: $$(24/20)^2 = \frac{32·x + 44(1-x)}{28}$$ You finally obtain: $x = 1/3$ and $y = 2/3$.

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  • $\begingroup$ @Martin-マーチン♦ It seems that you got wrong the result for the molar fraction. Solving the system yields $x=\frac{23}{75}=0.30667$ and $1-x=0.69333$ which given as a percentage would be $69.3\%$ of $CO_{2}$ which according to my book is the right answer. $\endgroup$ – Chris Steinbeck Bell Feb 28 at 2:48
  • $\begingroup$ @Martin-マーチン♦ The reason why I was almost at the end was my confusion regarding if should I assume that the same volume of gas was for the mixture as well?. Because in the problem did not explicitly said this. Other than assuming that by mentioning Avogadro's conditions $22.4\times V$ of the nitrogen gas and $22.4\times V$ of the mixture. Should this rationale be used in all sorts of problems of this kind?. $\endgroup$ – Chris Steinbeck Bell Feb 28 at 2:52
  • $\begingroup$ @Chris I only slightly edited this answer, I don't even really know where the numbers come from or what they mean. Sorry, I cannot help you there. $\endgroup$ – Martin - マーチン Feb 28 at 10:36

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