March 28, 2024, 07:04:08 PM
Forum Rules: Read This Before Posting


Topic: Can be found the composition of a mixture given an effusion through a pinhole?  (Read 1288 times)

0 Members and 1 Guest are viewing this topic.

Offline Blueberries116

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-1
  • Gender: Male
  • Freelance scientist and learner
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 [itex](N_2)[/itex] pass through a pinhole in [itex]20\,s.[/itex] Find the composition of a mixture between [itex]O_2[/itex] and [itex]CO_2[/itex] if this mixture is let pass through the same slit and the measured time for this effusion is [itex]24[/itex] seconds and of Avogadro's conditions.

[itex]\begin{array}{ll}
1.&\textrm{45.5%}\,O_2\\
2.&\textrm{69.3%}\,CO_2\\
3.&\textrm{12.4%}\,O_2\\
4.&\textrm{69.3%}\,O_2\\
4.&\textrm{45.5%}\,CO_2\\
\end{array}[/itex]

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:

[itex]\frac{v_{N_{2}}}{v_{mixture}}=\sqrt{\frac{F.W_{mixture}}{F.W_{N_2}}}[/itex]

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

[itex]\frac{v_{N_{2}}}{v_{mixture}}=\sqrt{\frac{\chi_{O_2}\cdot F.W\,O_2+\chi_{CO_{2}}\cdot F.W\,CO_{2}}{F.W_{N_2}}}[/itex]

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. Help please!
Have a nice day!

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
What is sum of partial volumes of oxygen of carbon dioxide?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Blueberries116

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-1
  • Gender: Male
  • Freelance scientist and learner
What is sum of partial volumes of oxygen of carbon dioxide?

It is not given, the problem only states "a certain volume of nitrogen gas", therefore with that given clue and the time for this gas and the mixture of that of the others referring to oxygen and carbox dioxide, is it possible to relate it with the composition of such mixture?. Can this problem be solved?.  ???
Have a nice day!

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
What if I tell you "a certain volume" means exactly 1 mL? Would it help?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Quote
But the problem arises on what volume should be used to establish an equation from where to find the composition of the mixture.
You don't need to know this. You simply need to assume that the effused volumes of nitrogen and mixture are the same. (The question doesn't say this, but you can't solve it otherwise.) Then the effusion times give you the relative rates of effusion, from which you can work out an effective molecular weight for the mixture, and hence the composition.

But I wonder if there is an error in the question. They seem to assume (at least, this assumption leads to one of the optional answers) that the effective MW is the arithmetic mean, given by
M = x*32 + (1-x)*44, where x is the mole fraction of oxygen.
However, this does not seem to me to be right. From the additivity of the effusion rates of O2 and CO2, I would expect the effective MW to be given by
1/sqrt(M) = x/sqrt(32) + (1-x)/sqrt(44)
This gives a different value for x, which does not correspond to any of the answers. (The composition by the first assumption is very close, but not identical, to the composition I calculate for the effused mixture.)

Sponsored Links