July 06, 2020, 01:03:07 PM
Forum Rules: Read This Before Posting

### Topic: Graham's Law: Deals in moles, not individual molecules?  (Read 1601 times)

0 Members and 1 Guest are viewing this topic.

#### djvan

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### Graham's Law: Deals in moles, not individual molecules?
« on: July 23, 2013, 03:55:15 PM »
I just watched a video that said that Graham's Law measures the average rate of gas molecules of one type in comparison to the average rate of gas molecules molecules of another type.  In other words, a single molecule of a heavier MW (let's say CO2) could have a higher rate than that of a lighter weight MW (let's say O2).  However, the average rate of all of the O2 molecules within the gas, will have a higher rate in comparison to the CO2 molecules.

Is my thinking correct?  If so, I have another question:

An MCAT prep book has the following question:

"Equal molar quantities of oxygen and hydrogen gas were placed in container A under high pressure.  A small portion of the mixture was allowed to effuse for a very short time into the vacuum in container B.  Which of the following is true concerning the partial pressures of the gases at the end of the experiment?"

I assume that Graham's law is to be put to use in this situation.  However, If only a small amount of molecules were allowed to effuse into container B, how can we assume that the molecules that did effuse followed Graham's law, which is based off of the average diffusion of ALL (all of one type, as in O2) molecules within a gas.  Also, at high pressure, don't gases stop behaving ideally?  Isn't Graham's law based on ideal gases?

The correct answer to the question is: "The partial pressure of hydrogen in container B is approximately four times as great as the partial pressure of oxygen in container B".  This answer is clearly obtained from Graham's law, which, when calculated, you see that H2 has a rate 4 times that of O2.

Technically, can Graham's Law be applied to this question?

Thanks!

#### magician4

• Chemist
• Full Member
• Posts: 567
• Mole Snacks: +70/-11
##### Re: Graham's Law: Deals in moles, not individual molecules?
« Reply #1 on: July 23, 2013, 04:59:56 PM »
Quote
(...) Graham's Law measures the average rate of gas molecules of one type in comparison to the average rate of gas molecules molecules of another type.  In other words, a single molecule of a heavier MW (let's say CO2) could have a higher rate than that of a lighter weight MW (let's say O2).  However, the average rate of all of the O2 molecules within the gas, will have a higher rate in comparison to the CO2 molecules.
Is my thinking correct?
how can a single molecule probably have a "rate"? it could have an individually higher velocity, yes - but once it's gone it's gone.
asides from that, I think I know what you wanted to express, and that seems ok to me

Quote
However, If only a small amount of molecules were allowed to effuse into container B, how can we assume that the molecules that did effuse followed Graham's law, which is based off of the average diffusion of ALL (all of one type, as in O2) molecules within a gas.
it isn't necessary that all molecules effuse: we just need enough molecules to effuse to fulfill the Maxwell-Bolzmann distribution in a meaningful way.

For example : if I had 1 mole of a gas at a pressure of 1 bar in A (and "zero" in B), and would allow 10-15 mole of this gas to escape to B, I'd still have such a pressure difference of $\approx$ 1 bar A-B , that in any meaningful way I could claim the difference to be 1 bar still.
nevertheless, 10-15 mole still is in the ballpark of 6*108 particles, and that by far still satisfies my personal understanding of "good enough for a Maxwell-Bolzmann-distribution in B"  (if B wasn't too large)
... and all this, on the other hand,  is good enough to fulfill Graham's law if we had two different gases to play with

Quote
Also, at high pressure, don't gases stop behaving ideally?
eventually they'll do, yes
... but to me, that seems to be beside the point here : we're not to think about such high pressures in context with the problem posed to you, if my reading was correct.
Quote
Technically, can Graham's Law be applied to this question?
yes it can, for the reasons I hope to have shown above

regards

Ingo
There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.