Solving Gas Mixture Problem: Calculate Number of Moles

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The discussion centers on a gas mixture problem involving helium, argon, and neon, where participants analyze the ambiguity of the question. It is established that the number of moles can only be calculated for argon, given its partial pressure of 1.10 atm, while helium's volume does not provide sufficient information for calculation. The participants agree that the 2.75 L and 25ºC apply to the entire mixture, complicating the calculations for the other gases. There is confusion regarding the instructor's claim that moles could be calculated for each gas, as the data supports only argon. Ultimately, the consensus is that the problem is more of a brain teaser than a straightforward ideal gas question.
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Homework Statement



A gaseous mixture contains 2.75 L of helium, some argon at 1.10 atm, and some neon at 25ºC. For which gas can the number of moles be calculated? Calculate the number of moles present of that gas.
(This is exactly what the problem asks.)

Homework Equations



PV = nRT

The Attempt at a Solution



I'm trying to help some chemistry students with this problem. Either the problem statement is a little ambiguous, or I'm missing something simple.
I believe the helium would occupy 2.75 L by itself, the 1.10 atm of argon is the partial pressure of the mixture for just argon, and the 25ºC (or 298 K) applies to the whole mixture. So far is this correct?

This is what I get next:

PHe(2.75 L) = nHeR(298 K)

(1.10 atm)VAr = nArR(298 K)

PNeVNe = nNeR(298 K)

but I can't see how you could calculate the moles of a gas in the mixture (but I do believe the neon can't be found, basically no information for that gas).
 
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No one? I really dislike this problem now. :biggrin: :devil:
 
the volume of 1 mol of gas is 22.4 L, so you have .123 mol of helium
 
Light bulb said:
the volume of 1 mol of gas is 22.4 L, so you have .123 mol of helium

That's volume at STP, it doesn't hold in general.

Question is ambiguous and wording is tricky. The only correct answer is "for none".

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methods
 
Did you translate this out of some other language?
 
The book is in English and I typed it exactly as it was written.

Whether or not the pressure and/or volume applied to one specific gas or the whole mixture, I think there's still not enough information...
 
Then it's helium ... the gas for which u can compute. It's a gas mixture so if one is at a certain preassure temp, then all are at the same etc...

Basically how many moles are in a 2.75L helium bottle at 1.1atm and 25'C.

PV=nRT be quick! :P
 
Lok said:
It's a gas mixture so if one is at a certain preassure temp, then all are at the same etc...

No. If you have a mixture each gas has its own pressure (called partial pressure), and total pressure is sum of partial pressures.

After rereading the question I think 1.1 atm for argon is the partial pressure, so number of moles of argon can be calculated, asssuming given volume and temperature (which seems to be logical). But it is not an ideal gas question, it is a brain teaser.

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I agree with your conclusion that argon is the gas that moles can be calculated for. It is an ideal gas question in that the question asks for which gas can moles be calculated using \newcommand{\chem}[1]{\ensuremath{\mathrm{#1}}}<br /> \chem{n=\frac{PV}{RT}}<br />

It is not straight forward but it is indeed an ideal gas question. Interesting!
 
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I think what the author had in mind was that the 2.75 L is the volume for all three gases, as well as 25ºC for the gases. Only the argon has a pressure, so the moles of argon only can be calculated.

The funny thing (well, not really) is that the instructor said the moles could be calculated for each gas and each one has the same number of moles...
 
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