# Gas mixture problem

1. Nov 5, 2009

### Bohrok

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

PV = nRT

3. 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).
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 6, 2009

### Bohrok

No one? I really dislike this problem now.

3. Nov 6, 2009

### Light bulb

the volume of 1 mol of gas is 22.4 L, so you have .123 mol of helium

4. Nov 7, 2009

### Staff: Mentor

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

5. Nov 7, 2009

### Lok

Did you translate this out of some other language?

6. Nov 7, 2009

### Bohrok

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...

7. Nov 9, 2009

### Lok

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

8. Nov 9, 2009

### Staff: Mentor

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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9. Nov 18, 2009

### _Molly_

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 $$\newcommand{\chem}[1]{\ensuremath{\mathrm{#1}}} \chem{n=\frac{PV}{RT}}$$

It is not straight forward but it is indeed an ideal gas question. Interesting!

10. Nov 18, 2009

### Bohrok

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...