# How to find Pressure and Root Mean Square Velocity

1. Sep 18, 2008

### Newtime

1. The problem statement, all variables and given/known data
The average kinetic energy of a 2.35g sample of argon gas in a 7.00L container is 2.58E-22 j/atom.

2. Relevant equations

a) What is the pressure of the gas?
b) What is the rot mean square velocity of teh argon atoms under these conditions?

3. The attempt at a solution

I tried several different formulas none of which helped at all. For all of them it seems I still have on variable for which i cannot solve. For isntance in the KE=.5mv^2 equation for b) my answer does not match the answer in the book when I isolate v and take the root. For part a) the question does not give the temperature so I tried rearranging several equations to no avail. Any help is appreciated, thank you.

2. Sep 18, 2008

### LowlyPion

Welcome to PF.

Have you examined
http://en.wikipedia.org/wiki/Temperature#Temperature_in_gases

3. Sep 18, 2008

### Newtime

Thanks, and yes I did look at it. The equation towards the midde of the page looked like it would help - and maybe I did it wrong - but after finding the temperature in Kelvin, then trying to plug that in along with the other information into the Pv = nRT eq. I still ended up with the incorrect answer. Any ideas?

4. Sep 18, 2008

### LowlyPion

You've taken the average kinetic energy to be times the number of atoms in your sample?

All 2.35/39.95 x 6.023x1023 of them right?

5. Sep 18, 2008

### Newtime

Yes I have and then to find the velocity in b) i divided that by (.5*m) then took the square root. The answer still comes out as wrong. And as far as part a) goes I'm still completely lost. Without a temperature I don't know what to do. I think it has something to do with the equation for most porbable velocity but I'm not sure.

6. Sep 18, 2008

### LowlyPion

7. Sep 18, 2008

### Newtime

The Maxwell–Boltzmann distribution is covered very briefly in my textbook; only about a paragraph and none of those equations are in the book. I just feel like I'm totally missing something in this problem. It seems so simple but everything I try is incorrect. I know all the theory and the equations that should be used to solve this problem but like I said I just feel like I'm missing something. Part a) I don't know where to start still and part b) is annoying me because it's such a very simple problem yet everything I do is wrong.

8. Sep 18, 2008

### yaychemistry

Hello,
I would actually suggest this wikipedia article:
http://en.wikipedia.org/wiki/Kinetic_theory

Also, ~$$10^{-22}\textrm{J}$$ is probably what one could expect for a kinetic energy of a single atom, thus I would expect that you need to do what you were doing before (e.g. divide 2.58E-22 J by 0.5*m and then take the square root), but double check that your units are correct. This relationship might help
$$1 \textrm{Joule} = 1 \frac{\textrm{kg}^2 \textrm{m}^2}{\textrm{s}^2}$$ where kg = kilograms, m = meters, and s = seconds. Thus if you divide the mass of 1 argon atom and make sure its in kilograms per atom (the periodic table gives it in grams per mole of atoms so you'll have to convert) your final answer for rms velocity will be in meters per second.

Last edited: Sep 18, 2008
9. Sep 18, 2008

### Newtime

But in doing that, once I take the root, aren't I left with units in (root(kg)m)/s ? Or should I disregard the kg? Wen I do this I still get the wrong answer for part b. I also tried part a using a few of the different eq.'s in that wikipedia article again ending up with the wrong answer. Would it be possible for someone to work this problem through or at least partially so and then I would do it and check to see if that is correct. If this problem appears on my upcoming test I want to know how to do it.

10. Sep 18, 2008

### yaychemistry

Ah, you caught a typo of mine!
$$1 \textrm{Joule} = 1 \textrm{kg}\frac{\textrm{m}^2}{\textrm{s}^2}$$

Sure, what is the book's answer so I can work it out and check it against the book? Also, the book could have the wrong answer, you never know how well they check their answers...

It could also help to post the work you're doing as well to see if we can spot any other errors.

11. Sep 18, 2008

### Newtime

Also, in the problem itself it says "Argon gas," and I assumed Ar meant Ar. But when a problem says something like Oxygen gas I know it's O2 (2 as a subscript) so is Ar something like Ar2 or Ar3? This would clearly alter my results and hopefully give the correct answer.

12. Sep 18, 2008

### Newtime

I just realized I typed book, I meant internet. This assignment is online and you get 99 tries per question but the answer isn't revealed to you until you get it right. This is partly why it's so frustrating. And there is a chance the online homework is wrong because I've already seen one type within it. As far as posting the work I'm doing, I've tried this problem 19 times using 19 separate methods, after the first 2 or 3 I was pretty much just trying anything and everything so I'm not sure if posting all my work will help though I can type in words what I think would be the correct steps if you want.

13. Sep 18, 2008

### yaychemistry

Argon gas should just be Ar.

For the online form do you know how close you have to get the answer? e.g. if the answer is 88.01 and you type 88.012 would you get it wrong? It might just be a matter of significant digits, and does it specify what units to give the answer in?

If its troublesome to post all the work, post the numbers that you have got using the method(s) that made the most sense to you.

14. Sep 18, 2008

### Newtime

The site is pretty forgiving, I've been a few digits off before and it counts it as correct. It also asks for part a to be given in atm and part b to be given in m/s. As far as some of my answers goes most of my answers for part a are below 1 atm with the exception of a few that made no sense which were xE-22 atm. For part b my answers are all over the place ans since this is the first problem of this type I have ever worked I'm not sure what a reasonable answer would be.

15. Sep 20, 2008

### Newtime

Bump. I've tried numerous other things with no avail.