# Pressure and Temperature question: Ideal Gas Laws

1. Dec 28, 2011

### shyguy79

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

Find the pressure and temperature when given the following data on helium gas:

Volume (V) = 0.10 m^3 ** Helium mass (m) = 4.0 amu ** Number of atoms (N) = 3.0 x 10^24

Assuming PV = 1 then P=1/V then Pressure = 1.0 / 0.10m^3 = 10Pa

2. Relevant equations
PV = NkT arranged for T = PV/Nk OR PV = nRT arranged for T = PV/nR

3. The attempt at a solution
Mm = Mr x 10-3 = 4.0 x 10-3 so the mass is 4.0 x 10^-3 / 6.02 x 1023 = 6.6 x 10^-27

The temperature T is then:
T = PV/Nk
T = P x 0.10m^3 / 3.0 × 10^24 mol−1 x 1.381 × 10^−23 J K−1

But I have no idea how to get Pressure - Ive tried using Boyles law PV = constant but get lost

Any pointers would be deeply appreciated

2. Dec 29, 2011

### Staff: Mentor

One mole of an ideal gas will occupy a volume of 22.4 liters at STP (Standard Temperature and Pressure, 0°C and one atmosphere pressure).

Though I think you still are lacking some information.

3. Dec 29, 2011

### shyguy79

Thanks for the reply, that's all I've been given - been told Charles Law and Botles Law may come in handy but don't know how!?

4. Dec 29, 2011

### Staff: Mentor

Charles' and Boyle's Laws are just special cases of the general gas law.

You are looking at: $PV \div T$ = constant
but I think you can only find the quotient $P \div T$

5. Dec 29, 2011

### technician

You are given the number of atoms = 3 x 10^24 which = 5moles (n)
If you substitute this into PV = nRT you can get T but the answer is not sensible ?????
I can't see why you have been given the mass of He atom = 4 amu !!!

6. Dec 29, 2011

### Staff: Mentor

It's possible that one might have to resort to the isentropic relationships for an ideal gas.

7. Dec 29, 2011

### technician

I think I have just realised that shyguy is assuming that PV=1.... it is not a given in the question....I don't know any more to add!!!!!!

8. Dec 29, 2011

### shyguy79

Yes, but working out the math the pressure would be 10 Pascals and the Temperature would 0.0012 Kelvin - and that make hardly any sense!

9. Jan 4, 2012

### Staff: Mentor

Does your textbook provide the answer to this?

10. Jan 4, 2012

### shyguy79

No, it's an assignment question but reading through the full wording it appears that the peak velocity is 1100m/s

11. Jan 4, 2012

### Staff: Mentor

Is it really late at night where you are? :uhh:

Velocity does not enter into the question. Well, only tangentially.

12. Jan 5, 2012

### mullac phie

13. Jan 5, 2012

### shyguy79

Thank you!! Just what I was looking for!