# Ideal Gas Law and pressure in a holder

1. Feb 18, 2012

### JSGandora

1. The problem statement, all variables and given/known data
What is the pressure inside a $50.0$ L container holding $105.0$ kg of argon gas at $20^\circ$ C?

2. Relevant equations
Ideal Gas Law: PV=nRT

3. The attempt at a solution
From the ideal gas law, I get
$P=\frac{nRT}{V}=\frac{\frac{105.0\times 10^3g}{36g/mol}\times 0.08214\frac{L\cdot atm}{mol\cdot K}\times 293^\circ K}{50.0 L}=25258atm$
which seems much too large. Am I doing something wrong?

Last edited: Feb 18, 2012
2. Feb 18, 2012

### SpecialKM

Nope! Your answer is quite reasonable, and it seems like nothing looks wrong in your setup of P= nRT/V and all the units match, your value for R is correct and you've converted C to K. Unless you made a calculating mistake, everything looks correct to me.

3. Feb 18, 2012

### JSGandora

Oh wow, thanks! It's hard to believe the massive pressure inside, which was why I was skeptical. Thanks again.

4. Feb 18, 2012

### SpecialKM

On second glance, your molar mass for Argon seems to be off, also, you've made a calculation error. Try again and tell me what you got. (I've worked it out this time).

5. Feb 18, 2012

### JSGandora

Oh, should it be $39.95\times 2$ since it's diatomic?

6. Feb 18, 2012

### SpecialKM

Nono, on the work you've showed in your original post, it looks like you've put 36 g/mol.

7. Feb 18, 2012

### JSGandora

But the molecular mass of Argon is 39.95 isn't it? I'm just substituting the 36 for 39.95x2.

8. Feb 18, 2012

### SpecialKM

By looking at the problem you've presented it should look like this:

P = [(105000/39.95)(0.08214)(293)]/(50)
P = 1265.1 atm
P = 1.27 x 10^3 atm

I've left out the units for simplicity, but you should write them in yourself.

9. Feb 18, 2012

### JSGandora

Oh, yeah. For some reason I thought argon was diatomic...