Ideal Gas Law and pressure in a holder

[JSGandora]

Homework Statement

What is the pressure inside a $50.0$ L container holding $105.0$ kg of argon gas at $20^\circ$ C?

Homework Equations

Ideal Gas Law: PV=nRT

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?

 

[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.

 

[JSGandora]

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

 

[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).

 

[JSGandora]

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

 

[SpecialKM]

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

 

[JSGandora]

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

 

[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.

 

[JSGandora]

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