Ideal Gas Law and pressure in a holder

  • Thread starter JSGandora
  • Start date
  • #1
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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
[itex]
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
[/itex]
which seems much too large. Am I doing something wrong?
 
Last edited:

Answers and Replies

  • #2
52
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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
95
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Oh wow, thanks! It's hard to believe the massive pressure inside, which was why I was skeptical. Thanks again.
 
  • #4
52
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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
95
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Oh, should it be [itex]39.95\times 2[/itex] since it's diatomic?
 
  • #6
52
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Nono, on the work you've showed in your original post, it looks like you've put 36 g/mol.
 
  • #7
95
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But the molecular mass of Argon is 39.95 isn't it? I'm just substituting the 36 for 39.95x2.
 
  • #8
52
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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
95
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Oh, yeah. For some reason I thought argon was diatomic...
 

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