hydrogen gas liquid equation


by chemister
Tags: equation, hydrogen, liquid
chemister
chemister is offline
#1
Mar22-05, 07:35 PM
P: 21
If 277 L of LIQUID hydrogen has a mass of 19.7 kg, what volume will this amount of hydrogen occupy as a GAS at 25 degrees celcius and 1 atm?



Is there some kind of equation that I can use to make this problem easier?

THANKS!
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Cesium
Cesium is offline
#2
Mar22-05, 07:42 PM
P: 274
PV = nRT

You'll use this equation for almost every gas problem.

R (ideal-gas constant) = 0.0821 L-atm/mole-K
V = volume (L)
P = pressure (atm)
n = amount (moles)
T = temperature (K)

V = nRT/P

You need to convert the 19.7 kg H2 to moles H2 and the 25 C to Kelvin.

If you need more help, go ahead and ask.
chemister
chemister is offline
#3
Mar22-05, 07:51 PM
P: 21
Do you have to take into account the hydrogen beginning as a LIQUID and ending as a GAS?

Cesium
Cesium is offline
#4
Mar22-05, 07:54 PM
P: 274

hydrogen gas liquid equation


No because when you go from a liquid to a gas, the number of moles of H2 will still stay the same and therefore you can predict the volume.
chemister
chemister is offline
#5
Mar22-05, 08:01 PM
P: 21
When converting 19.7 Kg to moles would you use H2 (being 2.0158g) or just H (being 1.0079g)?
Cesium
Cesium is offline
#6
Mar22-05, 08:03 PM
P: 274
Good question. Always use the diatomic molecule...so H2.
chemister
chemister is offline
#7
Mar22-05, 08:07 PM
P: 21
Thanks for your help. So if i have done things right my answer should be close to 2.39 X 10^5 L
Cesium
Cesium is offline
#8
Mar22-05, 08:12 PM
P: 274
Yes I got a similiar answer. By the way, what value are you using for R? 0.0821 or something more precise?
chemister
chemister is offline
#9
Mar22-05, 08:15 PM
P: 21
For this problem I just used 0.0821, but usually I use 0.082057


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