Posted by **Lori** on Thursday, September 13, 2012 at 7:38pm.

15.0 moles of gas are in a 6.00L tank at 298.15K. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300 L^2*atm/mol^2 and b=0.0430 L/mol.

I found the idea gas pressure to be 61.2 atm by using the ideal gas law. However I can't seem to get the right value for the pressure of methane. Below are my workings, can anyone see what I'm doing wrong?

P= [(nRT)/(V-nb)] - [(an)^2/V^2]

P=[(15 mol*0.08206*298.15K)/(6.00L-(15mol*0.0430)] -[(15mol*2.3)^2/(6.00L)^2]

P=68.53-33.06

P=35.47 atm

- Chemistry -
**DrBob222**, Thursday, September 13, 2012 at 10:02pm
In my text the second term is an^2 and not (an)^2.

68.53 is ok.

The second term is -[(15)^2 x 2.3]/36 = ?

- Chemistry -
**Jalynn**, Thursday, December 6, 2012 at 11:38pm
what is the absolute temperature of the gas at which 3.33 10-3 mol occupies 440. mL at 708 torr

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