12.0 moles of gas are in a 3.00L tank at 23.1?C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2?atm/mol2 and b=0.0430 L/mol.
Given, moles = n = 12.0 mol
Volume = v = 3.00 L
Temperature = T = 23.1 degree C = 296.1 K
Ideal gas equation => PV = nRT
P = nRT /V
P = (12.0) ( 0.0821) (296.1) / 3.00
P = 97.24 atm.
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Given,
Vanderwaal constants
a=2.300 L2 atm/mol2 and
b=0.0430 L/mol.
moles = n = 12.0 mol
Volume = v = 3.00 L
Temperature = T = 23.1 degree C = 296.1 K
[P + (n^2a/V^2)] [V-nb] = nRT
[P + (144 x 2.300) / 9] [ 3 - 0.516] = 12 x 0.0821 x 296.1
[P + 36.8] - 2.484 = 291.7
P = 257.40 atm
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Difference = P vanderwaal - P ideal
= 257.40 - 97.24
= 160.16 atm
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