If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation) ? For argon, a=1.345 (L2⋅atm) / mol2 and b=0.03219L/mol.

Question

Pablo Burns

Chemistry

1 January, 23:03

If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation) ? For argon, a=1.345 (L2⋅atm) / mol2 and b=0.03219L/mol.

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Answers (1)

Know the Answer?

Kash Andrews · Answer 1

The answer to your question is:

Explanation:

Ideal gases

PV = nRT

P = nRT / V

P = (1) (0.082) (303) / 0.5

P = 49.7 atm

Real pressure

P = V² (nRT) / (V - nb) - an²

P = (0.5) ² (1) (0.082) (303) / (0.5 - (1) (0.03219) - (1.345) (1) ²

P = (6.2115) / (0.4678) - 1.345

P = 13.28 - 1.345

P = 11.93 atm