If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘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

Chemistry

Lilibeth

Today, 14:37

If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘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)

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Julius Fritz · Answer 1

Hey There!

ideal gas law:

PV = nRT

P = nRT/V

P = (1*0.082) (18+273) / (0.5) = 47.724 atm

For VDW:

(P + a (n/V) ²) (V - nB) = nRT

P = nRT / (V - nB) - a (n/V) ²

P = 1*0.082 * (18+273) / (0.5-1*0.03219) - 1.345 * (1/0.5) ² = 45.62

P = 45.62 atm

Pdif = P2-P1 = 47.724 - 45.62 = 2.104 atm

If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘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.

If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘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.