If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C , what...

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If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C , what...

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.

Express your answer to two significant figures and include the appropriate units.

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Answer 1

Answer #1

Ideal gas equation-

PV = nRT

So,

P_ideal = nRT/V = 1mol*0.0821Latm/molK*303K / 0.500L

P_ideal = 49.75atm

Van der waals equation-

(P + n²a/V²)(V-nb) = nRT

P = [nRT/(V-nb)] - n²a/V²

P=[1mol*0.0821Latm/molK*303K/(0.5L-1mol*0.03219L/mol)

- 1mol²*1.345L²atm/mol²/0.5L*0.5L

P_real = 24.876Latm/0.46781L - (1.345L²atm/0.25L²)

P_real = 53.175L - 5.38atm

P_real = 47.79atm

Difference = P_ideal - P_real = 49.75atm - 47.79atm

Difference = 1.96atm = 2.0atm

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Answer 2

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