If 1.00 mol of argon is placed in a 0.500-L container at 19.0 ∘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

Haleigh Johns

27 August, 15:52

If 1.00 mol of argon is placed in a 0.500-L container at 19.0 ∘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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Darian Arnold · Answer 1

41.083 atm 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.

Explanation:

Data given for argon gas:

number of moles = 1 mole

volume = 0.5 L

Temperature = 19 degrees or 292.15 K

a = 1.345 (L2⋅atm) / mol2

b = 0.03219L/mol.

R = 0.0821

The real pressure equation given by Van der Waals equation:

P = (RT : Vm-b) - a : Vm^2

Putting the values in the equation:

P = (0.0821 x 292.15) : (0.5 - 0.03219) - 1.345: (0.5) ^2

= 23.98:0.4678 - 1.345 : 0.25

= 51.26 - 5.38

= 45.88 atm is the real pressure.

The pressure from the ideal gas law

PV = nRT

P = (1 x 0.0821 x 292.15) : 0.5

= 4.797 atm

the difference between the ideal pressure and real pressure is

Pressure by vander waal equation - Pressure by ideal gas law

45.88 - 4.797

= 41.083 atm. is the difference between the two.