Calculate the pressure of Ar at 300 K when its molar volume (V/n) is only 1.42 L

Question

Calculate the pressure of Ar at 300 K when its molar volume (V/n) is only 1.42 L/mol using the van der Waals equation of state. For Ar, the van der Waals parameters are a = 1.355 bar dm6/ mol2 and b = 0.0320 dm3 /mol.

            In Chem 457, you will learn that the van der Waals parameters (a and b) can be combined to calculate the well depth (depth of attraction) in the Ar-Ar interaction potential (i.e., depth of the minimum in a plot of energy versus separation distance for two Ar atoms, like Fig. 1.7). This well depth turns out to be ~1 kJ/mol for the values of a and b given in part (a) of this problem. In contrast, the strength of a typical C-H bond in organic molecules is ~400 kJ/mol, 400-fold bigger! This also represents the depth for a similarly-shaped attractive well to Fig. 1.7, but instead for the C--H separation which is held together by this much stronger covalent-bond attraction.

            Sometimes well depths and bond energies are given in units of eV, commonly used in quantum mechanics. It’s useful to remember that one eV corresponds to ~100 kJ/mol (actually 97.0 kJ/mol).

Explanation / Answer

we know that gas equation for the normal gas

P = [nRT/ V-nb] - [n2 a / V2]

from the problem

V = 1.42 L, n = 1 mol, T = 300K

vander wall constants a = 1.355 bar dm^6 mol-2 , b = 0.0320 dm3 mol-1

and we know the gas constant value R = 0.08314 L bar K-1 mol-1.

plug in everything to get the pressure

P = [1 x 0.08314 x 300/ 1.42-1 x 0.0320] - [12 1.355 / 1.422]

P = 17.9697 - 0.67212

P = 17.297 atm

pressure using the van der waals equation is less than the pressure using the ideal gas equation, the attractive part of the potential dominates

where as if it is reverse then repulsive forcess

calculate the pressure using the ideal gas

PV = nRT

P = nRT/V

P= 1 x 0.08314 x 300/1.42

P = 17.565 atm

so pressure using the van der waals equation is less than the pressure using the ideal gas equation, the attractive part of the potential dominates

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