Question

If 1.00 mol of argon is placed in a 0.500-Lcontainer at 26.0 ∘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)/mol2and b=0.03219L/mol.

Express your answer to two significant figures and include the appropriate

Answer 1

The Van der Waals equation is a description of real gases, it includes all those interactions which we previously ignore in the ideal gas law.

It considers the repulsion and collision, between molecules of gases. They are no longer ignored and they also are not considered a"point" particle.

The idel gas law:

PV = nRT

P(V/n) = RT ; let V/n = v; molar volume

P*v = RT

now, the van der Waals equation corrects pressure and volume

(P+ a/v^2) * (v - b) = RT

where;

R = idel gas law; recommended to use the units of a and b; typically bar/atm and dm/L

T = absolute temperature, in K

v = molar volume, v = Volume of gas / moles of gas

P = pressure of gas

Knowing this data; we can now substitute the data

given

a = 1.345

b = 0.03219

(P+ a/v^2) * (v - b) = RT

v = V/mol = 0.5/1 = 0.5

(P + 1.345/(0.5^2)) * (0.5 - 0.03219) = 0.082*(26+273)

P = 0.082*(26+273)/ (0.5 - 0.03219) - 1.345/(0.5^2)

P = 47.030 atm

for ideal gas:

PV = nRT

P = nRT/V = (1*0.082*(26+273))/(0.5) = 49.036

dV = Videal - Vreal = 49.036-47.030 = 2.006 L

2 sig fig --> 2.0 L