Question

Calculate the pressure exerted by Ar for a molar volume 0.45 L at 200 K using the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar dm⁶ mol^-2 and 0.0320 dm³mol^-1, respectively. Please write your answer (unit: bar) with 2 decimals, as 12.23. Please do not add unit to your answer.

Find compression factor Z for this problem

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

For argon

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

(P+ (1.355 ) / (0.45^2)) * (0.45- 0.0320 ) = 0.08314 bardm3/molK *(200)K

(P+ 6.691) =  0.08314*(200)/ (0.45- 0.0320 )

P = 39.7799-6.691

P = 33.0889 bar

P = 33.09 no units

Z for this problem:

Pr = P/Pc = 33.09 /48.7459 = 0.67882

Tr = T/(Tc) = (200)/(-122+273.15) = 1.323

From Z graph:

Z = 0.87 APPROX, fom graph