Question

Calculate the pressure exerted by benzene for a molar volume of 2.4 L at 600. K using the Redlich-Kwong equation of state:
P=RTVm−b−aT√1Vm(Vm+b)=nRTV−nb−n2aT√1V(V+nb)
The Redlich-Kwong parameters a and b for benzene are 452.0 bar dm6 mol−2 K1/2 and 0.08271 dm3 mol−1, respectively.

Is the attractive or repulsive portion of the potential dominant under these conditions?

Answer 1

[* The Redlich-Kwong equation is wrongly typed and also the unit of 'a' is wrong]

2.4 L = 2.4dm³mol^-1

P= (RT/V_m -b) - a/T * 1 / V_m(V_m + b)

= (8.314 x 10^-2 bar dm³ mol^-1K^-1 x 600 K / 2.4 dm³mol^-1-0.08271 dm³mol^-1) - (452.0 bar dm³mol^-2 K^1/2/ 600K)* 1/2.4 dm³mol^-1x (2.4 dm³mol^-1 - 0.08271dm³mol^-1)

= 49.884 / 2.31729 bar - 452.0 /136.201037 bar

= 21.5268 bar - 3.3186 bar

=>P= 18.2 bar

Now, P_ideal= RT/V = 8.314 x 10^-2 L bar mol^-1K^-1 x 600K / 2.4 L = 49.884 /2.4 bar = 20.8 bar

Since P < P_ideal , the attractive part of the potential dominates.