Question

What is the molar volume of n-hexane at 660K and 91 bar according to (a) the ideal gas law and (b) the van der Waals equation? For the latter, first use the equations at the end of the CHE346 class notes on van der Waals gases to determine values of the constants a and b in the van der Waals equation from the critical constants for n-hexane, Tc = 507.7 K and Pc = 30.3 bar.

Answer 1

n-Hexane, P=660K , P = 91bar

Vm = ?

a) PV = nRT

PVm = RT where Vm = V/n = molar volume

Vm = RT/P

R = 83.14472 L*mbar*K^-1*mol^-1

Vm = 83.14472 L*mbar*K^-1*mol^-1 * ( 1bar/1000mbar) * (660 K) / 91bar = 0.6030 L/mol

b) V_mc = 3b; Pc = a/27b²; Tc = 8a/27bR

[P+a/V_m²][V_m-b] = RT

V_m³-[b+RT/P]V_m² + V_m a/P - ab/P = 0 a cubic equation that have to solve in order to obtain Vm

b = V_mc /3;   

Tc = 507.7 = 8a/27bR a = 507.7*27*bR/8

Pc = 30.3 = a/27b² a= 30.3 *27*b²

507.7*27*bR/8 = 30.3 *27*b²

507.7*27*R/8 = 30.3 *27*b

b = 507.7*27*R/(8*30.3 *27)

Vmc =3b = 3 [507.7*27*R/(8*30.3 *27)]

Vmc = 0.5224 Lmol^-1

a =  30.3 *27*b² = 30.3 * 27 * (0.5224)² = 223.26

V_m³-[0.5224+ 83.14472E-3 * 660/91]V_m² + V_m 223.26/91 -223.26*0.5224/91 = 0

Ax³ + Bx² + Cx + d = 0

A=1

B=-1.1254

C = 2.4534

D = 1.2816

X = 0.60 = Vm