Question

Derive an expression for reversible work performed on/by a Redlich-Kwong gas. Use your expression to calculate and compare w_rev for reversible compression of carbon dioxide and methane gases from 10.0 L to 1.0 L at 25^oC. Which gas requires more energy to be compressed? What molecular property contributes most to the difference in work required?

Answer 1

The equation of state of a Redlich -Kwong gas is

The work done from V1 to V2 is given by

Using the fact that integral of 1/x is log x,

The constants are given by

a= 0.42748 * R²T_c^2.5 /P_c

b = 0.08664 * RT_c /P_c

R = 8.314 J/KM

V₂ = 1.0

V₁ = 10.0

T = 25 C = 298 K

Taking n = 1,

For carbon di oxide,

T_c = 304.25 K

P_c = 7.39 MPa

So, W = -5705.22 J

For methane,

T_c = 190.55 K

P_c = 4.495 MPa

So, W = -5705.06 J

Carbon di oxide requires more energy to compress per mole.

The quantities that this depended on are critical pressure and critical temperature. With more dependence on temperature