Question

Consider the compression at 15.6 C of 0.35 mol of H2O(g) (assumed ideal) from 0.10 atm to 1.00 atm. Calculate the values of ΔG when the process is carried out (a) isothermally and reversibly and (b) isothermally and irreversibly under a constant external pressure of 1.00 atm.

Answer 1

PV = nRT
V = nRT/ P
V₁ = 0.35 x 0.082 x 288.6 / 0.1

      = 82.83 L = 0.08282 m³
V₂ = 0.35 x 0.082 x 288.6 / 1

      = 8.283 L = 0.008282 m³
(a) isothermally and reversibly

W    = - G

       = - nRT ln (V₂/V₁)

      = 0.35 x 8.314 x 288.6 x ln (0.008282/0.08282)

        = 1933.7 J

ΔG = - 1933.7 J
(b) isothermally and irreversibly under a constant external pressure of 1.0 atm.

W = - ΔG

    = - p(V₂ - V₁)

    = 1.01 x 10^5 x (0.008282 - 0.08282)

    = 7528.3 J

ΔG = - 7528.3 J