The pressure of methane in the gas phase is 72.9 bar when T = 273.15 K and V_m =

Question

The pressure of methane in the gas phase is 72.9 bar when T = 273.15 K and V_m = V/n = 0.250 dm^3 mol^-1. V_m is the molar volume. The critical parameters for methane are T_c = 190.53 K, V_m, c = 0.09900 dm^3 mol^-1, and P_c = 45.980 bar. V_m, c denotes the molar volume at the critical point. For water, the values of the critical parameters are T_c = 647.126 K, V_m, c = 0.05595 dm^3 mol^-1, and P_c = 220.55 bar. Find values of T and V_m for which the pressure of water in the gas phase is predicted to be 349.68 bar, based on the Law of Corresponding States.

Explanation / Answer

According to van der Waals, the theorem of corresponding states indicates that all fluids, when compared at the same reduced temperature and reduced pressure, have approximately the same compressibility factor.

Reduced pressure of methane, Pr1 = P1/Pc1 = 72.9 bar / 45.98 bar = 1.5855

Reduced pressure of water, Pr2 = P2/Pc2 = 349.68 bar / 220.55 bar = 1.5855

Pr1 = Pr2

Reduced temperature of methane, Tr1 = T1/Tc1 = 273.15 K / 190.53 K = 1.4336

Based on law of corresponding states, Tr1 = Tr2

Reduced temperature of water, Tr2 = T2 / Tc2 = T2 / 647.126 = 1.4336

Temperature of water, T2 = 647.126 * 1.4336 = 927.74 K

Z = P * Vm / (R T)

Z1 = Z2

P1 * V1 / T1 = P2 * V2 / T2

72.9 * 0.25 / 273.15 = 349.68 * V2 / 927.74

Molar volume of water, V2 = 0.177 dm3 mol-1

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