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 law of corresponding states if we compare two gases at equivalent states relative to their critical points they should have the same properties.

Here reduced pressure for methane and water is

Pr (methane) = 72.9/45.98= 1.585

Pr(water ) = 349.68/220.55 = 1.585

As Pr(methane )= Pr(water), other propeties should also be same

Now ,

Vm/ Vmc (methane) = Vm/ Vmc (water)

or 0.250/0.099 = Vm/ 0.05595

or Vm = 0.1413 dm3mol-1

similarly,

T/Tc (methane) = T/Tc(water)

or 273.15/190.53 = T/647.126

or T = 927.47 K

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