At 25 degree C and 1 bar partial pressure the solubility of

At 25 degree C and 1 bar partial pressure, the solubility of ethane in water is very small: the equilibrium mole fraction is x_C2H6 = 0.33 times 10^-4. What is the solubility of ethane at 25 degree C when the partial pressure is now 35 bar? Assume that at 25 degree C, the compressibility factor of ethane is given by the empirical relation, z = 1 - 7.63 times 10^-3 P - 7.22 times 10^-5P^2, where P is in bar. Also, at 25 degree C, the saturation pressure of ethane is 42.07 bar and that of water is 0.0316 bar. Oil reservoirs below ground frequently are in contact with underground water, and in connection with an oil drilling operation. Suppose you need to calculate the solubility of water in a heavy oil at the underground conditions: temperature, 140 degree C; pressure, 410 bar. Experiments at 140 degree C and 1 bar indicate that the solubility of steam in the oil is, x_1 = 3.5 times 10^-3, where x_1 is the mole fraction of steam. Assume Henry\'s Law in the form of f_1 = H (T)x_1, where H(T) is a constant, dependent only on the temperature and f_1 is the fugacity of H_2O. Also assume that the vapor pressure of the oil is negligible at 140 degree C. Use steam table for data needed for H_2O.

Solution

solution: here above problem 4 is solved by henry law as it states that solubility is proprtional to partial pressure of solution,

hence

x=Hxp*P

1) for p= 1 bar, x=.33*10^-4

we get

Hxp=.33*10^-4/10^5=3.3*10^-10 mol/m3 Pa

2) for second step

P=35 bar and Hxp=3.3*10^-10 mol/m3Pa

we get

x=Hxp*P=3.3*10^-10*35*10^5=1.155*10^-3

it is solubility of ether at 25 c and 35 bar partial pressure

3) compression factor is ratio of gas molar volume to tdeal gas molar volume

z=vg/Vi

where Vi=22.414 L

here value of z=1-7.6*.001*35-7.22*.00001*35^2=.644505

hence Vg=z*vi=.644505*22.414=14.44593 l

5) hence voume of ethane mixed is=x*Vg=1.155*10^-3*14.44593=.01668lit