Question

The pressure in a vessel that contains only methane and water at 70^oC is 10.0 atm. At the given temperature, the vapor pressure of pure water is 0.3075 atm, and the Henry’s Law constant for methane in water is 6.66 x 10⁴ atm.

(a) Using Raoult’s Law, estimate the partial pressure of water and methane in the vapor phase. You may assume that the mole fraction of water in the liquid phase is close to one.

(b) Using your answer from part (a), estimate the mole fraction of methane in the liquid phase using Henry’s Law. Was the assumption that you made in part (a) (e.g., that the mole fraction of water in the liquid phase is close to 1) a good assumption?

Answer 1

Pressure in vessel = 10 atm

Vapor pressure of water at 70 oC = 0.3075 atm

Henry’s Law constant for methane in water = 6.66 x 10⁴ atm/mole fraction.

(a). Since mole fraction of water in liquid phase is nearly 1.

So, putting Xw = 1

Partial pressure of water Pw = Xw * P^o_w

= 1 * 0.3075

= 0.3075 atm

Partial pressure of methane Pm = Xm * K_H   ........(1)

(b). Mole fraction of water = 1 - Xm

Partial pressure of water = (1 - Xm) * 0.3075

Total pressure is given as:

P = Pm + Pw

10 = Xm * K_H + (1 - Xm) * 0.3075

10 = Xm * 6.66 x 10⁴ + (1 - Xm) * 0.3075

10 = Xm * 6.66 x 10⁴ + 0.3075 - 0.3075 * Xm

Xm = 1.46x10^-4  

Hence, mole fraction of methane = 1.46x10^-4  

Putting in equation (1):

Pm = 1.46x10^-4 * 6.66 x 10⁴

Pm = 9.7236 atm

Hence, partial pressure of methane = 9.7236 atm