Question

# a) the pressure of pure water vapor at 25 ° C equals 23.758, if the total...

a) the pressure of pure water vapor at 25 ° C equals 23.758, if the total pressure on the liquid is only due to water vapor, calculate the vapor pressure of water at 25 ° C if enough gaseous oxygen is added Steam to give a total pressure of 1,000atm. Consider that oxygen does not dissolve in water.

b) Calculate the molar fraction of oxygen dissolved in the water under the conditions of part a. The constant for Henry's law in this is almost k = 985 torr. Calculate the effect of this oxygen dissolved in the vapor pressure.

(a) As the pressure is only due to water vapor, the vapor pressure of water = 23.758/ 1000 = 0.023758 atm

(b) For water, assume Raoult's law:

where Ptotal = 1000 atm, Pvapor = 0.023758 atm

For oxygen, Henry's Law

where Ptotal = 1000 atm, H2 = 985 torr = 1.296 atm

Adding equation 1 and 2, and substituting following: y1 +y2 =1, x1 = 1-x2

Results in expression:

x2 = 785.995

The result is very weird. I think the value of Henry's constant should be >1000 atm.

The dissolution of oxygen in water will increase the water vapor pressure.

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