Question

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.

Answer #1

(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 P_{total} = 1000 atm, P_{vapor} = 0.023758
atm

For oxygen, Henry's Law

where P_{total} = 1000 atm, H_{2} = 985 torr =
1.296 atm

Adding equation 1 and 2, and substituting following:
y_{1} +y_{2} =1, x_{1} =
1-x_{2}

Results in expression:

x_{2} = 785.995

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

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