How do you calculate the partial pressure of O2 in the air if the atmospheric pressure is 370 mmHg, PH2O is 23 mmHg, and concentration of Oxygen is 20.93%?

1 Answer
Truong-Son N.
May 15, 2018

#P_(O_2) = "72.6 torr"#

The atmosphere contains water vapor, from which we must subtract the water. We assume water vapor is an ideal gas, and thus Dalton's law of partial pressures applies:

#P_"wet air" = P_(H_2O ) + overbrace(P_(O_2) + . . . )^(P_"dry air")= "370 mm Hg"#

We lumped it as

#P_"dry air" = P_(O_2) + . . . # except for #P_(H_2O)#,

since obviously, #P_("wet air") = P_"dry air" + P_(H_2O)#.

So, at approx. #24.5^@ "C"# at which #P_(H_2O) = "23 mm Hg"#,

#P_"dry air" = overbrace("370 mm Hg")^(P_"wet air") - overbrace("23 mm Hg")^(P_(H_2O)) = "347 mm Hg"#

From the same principles,

#P_(O_2) = chi_(O_2(v))P_"dry air"#

where #chi_(O_2(v))# is the mol fraction of #O_2# gas in the vapor phase above the surface of the water.

Therefore, the partial pressure of #O_2# is simply:

#color(blue)(P_(O_2)) = overbrace(0.2093)^(chi_(O_2(v))) cdot overbrace("347 mm Hg")^(P_"dry air")#

#=# #color(blue)("72.6 mm Hg")#

What is this in torr?

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