Posted by **Dave** on Monday, January 1, 2007 at 3:45pm.

A sample of gas collected over water at 42 degrees C occupies a volume of one liter. The wet gas has a pressure of .986 atm. The gas is dried and the dry gas occupies 1.04 L with a pressure of 1.00 atm at 90 degrees C. Using this information, calculate the vapor pressure of water at 42 degrees C.

I was going to approach this problem by finding the pressure of the gas when dry at 42 degrees C, and then do

P[dry]= P[total] - P[water pressure]

but I realized this would get me nowhere as I didn't know the total pressure. Is there a better way to do this problem?

Calculate total mols H2O and gas @ 42

^{o} C from PV = nRT.

Calculate mols dry gas from the second set of conditions @ 90

^{o} C.

The difference in mols should be the mols H2O vapor and that put back into the PV=nRT at 42 should give P of water at 42.

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