posted by Bree .
A sample of gas collected over water at 42 C occupies a volume of one liter. The wet gas has a pressure of 0.986 atm. The gas is dried, and the dry gas occupies 1.04 L with a pressure of 1.00 atm at 90 C. Using this information, calculate the vapor pressure of water at 42 C.
What I have so far:
n=PV/RT = (1.00 atm)(1.04L)/(0.0821)(363.15K) = 1.04/29.81 = 0.035 mol
I think you are right so far although I wouldn't round to 2 s.f. as you did.
I think we know Ptotal = Pdry gas + PH2O = 0.986 atm.
Then we should be able top calculate the P of dry gas at 42 C by using n = 0.0349 moles and substituting the conditions at 42 C. Substitute this Pgas into the Dalton equation above and calculate PH2O. I get something like 63 mm. I found a site on the Internet that listed PH2O @ 42 C as 61.5 mm. Check my thinking.