Asked by Dave
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<sup>o</sup> C from PV = nRT.
Calculate mols dry gas from the second set of conditions @ 90<sup>o</sup> 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.
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<sup>o</sup> C from PV = nRT.
Calculate mols dry gas from the second set of conditions @ 90<sup>o</sup> 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.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.