Ideal gases A and B are in separate bottles separated by a valve. Gas A has a pressure of 329 torr and a volume of 118 cc. Gas B has a pressure of 411 torr and volume of 216 cc. The valve is opened and the temperature is held constant. What is the partial pressure of gas A?
How would you do this?
2007-04-11 15:21:51 · 5 answers · asked by xxbabiesue123xx 2 in Science & Mathematics ➔ Chemistry
Very carefully.
At start PaVa=na RT
and PbVb=nb RT
where "a" refers to gas A and "b" to gas B.
After they are mixed,
PtVt= (na+nb) RT
You can show that Pt= (PaVa+PbVb)/(Va+Vb)
Then the partial pressure of A is
(329x118)/ (118+216)
and for B (411x216)/(334)
2007-04-11 15:42:31 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
Whatever the # of molecules of Gas A that were in the original flask, they are still there in the combined flask. Except before it had a volume of 118 cc and now it is in a volume of (118 + 216) cc. So use Boyle's Law to calculate the new pressure of A based on the larger volume and you have your answer.
2007-04-11 15:33:05 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
that's Dalton's regulation of partial pressures. The stress exerted by skill of the guy gas is as though the gas replaced into occupying the quantity by myself. entire vloume is a million + 2 + 2 +a million = 6 contraptions. The partial stress of Ne is two x a million.5/6 = 0.5 Partial stress of Neon ( or helium ) is 0.5 atm.
2016-12-09 00:43:25 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The pressure of A is reduced because of the greater volume made available. It's a simple PaVa = PbVb problem. Or, look at it this way: moles A (condition a) = moles A (condition b)
moles = PV / RT R and T cancel out.
2007-04-11 15:42:51 · answer #4 · answered by Robert J 2 · 0⤊ 0⤋
329x118 =pp x (118+ 215)
2007-04-11 15:40:55 · answer #5 · answered by Anonymous · 0⤊ 0⤋