mak10 Posted November 16, 2004 Posted November 16, 2004 flask X contains 1 dm3 of helium at 2 kPa pressure and flask Y contains 2 dm3 of neon at 1 kPa pressure. if the flasks are connected at constant temperature, what is the final pressure? any ideas on how to solve this?? -mak10
swansont Posted November 16, 2004 Posted November 16, 2004 At constant T, PV is also constant. In a mixture, partial pressures add to be the total pressure.
mak10 Posted November 16, 2004 Author Posted November 16, 2004 partial pressures added gives 3 kPa... but the correct answer is different. -mak10
jdurg Posted November 16, 2004 Posted November 16, 2004 Correct. That's because the overall volume has changed. So if you use P1V1 = P2V2 you can calculate the pressure of each gas at the new volume. Then you add up the partial pressures and you have your answer. (I got 4/3 kPa.)
mak10 Posted November 16, 2004 Author Posted November 16, 2004 i still dont get it. whats the new volume? how did you get it? your answer is correct.... it is 4/3 kPa... but can you show me, briefly, the exact steps you took to get this answer? am confused... thanks jdurg! -mak10
jdurg Posted November 16, 2004 Posted November 16, 2004 i still dont get it. whats the new volume? how did you get it? your answer is correct.... it is 4/3 kPa... but can you show me' date=' briefly, the exact steps you took to get this answer? am confused... thanks jdurg! -mak10[/quote'] Sure. According to your initial post, the total volume will be 3 dm^3. This is what you get by adding the volume of gas 1 to the volume of gas 2. Since the flasks are connected, the overall volume will be the sum of the volume of the two flasks. To get the pressures, I just used P1V1 = P2V2 and figured out the new pressure of each gas. Gas X; (2kPa)(1 dm^3) = P2(3 dm^3). P2 = 2/3 kPa. Gas Y; (1kPa)(2 dm^3) = P2(3 dm^3). P2 = 2/3 kPa. So the partial pressure of each gas is 2/3 kPa. The total pressure is the sum of the partial pressures, so you just add the two partial pressures and get 4/3 kPa.
mak10 Posted November 16, 2004 Author Posted November 16, 2004 ingenious... thanks a lot jdurg! -mak10
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