Two vessels containing 4dm^3 of argon and 10dm^3 of water vapour respectively at 150 degree celsius and 101 kPa were connected together.The connected vessels were allowed to cool to room temperature.What was the final pressure in the connected vessel?
Ans:20.3kPa
i used p1v1 + p2v2 = p3v3,couldn't get the answer..
Originally posted by Bigcable22:Two vessels containing 4dm^3 of argon and 10dm^3 of water vapour respectively at 150 degree celsius and 101 kPa were connected together.The connected vessels were allowed to cool to room temperature.What was the final pressure in the connected vessel?
Ans:20.3kPa
i used p1v1 + p2v2 = p3v3,couldn't get the answer..
Use PV=nRT to find the number of moles of argon only (before connecting the vessels).
After connecting the vessels (ie. new combined volume), apply PV=nRT with n = number of moles of argon only.
Solve for P.
why dont i have to take into consideration of initial volume h20 vapour?
Originally posted by Bigcable22:why dont i have to take into consideration of initial volume h20 vapour?
Because it doesn't matter how much water vapour you initially had. Since at rtp (after you connect the vessels), it'll be a liquid and have zero vapour pressure, and won't affect your final pressure. Only the argon gas (ie. the only remaining gas in the combined containers) will exert a pressure.
We can still have water vapour at rtp.
Question is insufficient in that sense.
Henry's Law?