Two cylinders at 27C are connected by a closed stopcock ystem. One cyclinder contains 2.4 L of hydrogen gas at o.600 atm; the other cylinder contains 6.8L of helium at 1.40 atm. Assume valve take up no room.

Question

what is the total pressure when the valve is open ?
I don't know how to find the total pressure
(0.600 atm)(L)/(0.0821)(300K)=5.846x10^-2
n=(1.40 atm)(6.8L)/(0.0821)(300k)=3.865x10^-1
Total pressure =(n1+n2..)(RT/V)

Answer 1

I responded to your original post and asked, "what is your question?"

Answer 2

Isnt PV a constant in both continers?
Then the sum of P1V1 + P2V2 is a constant.

So PfinalVfinal=P1V1+P2V2
pfinal= (p1v1+P2v2)/(v1+V2)
=(.6*2.4 + 1.4*6.8)/9.2 atm
= 1.19 atm Compare that to what you got.

Answer 3

My question is how do I find the total pressue, I found the number of moles for each situation do i add them and plug it into the Total pressure =(n1+n2..)(RT/V)
This is what I have is this correct, if so then I do not know what to do after ths.
(0.44496)(27)(0.0821)/V(I do not know what to plug in here

Answer 4

for V, the sum of the initial volumes.

Answer 5

Yes, you are doing it ok although it's a little longer than the way Bob Pursley has worked it.
P = nRT/V =
n = n1 + n2 = ok
R is ok.
T must be changed to 300 K.
V = 2.4 L + 6.8 L = 9.2 L.
And the answer comes out to 1.19 atm as Bob Ps post indicates.