Gas Law Problem: Finding Final Pressure in Two Connected Containers

[bcjochim07]

Homework Statement

Containers A and B hold the same gas. The volume of B is four times the volume of A. The two containers are connected by a thin tube (negligible volume) and a valve that is closed. The gas in A is at 300 K and pressure 1.0 * 10^ Pa. The gas in B is at 400 K and 5.0 * 10^5 Pa. Heaters will maintain the temperatures of A and B even after the valve is opened. After the valve is opened, gas will flow one way or the other until A and B have equal pressure. What is this final pressure?

Homework Equations

The Attempt at a Solution

The part with the two different temperatures confuses me. Here's what I tried:
Maybe this can be considered a constant temperature process

P1V1=P2V2 Va = Volume of A Vb = 4Va

(1.0 * 10^5 Pa)(Va) = (P2)(5Va)
P2= 20000 Pa ---- I am quite sure this is not correct because if I choose the other gas, I get a different answer. Also, I think the number of moles is changing because the gas flows between the compartments. Could somebody give me a hint as to how to work in all of these factors into an equation? Thanks.

 

[Borek]

At equilibrium:

1. Pressure is identical in both tanks.
2. Amount of gas is identical to initial.

This gives two equations in two variables. None of them is pressure, but once you know how much gas is in each tank, calculating pressure is a breeze.

Could be it can be done much easier, that's my first idea.

 

[bcjochim07]

I've been working on this, but still can't figure out what to do:

nb=moles in b initially na=moles in a initially

(5*10^Pa)(4Va) = nb(8.31)(400K) nb= 601.68Va
(1*10^5)(Va)=na(8.31)(300K) na= 40.11Va

nb= 15na, so initially there are 15na moles

(4/5)(16na) = 12.8na the moles that go into B
(1/5)(16na) = 3.2na the moles that go into A

(Pa)(Va)=(3.2na)(8.31)(300K)
(Pb)(Vb)=(12.8na)(8.31)(400K)

but if I solve for the pressures and set them equal, all the variables cancel

 

[alphysicist]

I've been working on this, but still can't figure out what to do:

nb=moles in b initially na=moles in a initially

(5*10^Pa)(4Va) = nb(8.31)(400K) nb= 601.68Va
(1*10^5)(Va)=na(8.31)(300K) na= 40.11Va

nb= 15na, so initially there are 15na moles

(4/5)(16na) = 12.8na the moles that go into B
(1/5)(16na) = 3.2na the moles that go into A

(Pa)(Va)=(3.2na)(8.31)(300K)
(Pb)(Vb)=(12.8na)(8.31)(400K)

but if I solve for the pressures and set them equal, all the variables cancel

After the valve is opened, the number of moles in each tank will change (but the total number of moles will be the same). You found that initially the total moles are equal to $$n_{\rm total}= 16 n_{a,i}$$; what will this expression be at the end?

Also, note that for each tank (separately) the quantity $P/n$ must be constant (since V and T are constant for each tank). Using these should give you the answer; what do you get?

 

[bcjochim07]

Ok let me try it:

for tank A

p1/n1 = p2/n2 (1.0*10^5Pa)/(na) = (P2/3.2na) P2= 320,000 Pa

(5.0 * 10 ^5Pa)/(15na) = (P2/12.8na) P2 = 426667Pa

Why didn't these come out the same?

 

[alphysicist]

The numbers you got for the final number of moles (3.2 and 12.8) are not correct. You got those numbers by assuming that the tank with four times the volume would end up with four times the number of moles, but that's not true since their temperatures are different.

Here is what you know:

$$\begin{align} n_{\rm total}&= 16 n_{a1}\nonumber\\ \frac{P_{a1}}{n_{a1}} &= \frac{P_{a2}}{n_{a2}} \nonumber\\ \frac{P_{b1}}{n_{b1}} &= \frac{P_{b2}}{n_{b2}}\nonumber \end{align}$$

and you can also use $PV=nRT$ for any tank at anyone point in time.

By setting $P_{a2}=P_{b2}$ you can find out how $n_{\rm total}$ and $n_{a2}$ are related. Once you have that, you can calculate $P_{a2}$.