# Identical ideal gas in two different volume containers

1. Mar 28, 2013

### keltik

1. The problem statement, all variables and given/known data

Container A in Fig. 19-22 holds an ideal gas at a pressure of $5*10^5 [Pa]$ and a temperature of $300 [K]$.

It is connected by a thin tube (and a closed valve) to container B, with four times the volume of A.

Container B holds the same ideal gas at a pressure of $1*10^5 [Pa]$ and a temperature of $400 [K]$

The valve is opened to allow the pressures to equalize, but the temperature of each container is maintained.

What is then the pressure?

(I hope you can make up the problem-figure by yourself, if not i can post a picture.)

2. Relevant equations
only the "ideal gas law":

$P*V=n*R*T$

3. The attempt at a solution

When the state is "valve closed" then the following values hold:
Volume: $V_{A} = \frac{V_{B}}{4} [m^3]$
Temperature: $300 [K]$
Pressure: $5*10^5 [Pa]$

Volume: $V_{B} = 4*V_{A}$
Temperature: $400 [K]$
Pressure: $1*10^5 [Pa]$

Ok, setting up the "ideal gas law" for container A gives equation:
$P_{A}*V_{A}=n_{A}*R*T_{A}$

Plugging in Temperature, Pressure and the gas constant R, yields:
$5*10^5*V_{A}=n_{A}*8.3*300$

Solving that for Moles in container A, yields:
$n_{A}=200.8*V_{A}$

Doing exactly the same for container B, yields:
$n_{B}=30.1*V_{B}$

in that equation exchanging $V_{B}$ with $4*V_{A}$ gives:
$n_{B}=30.1*4*V_{A}$, which is in turn:
$n_{B}=120.4*V_{B}$

A sanity check on these Mole-numbers reveals that these values could be correct, since there is a higher pressure in container A with the lower Volume than there is pressure in container B. So these numbers make sense. But i still dont have any values for it, so i decided to look at the "valve opened and equalized"-state:

Equalized means that container A gives container B a certain amount of his Molecules, since it has more pressure (although the temperature is higher in container B, it cannot outweigh the overpressure of container A)

This happens isothermically, which means $T_{A} = 300 [K]$ and $T_{B} = 400 [K]$ remain as they are.

My forecast is that in container A the pressure is going to be less and in container B the pressure is going to be higher than in their respective "valve is closed"-state, but i cant give you numbers.

Also total volume is now:
$V_{C} = V_{A} + V_{B}$

And the total number of molecules is:
$n_{C} = n_{A} + n_{B}$

But plugging those values in does not cancel something out (or i was too stupid for it?), thus yielding no concrete values.

Can someone give me one or two hints?

Thanks for your time. (I know i could look into the solutions manual, but that would be too easy....)

2. Mar 28, 2013

### Staff: Mentor

I would still keep both containers as separate parts in the equation, as their temperature is different.
Assuming n*VB moles flow from A to B, can you calculate the pressure in A and B?

3. Mar 28, 2013

### Staff: Mentor

What you are missing is that in the final state, the pressures in the two chambers will be the same.

4. Mar 29, 2013

### keltik

That is correct, i didnt write that down, but i am aware of that.

5. Mar 29, 2013

### keltik

Thank you guys i looked into the solutions manual for hints.

The "trick" basically is:
- i know that the total molecules will not change (no leakage) after equalizing, so i can write:

$n_{total, closed} = n_{total, equalized}$

which can be written as:

$n_{A,closed} + n_{B,closed} = n_{A,equalized} + n_{B,qualized}$

the rest is "use the ideal gas law, plug everything in and insert it here, cancel out", out comes the "equalized pressure".

Last edited: Mar 29, 2013
6. Dec 11, 2014

### Ahamed

Hi, Can you tell me which text book does this question comes from?

7. Dec 11, 2014

### Staff: Mentor

Welcome to physicsforums Ahamed.
This thread is 1.5 years old and keltik didn't make posts since then. I'll close the thread, for the unlikely case that he will ever see that he can send you a private message.

8. Dec 11, 2014

### Staff: Mentor

Hi Ahamed,

Google "Container A in Fig. 19-22 holds" (in quotes), and you will get lots of hits to the problem... :-)