# Ideal gas in two containers

1. Aug 21, 2004

### Zorodius

I'm having trouble with this problem.

Here's what I know:

Since the gas is ideal, we can describe the initial state of the first container with:

$$p_1 V = n_{1i} R T_1$$

Where p1 is the pressure in the first container, V is the volume of the first container, n1i is the initial number of moles present in the first container, R is the gas constant, and T1 is the temperature in the first container.

The same applies for the second container, with:

$$4 p_2 V = n_{2i} R T_2$$

Same meanings for the constants as before, noting that 4V is the volume of the second container.

Then, after the valve has been opened, the pressures in both containers will be the same, and for the first container:

$$p_f V = n_{1f} R T_1$$

for the second container:

$$4 p_f V = n_{2f} R T_2$$

as you might guess, I'm using pf for the final pressure, n1f as the final number of moles in the first container, and n2f as the final number of moles in the second container.

I also know that the total number of moles in the two containers is invariant, so:

$$n_{1i} + n_{2i} = n_{1f} + n_{2f}$$

That's all I have been able to come up with With five equations and six unknowns (n1i, n2i, n1f, n2f, V, pf), this is not a story with a happy ending. As far as I know, the problem is insoluble with the information I have here. There must be something else I'm supposed to realize about the problem, but I'm drawing a blank.

Can I get a hint here?

2. Aug 21, 2004

### maverick280857

Okay so you have

$$n_{1i} + n_{2i} = n_{1f} + n_{2f}$$

now substitute for $$n_{1i}$$ using

$$p_1 V = n_{1i} R T_1$$

and for $$n_{2i}$$ using

$$4 p_2 V = n_{2i} R T_2$$

You have two more equations in your post correspoding to the values of the final number of moles in each container. Go ahead and substitute them too. Note that V and R cancel on both sides. So from what I see, you get a simpler equation from which you can find $$p_{f}$$. (You know T1, T2, p1(initial), p2(initial).

The key is to eliminate the dependent variables and get a simpler relationship which gets rid of everything you do not know. This is similar to setting up a constraint (in, say, mechanics) when you set up equations which involve geometric parameters such as the radius of pulleys, length of ropes and so forth (these are the things you do not know but they do not enter the final equations once you differentiate in mechanics :-)).

Hope that helps...

Cheers
Vivek

EDIT: In short, don't consider this problem as involving a system of equations separate from each other (independent). They are in fact quite dependent.

3. Aug 21, 2004

### Staff: Mentor

Write expressions for $n_1$ and $n_2$ in terms of P, V, & T. Then write expressions for $\Delta n_1$ and $\Delta n_2$ in terms of $\Delta P$. Set $\Delta n_1 = - \Delta n_2$ and you'll get one equation in one unknown (if you see where I'm going with this).

4. Aug 22, 2004

### maverick280857

Doc,

$$n_{1i} + n_{2i} = n_{1f} + n_{2f}$$ and the subsequent substitutions imply the same thing (said differently of course)

Vivek

5. Aug 22, 2004

### Zorodius

Thanks a lot for the replies!

This has revealed a significant hole in my understanding - I previously thought that having six unknowns and five equations in a problem was never something that could be solved. Clearly, I was mistaken. Could I get an explanation of when it's possible to solve a problem like that and when it isn't - or could someone suggest some search terms that would lead me to information on this, so I could find out for myself?

6. Aug 22, 2004

### Staff: Mentor

Well, if you were trying to solve for all six unknowns you'd have a problem. But you're not. All you need to find is the final pressure.

Follow the advice that Vivek and I gave and you'll see that "unknowns" will drop out and you'll be able to solve for the pressure.