Ideal gas in two containers

[Zorodius]

I'm having trouble with this (http://img.photobucket.com/albums/v165/Zorodius/20-17.jpg) 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 :frown: 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?

[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.

[Doc Al]

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).

[maverick280857]

Doc,

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

Adios
Vivek

[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?

[Doc Al]

That's all I have been able to come up with :frown: With five equations and six unknowns (n1i, n2i, n1f, n2f, V, pf), this is not a story with a happy ending.
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.