Callen Thermodynamics 2.8-2 matter flow equilibrium

[It's me]

Homework Statement

A two component gaseous system has a fundamental equation of the form
$$S=AU^{1/3} V^{1/3} N^{1/3} + \frac{BN_1N_2}{N}$$ where $$N=N_1+N_2$$
and A and B are positive constants. A closed cylinder of total volume 2V_0 is separated into two equal subvolumes by a rigid diathermal partition permeable only to the first component. One mole of the first component at a temperature T_l is introduced in the left-hand subvolume and a mixture of 1/2 mole of each component is introduced in the right-hand subvolume at a temperature T_r.

Find the equilibrium temperature T_e and the mole numbers in each subvolume when the system has come to equilibrium, assuming that $$T_r=2T_l=400K$$ and that $$37B^2=100A^3V_0$$. Neglect the heat capacity of the walls of the container.

Homework Equations

$$\frac{1}{T}=(\frac{\partial S}{\partial U})$$
$$\frac{P}{T}=(\frac{\partial S}{\partial V})$$
$$\frac{-\mu_1}{T}=(\frac{\partial S}{\partial N_1})$$

The Attempt at a Solution

I am trying to use the definitions of the equations of state and conservation of internal energy and chemical potential in order to find relationships that will allow me to solve for N_1l, N_1r, and T_e. This is how far I've gone:

$$\frac{1}{T}=(\frac{\partial S}{\partial U})=\frac{1}{3}AU^{-2/3}V^{1/3}N^{1/3}$$
$$\Rightarrow U=(\frac{ATV^{1/3}N^{1/3}}{3})^{3/2}$$
$$\frac{-\mu_{1l}}{T}=(\frac{\partial S}{\partial V})=\frac{1}{3}AU^{1/3}V^{1/3}(N_{1l}+N_{2l})^{-2/3}+\frac{(N_{1l}+N_{2l})BN_{2l}-BN_{1l}N_{2l}}{(N_{1l}+N_{2l})^2}=\frac{1}{3}AU^{1/3}V^{1/3}N_{1l}^{-2/3}$$
$$\frac{-\mu_{1r}}{T}=(\frac{\partial S}{\partial V})=\frac{1}{3}AU^{1/3}V^{1/3}(N_{1r}+N_{2r})^{-2/3}+\frac{(N_{1r}+N_{2r})BN_{2r}-BN_{1r}N_{2r}}{(N_{1r}+N_{2r})^2}$$
$$\mu_{1l}=\mu_{1r}$$

I really don't know how to proceed from here and how to use the relation given at the end of the problem. Any help would be greatly appreciated.

 

[TSny]

Hello, and welcome to PF!

So far, your equations look good to me. The last term on the right side of your next to last equation will simplify somewhat.

Looks like you're in for some fairly complicated algebra to solve for the equilibrium temperature $T_e$ and the final mole numbers $N_{1l}$ and $N_{1r}$. I think you will need a few more relations. As you noted, $\mu_{1l} = \mu_{1r}$ at equilibrium. What other quantity must be the same for the left and right compartments at equilbrium?

There are also a couple of conserved quantities that don't change during the approach to equilibrium. This should give you two more equations to work with.

 

[It's me]

At equilibrium $$U_l=U_r$$, right? And also $$N_{1l} + N_{1r}= 3/2$$ I haven't been able to use these relations though, I still get more variables than equations. Am I missing something?

 

[TSny]

At equilibrium $$U_l=U_r$$, right?

No, the energy on the left does not need to equal the energy on the right. What can you say about the final total energy of the system?

And also $$N_{1l} + N_{1r}= 3/2$$ I haven't been able to use these relations though, I still get more variables than equations. Am I missing something?

This equation is correct.

Can you say anything about the temperatures of the left and right compartments at equilibrium? Does that give you another relation?

 

[It's me]

Oh right, I meant to say $$U_i=U_f=U_l+U_r$$. $$T_l=T_r=T_e$$

 

[TSny]

Yes. That looks good. Conservation of total energy should lead to a relation between T_e, the initial temperatures, and the initial final number of moles on the left and right.

 

[Othin]

Oh right, I meant to say $$U_i=U_f=U_l+U_r$$. $$T_l=T_r=T_e$$

Am I missing something? Doesn't the last equation contradict the imposition that $T_r=2T_l ?$

 

[Othin]

Yes. That looks good. Conservation of total energy should lead to a relation between T_e, the initial temperatures, and the initial final number of moles on the left and right.

Wouldn't I need to know something about their heat capacity to find this relation?

 

[TSny]

Am I missing something? Doesn't the last equation contradict the imposition that $T_r=2T_l ?$

In post #5, $T_r$ and $T_l$ refer to the final equilibrium temperatures of the left and right compartments rather than the initial temperatures.

 

[TSny]

Wouldn't I need to know something about their heat capacity to find this relation?

No, I don't believe you do. There are 3 unknowns: $T_e$, and the final values of $N_{1l}$, and $N_{1r}$.

And you have three relations:
(1) $T_{l, final} = T_{r,final}$
(2) $\mu_{1l,final} = \mu_{1r,final}$
(3) Conservation of energy

 

[Othin]

No, I don't believe you do. There are 3 unknowns: $T_e$, and the final values of $N_{1l}$, and $N_{1r}$.

And you have three relations:
(1) $T_{l, final} = T_{r,final}$
(2) $\mu_{1l,final} = \mu_{1r,final}$
(3) Conservation of energy

Yes, after a little work I came to a system which is theoretically solvable, though far too big.

 

[TSny]

From post #1

The Attempt at a Solution

$$\frac{-\mu_{1l}}{T}=(\frac{\partial S}{\partial V})=\frac{1}{3}AU^{1/3}V^{1/3}(N_{1l}+N_{2l})^{-2/3}+\frac{(N_{1l}+N_{2l})BN_{2l}-BN_{1l}N_{2l}}{(N_{1l}+N_{2l})^2}=\frac{1}{3}AU^{1/3}V^{1/3}N_{1l}^{-2/3}$$
$$\frac{-\mu_{1r}}{T}=(\frac{\partial S}{\partial V})=\frac{1}{3}AU^{1/3}V^{1/3}(N_{1r}+N_{2r})^{-2/3}+\frac{(N_{1r}+N_{2r})BN_{2r}-BN_{1r}N_{2r}}{(N_{1r}+N_{2r})^2}$$

Just for the record, there is a typographical error above where the first $(\frac{\partial S}{\partial V})$ should be $(\frac{\partial S}{\partial N_{1l}})$ and the second $(\frac{\partial S}{\partial V})$ should be $(\frac{\partial S}{\partial N_{1r}})$

 

[TSny]

Yes, after a little work I came to a system which is theoretically solvable, though far too big.

I used Mathematica to get a numerical solution to one of the equations.