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## 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.