Equilibrium with respect to matter flow

[Useful nucleus]

Homework Statement

A two component gaseous system has a fundamental equation of the form
S=AU^1/3V^1/3N^1/3 + (BN₁N₂)/N
N=N₁+N₂
where A and B are positive constants. V is volume , S is entropy, U is internal energy, N is mole number. A closed cylinder of total volume 2V₀ 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 subvolume and a mixture of 1/2 mole of each component is introduced in the right subvolume at a temperature T_r.

Find the equilibrium T^e and the mole numbers in each subvolume.

Homework Equations

At equilibrium the chemical potentials of species 1 on right and left chambers are equal and also temperatures are equal.

The Attempt at a Solution

The main issue is that whenever I compute the chemical potential of species 1 in the left chamber or the right chamber and equate them

(∂S/∂N₁)_left=(∂S/∂N₁)_right

I end up getting 0 = 0. I expected to obtain an expression that helps in evaluate the amount of species (1) in the left and right chamber. Am I missing something? Help is appreciated.

 

[Useful nucleus]

Resolved, one has to be very careful in the differentiation and pay more attention to the sub/super scripts.

 

[Othin]

Did you also meet a huge system of equations while solving this, or did you solve it by hand?

 

[Othin]

Isn't there a formal inconsistency in this problem? Since there were no initial internal constants to be removed (The way I read it, looks like someone simply filled the cylinder with the two gases), how can, then, initial temperature and specially initial entropy be defined? On his second postulate it's stated that the entropy function is defined only for equilibrium states (which is also why the book needs to deal with quasi-static transformations). I think energy conservation will be needed to complete the exercise, but if entropy can't be defined at the instant t=0, neither can U, right?

 

[Useful nucleus]

Did you also meet a huge system of equations while solving this, or did you solve it by hand?

Yes using the temperature equation of state, conservation of total energy and equality of μ at equilibrium , I can get one equation in one unknown (for example the mole numbers in the right chamber at the final state) but the solution needs to be done "graphically" rather than by hand.

Isn't there a formal inconsistency in this problem? Since there were no initial internal constants to be removed (The way I read it, looks like someone simply filled the cylinder with the two gases), how can, then, initial temperature and specially initial entropy be defined? On his second postulate it's stated that the entropy function is defined only for equilibrium states (which is also why the book needs to deal with quasi-static transformations). I think energy conservation will be needed to complete the exercise, but if entropy can't be defined at the instant t=0, neither can U, right?

I agree with you that the wording of the problem is not very consistent with typical thermodynamic analysis. But I think the author here intended to mean that first the two chambers were filled with the gases at which point the champers were separated by adiabatic and non-permeable separation. Then this separation was replaced by a diathermal and permeable one (for one type of the gases). It would have been more instructive if reworded and possibly with a simpler fundamental equation.