- #1

baseballfan_ny

- 92

- 23

- Homework Statement
- Two identical systems ##S_1## and ##S_2## are both in thermal contact with a large reservoir and in diffusive contact with one another. For both systems, the free energy F is related to the particle number N by ##F = cN^2##, where c is an N-independent constant (the same constant

for both systems).

(a) A battery maintains a chemical potential difference ##\Delta = \mu_{2,ext} − \mu_{1,ext} > 0 ## between the two systems. In diffusive equilibrium, find the number ##N_1## of particles in ##S_1## and the number ##N_2## of particles in ##S_2##, expressed in terms of ##\Delta##, c, and the total particle number ##N = N_1 + N_2##.

(b) Now the battery is disconnected, and useful work is extracted isothermally as the particles flow slowly from ##S_1## to ##S_2## until diffusive equilibrium is reestablished. How much work is

extracted?

- Relevant Equations
- ## \mu = \frac {\partial F} {\partial N}_{T, V} ##

So I think I have the principles mixed up here because I'm getting kind of "circular" answers.

## N = N_1 + N_2##

##dN## = 0 bc/ particle number fixed so ##dN_1 = -dN_2##

##F = cN^2 = c(N_1 + N_2)^2##

In diffusive equilibrium, free energy would be minimized and chemical potentials equal...

$$ dF = \frac {\partial F} {\partial N_1}_{T, V} dN_1 + \frac {\partial F} {\partial N_2}_{T, V} dN_2 = 0 $$

$$ dF = (2cN_1 + 2cN_2)dN_1 - (2cN_1 + 2cN_2)dN_1 = 0 $$

$$ (N_1 + N_2) = (N_1 + N_2) $$ which is useless

I also tried proceeding from chemical potential equilibrium...

$$ \mu_1^{total} = \mu_2^{total} $$

$$ \mu_{1,int} + \mu_{1,ext} = \mu_{2,int} + \mu_{2,ext} $$

$$ \mu_{1,int} − \mu_{2,int} = \mu_{2,ext} − \mu_{1,ext} = \Delta $$

And then I believe ## \mu_{1, int} = \frac {\partial F} {\partial N_1}_{T, V} = (2cN_1 + 2cN_2)## and ## \mu_{2, int} = \frac {\partial F} {\partial N_2}_{T, V} = (2cN_1 + 2cN_2)## ... so I get ## 0 = \Delta##? I know I've got the principles mixed up somehow or another here...

## N = N_1 + N_2##

##dN## = 0 bc/ particle number fixed so ##dN_1 = -dN_2##

##F = cN^2 = c(N_1 + N_2)^2##

In diffusive equilibrium, free energy would be minimized and chemical potentials equal...

$$ dF = \frac {\partial F} {\partial N_1}_{T, V} dN_1 + \frac {\partial F} {\partial N_2}_{T, V} dN_2 = 0 $$

$$ dF = (2cN_1 + 2cN_2)dN_1 - (2cN_1 + 2cN_2)dN_1 = 0 $$

$$ (N_1 + N_2) = (N_1 + N_2) $$ which is useless

I also tried proceeding from chemical potential equilibrium...

$$ \mu_1^{total} = \mu_2^{total} $$

$$ \mu_{1,int} + \mu_{1,ext} = \mu_{2,int} + \mu_{2,ext} $$

$$ \mu_{1,int} − \mu_{2,int} = \mu_{2,ext} − \mu_{1,ext} = \Delta $$

And then I believe ## \mu_{1, int} = \frac {\partial F} {\partial N_1}_{T, V} = (2cN_1 + 2cN_2)## and ## \mu_{2, int} = \frac {\partial F} {\partial N_2}_{T, V} = (2cN_1 + 2cN_2)## ... so I get ## 0 = \Delta##? I know I've got the principles mixed up somehow or another here...