### Physics 9B Practice Final Problem #5 Again

Two volumes A and B of ideal gas with N

For each volumes of gas A and B, the multiplicity depends on the volume according to

ω

ω

with a and b independent of V

Question so kindly provided by Joe Kiskis

This is a stat mech problem, so to find the equilibrium, we need to maximize entropy.

Again, maximize entropy.

Third time, for the readers at home: Maximize entropy. S = k ln(ω), so S is maximum when ∂S = 0.

_{A}and N_{B}molecules respectively are separated by a piston that is free to move while the total volume V = V_{A}+ V_{B}is constant. Both gases are in contact with a heat reservoir at a fixed temperature T.For each volumes of gas A and B, the multiplicity depends on the volume according to

ω

_{A}= aV_{A}^{NA}ω

_{B}= bV_{B}^{NB}with a and b independent of V

_{A}and V_{B}. After the piston comes to equilibrium, what is the pressure in the gas? Express it in terms of k, V, N_{A}, N_{B}, T, a, and b. Do the problem using the given information and the ideas of statistical mechanics. There are at least a couple ways to do the problem. If you want to use the equation of state, you must derive it from the information given above and general results on the properties of entropy.Question so kindly provided by Joe Kiskis

This is a stat mech problem, so to find the equilibrium, we need to maximize entropy.

Again, maximize entropy.

Third time, for the readers at home: Maximize entropy. S = k ln(ω), so S is maximum when ∂S = 0.

- S
_{A}= k ln (aV_{A}^{NA}) - S
_{A}= k N_{A}ln(V_{A}) + k ln(a) - Repeat this for S
_{B}, but they both need to be in terms of one variable V, using the relation: - V = V
_{A}+ V_{B} - S
_{total}= S(V_{A}) + S(V−V_{A}) - ∂S⁄∂V
_{A}= ∂S(V_{A})⁄∂V_{A}+ ∂S(V−V_{A})⁄∂V_{A} - Realize that since V is constant, it's derivative is 0 and the second part can be rewritten.
- ∂S⁄∂V
_{A}= ∂S(V_{A})⁄∂V_{A}− ∂S(V_{B})⁄∂V_{A} - ∂S⁄∂V
_{A}= k N_{A}⁄V_{A}− k N_{B}⁄V_{B}= 0 - And to maximize entropy, this is all equal to 0.
- k N
_{A}⁄V_{A}= k N_{B}⁄V_{B} - At the same time, note that ∂S(V
_{A})⁄∂V_{A}= ∂S(V_{B})⁄∂V_{A} - This is important because ∂S⁄∂V|
_{U}= P⁄T - P
_{A}⁄T = P_{B}⁄T - Replug in V = V
_{A}+ V_{B}and solve for V_{A} - V
_{A}= V⋅N_{A}⁄[N_{A}+ N_{B}] - Now we have two equations for ∂S⁄∂V, set them equal:
- P
_{A}⁄T = k N_{A}⁄V_{A} - Plug in the solution for V
_{A}and solve for P - P = k N
_{A}T⁄[V⋅N_{A}⁄(N_{A}+N_{B})] - P = k T (N
_{A}+ N_{B})⁄V - Holy crap! That's the answer!!!

## Comments

## Post a Comment