[canonicalpartitionstudent]

With increasing T, population of state A decreases due to the exciting of the objects to state B and then C. Q doubles, P_A is reduced to half. The loss in population in state A is equal to the increase in state B and C. Population of B is (in the limit of high temperature) twice the population of state C.

In other words, with increasing T, objects will be excited to state B and to a lesser extent, to state C.
State A, the ground state, stays half populated in the limit of high temperature.

For the internal energy:

I found the same equation, with N equal to 1.
U is then equal to:

U = -1/Q.-Σ_i.E_i .e^-β.E_i

For S, I obtain:

S = U/T + k_b.ln(Q)

Correct?
Thank you very much.

[canonicalpartitionstudent]

I calculated U and S and I obtain these values:

For 2000 K:

β=0,060136181   
Q=10,53901093   
P_A=0,569313386
P_B=0,298395927   
P_C=0,132290687      
S=0,019928888 kJ.K^-1.mol^-1     
U=0,695267987 kJ.mol^-1

For 5000 K:

S=0,02036343 kJ.K^-1.mol^-1 
U=0,776705416 kJ.mol^-1

In the infinite temperature limit (10000000000 K):
S=0,020660662 kJ.K^-1.mol^-1
U=0,833333305 kJ.mol^-1

[Corribus]

For the populations it looks like you're doing a good job. Aside from the trends you correctly describe, here are a few other things to notice that might help you understand the meaning of Q.

At low temperature (as T 0), Q  6. And as T  ∞, Q  12. Notice that the ground state has a degeneracy of 6, and the total number of states is 12. The partition function is basically a description of how the available particles are partitioned among all the available states, weighted statistically by the temperature. At low temperatures, only the lowest energy states are available, and because of degeneracy, there are 6 of them. Q = 6. At the highest temperatures, the energies between the states become inconsequential (compared to background energy), and all states become effectively equally populated. Another way of looking at this is that the difference in energy between all the available states becomes so small compared to the amount of energy in the system, that they are all effectively degenerate. Since there are a total of 12 states available in the system, the effective degeneracy is 12, which is also the Q value. In a way, the Q value is a representation of the average number of states populated at a given temperature. This is why we determine the probability of populating a certain state as the ratio with Q in the denominator.

IMPORTANT: Many students make the mistake of assuming that at high temperature, only the upper states wills be populated, but this is not the case. At high temperature, all states in the system have equal likelihood of being populated.

You'll also notice that at high temperature, PA = 0.5, PB = 0.333 and PB = 0.167. These are values of 1/2, 1/3, and 1/6, respectively. If at high temperature all the available states have an equal likelihood of population, then the probability of an energy level being populated should be equal to the number of states in that energy level (the degeneracy) divided by the total number of microstates available in the system. States A, B, and C have 6, 4, and 2-fold degeneracy, respectively, given a total of 12 microstates available. So the probability of A, B, and C being populated when there is no "bias" do to the energy (i.e., at high temperature) are easily determined to be 6/12, 4/12, and 2/12... 0.5, 0.333, and 0.167!

For the internal energy, looks like you're still doing something wrong with the calculation, so let's work through it step by step.

If
[tex]U = - \frac {N}{Q} \frac {dQ}{d\beta}[/tex]
And
[tex]Q = 6 + 4e^{-4\beta} + 2e^{-6\beta}[/tex]
And N = 1 mole, then
[tex]U = \frac {16 e^{-4\beta} + 12 e^{-6\beta}}{6 + 4e^{-4\beta} + 2e^{-6\beta}}[/tex]
Is this what you got?

[canonicalpartitionstudent]

Thank you for your explanation! Very interesting!

For the calculation of the internal energy at 5000K, I found these results:

S=0,020646648   
U=2,192794286

I used an equation found in my course (see attachment). This is the same but I forgot the degeneracy to obtain 16 and 12.

[Corribus]

Great, you now have what I have.

Let me show you something else interesting. I know the equations look complicated and are sometimes hard to decipher, which makes the business of statistical mechanics seem very abstract. But actually, the concepts are rather simple and many things can be calculated, at least at the limits, without using equations. When you do it this way, the meanings of the equations start to become apparent.

Already we've looked at the populations, so let's look at the internal energy value.

The internal energy you're calculating is just the weighted average energy of the system per occupied state. Each microstate has a particular energy value associated with it. So the total internal energy is just equal to the sum of energy of each state multiplied by how many particles are in that state, divided by the total number of occupied states. This is basically what that equation is saying. At low temperature, everything is in the lowest energy state, which happens to have an energy of zero. So the average internal energy of the system per occupied state is also equal to zero: every particle in the system has an energy of zero, so the average energy of all particles is zero. At high temperature, we've already shown that the particles are equally distributed among all possibible states. There are 6 states with energy = 0, 4 states with energy = 4, and 2 states with energy = 6. 6 * 0 + 4 * 4 + 2 * 6 = 28. So there's a total of 28 kJ/mol of energy in the system at infinite temperature, on average, divided by 12 total populated states = 2.33 kJ/mol of average energy per populated state.

So in sum, all the internal energy is expressing is: what is the average energy per occupied state?

Anyway, this was a great problem that illustrates a lot of fundamental concepts in statistical mechanics. I'll have to remember it for the next time (if ever) I have to teach physical chemistry. Thanks for sharing it.

[canonicalpartitionstudent]

Great, claryfying information (!), and as you said, it's very logic!!

There was also another question:

"Discuss your result in view of the temperature dependence of chemical equilibria”

What should I say?


I'm really glad you helped me, thank you very much for your time and patience.
Besides this exercise I have 2 other questions, which I solved for 90%. If you are interested I can share it also.

[Corribus]

Well, that part of the question is kind of vague. I'd probably start by saying something about how temperature impacts chemical equilibria, and in particular relate it to entropy. For example, you've shown explicitly here that raising the temperature results in an increase in entropy, so raising the temperature in a system already in equilibrium would tend to shift the equilibrium towards products with higher entropy.

But yeah, I'm not sure exactly what the question is after, here.