Question of Statistical Thermodynamics (Boltzmann Distribution)

[asdfTT123]

I apologize if this is the wrong thread but since this relates to thermo I figured this would be a good place to post this question. This is a problem that was assigned to us for physical chemistry but I can't find a good justification for one of the problems.

1. A system containing 38 particles has three equally spaced energy levels available. Two population distributions are A (18, 12, 8) and B (17, 14, 7) (populations listed lowest to highest energy).

a. Show that both distributions have the same energy.

This part is pretty easy. Since each energy level is listed lowest to highest and they are equally spaced we can assume the energy levels follow E = 1, 2, and 3 respectively. Therefore,

Energy of A = 18 * 1 + 12 * 2 + 8 * 3 = 66
Energy of B = 17 * 1 + 14 * 2 + 7 * 3 = 66

b. Is either distribution a Boltzmann distribution?

I think I know the answer for this but I need a good, solid justification. According to the equation for the weight of a system:

W = N!/n0!n1!n2!, The Boltzmann distribution is the distribution with the greatest weight while satisfying the conditions where the total number of molecules and total energy of the system remains constant.

Weight of A = 38!/18!12!8! = 4.23 x 10^15
Weight of B = 38!/17!14!7! = 3.35 x 10^15

I think A is the Boltzmann because the weight of the configuration is greater than that of B or any other valid configurations that meet the constraints that I can think of, but I don't have a good justification.

A way I can prove this is if I come up with an equation for dW = 0, showing that is when the weight is maximized, but I don't know how to do that while keeping the constraints.

If anyone can help me out by coming up with a mathematical justification or if you have anything else to add, I would greatly appreciate it...thanks!

 

[marcusl]

For A, look at the number of particles in each of the 3 states and compare their ratios to what you'd expect from the Boltzmann distribution. Do the same for B.

 

[asdfTT123]

Can you clarify? I tried doing B using the standard Boltzmann equation,

ni/N = e^(-BEi)/e^(-BE1) + e^(-BE2) + e^(-BE3), but you can't get a result because using the values of E I have e^(-BEi) approximates to about 0 resulting in an undefined answer.

 

[siddharth]

a. Show that both distributions have the same energy.[/B]

This part is pretty easy. Since each energy level is listed lowest to highest and they are equally spaced we can assume the energy levels follow E = 1, 2, and 3 respectively. Therefore,

Energy of A = 18 * 1 + 12 * 2 + 8 * 3 = 66
Energy of B = 17 * 1 + 14 * 2 + 7 * 3 = 66

That's only for the specific case when E=1,2 and 3. You need to show this for a general set of energy levels $E + n \Delta$, where $n = 0,1,2$. Also, I'm guessing that the question wants to know if the total mean energies are the same.

b. Is either distribution a Boltzmann distribution?

I think I know the answer for this but I need a good, solid justification. According to the equation for the weight of a system:

W = N!/n0!n1!n2!, The Boltzmann distribution is the distribution with the greatest weight while satisfying the conditions where the total number of molecules and total energy of the system remains constant.

Weight of A = 38!/18!12!8! = 4.23 x 10^15
Weight of B = 38!/17!14!7! = 3.35 x 10^15

I think A is the Boltzmann because the weight of the configuration is greater than that of B or any other valid configurations that meet the constraints that I can think of, but I don't have a good justification.

A way I can prove this is if I come up with an equation for dW = 0, showing that is when the weight is maximized, but I don't know how to do that while keeping the constraints.

If anyone can help me out by coming up with a mathematical justification or if you have anything else to add, I would greatly appreciate it...thanks!

I think you'll need more information. If you assume that the system is in thermal equilibrium with a heat reservoir, and the population distribution refers to the mean number of particles in each state, I think you'll still need to know the spacing between the energy levels.

 

[asdfTT123]

I think you are thinking too deeply into this. I'm sure the justification for pt A is fine as long as the energy levels are evenly spaced. I also did calculate the average energy. For pt B, I'm sure there's a justification with the information given. I'm looking for something statistical or math related to make the connection.

 

[marcusl]

If the energy levels are evenly spaced, let's say they are E1, E2=E1+dE, E3=E2+dE, then the population ratio N2/N1=exp(-dE/kT) must equal N3/N2=exp(-dE/kT), or in words the ratios are equal because the energy increases the same amount going from E1 to E2 as E2 to E3. Case A satisfies this, Case B does not. I think that's all you need to do.

 

[nrqed]

I apologize if this is the wrong thread but since this relates to thermo I figured this would be a good place to post this question. This is a problem that was assigned to us for physical chemistry but I can't find a good justification for one of the problems.

1. A system containing 38 particles has three equally spaced energy levels available. Two population distributions are A (18, 12, 8) and B (17, 14, 7) (populations listed lowest to highest energy).

a. Show that both distributions have the same energy.

This part is pretty easy. Since each energy level is listed lowest to highest and they are equally spaced we can assume the energy levels follow E = 1, 2, and 3 respectively. Therefore,

Energy of A = 18 * 1 + 12 * 2 + 8 * 3 = 66
Energy of B = 17 * 1 + 14 * 2 + 7 * 3 = 66

b. Is either distribution a Boltzmann distribution?

I think I know the answer for this but I need a good, solid justification. According to the equation for the weight of a system:

W = N!/n0!n1!n2!, The Boltzmann distribution is the distribution with the greatest weight while satisfying the conditions where the total number of molecules and total energy of the system remains constant.

Weight of A = 38!/18!12!8! = 4.23 x 10^15
Weight of B = 38!/17!14!7! = 3.35 x 10^15

I think A is the Boltzmann because the weight of the configuration is greater than that of B or any other valid configurations that meet the constraints that I can think of, but I don't have a good justification.

A way I can prove this is if I come up with an equation for dW = 0, showing that is when the weight is maximized, but I don't know how to do that while keeping the constraints.

If anyone can help me out by coming up with a mathematical justification or if you have anything else to add, I would greatly appreciate it...thanks!

You should be able to show for a Boltzmann distribution and equally spaced energy levels, the ratio of the occupation number for adjacent energy levels should be constant (so $$\frac{N_3}{N_2} = \frac{N_2}{N_1}$$, etc).