Solutions 15
- Page ID
- 204100
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Make a table listing the possible ways to get the listed energies by distributing the particles among the four energy levels. The columns represent the energies available to a particle. Each row is a configuration of particles. The numbers in a row are the number of particles of a certain energy. Each row therefore adds to a total of 4 particles. The number of configurations that result in a given energy will be the weightings in \(\Omega\) in \(S = k \ln \Omega\).
a) \(E = 3 \epsilon\), \(\Omega = 2\), \(S\) = 9.5699e-23
| \(0 \epsilon\) | \(1 \epsilon\) | \(2 \epsilon\) | \(4 \epsilon\) |
|---|---|---|---|
| 2 | 1 | 1 | 0 |
| 1 | 3 | 0 | 0 |
b) \(E = 4 \epsilon\), \(\Omega = 4\), \(S\) = 1.9140e-23
| \(0 \epsilon\) | \(1 \epsilon\) | \(2 \epsilon\) | \(4 \epsilon\) |
|---|---|---|---|
| 3 | 0 | 0 | 1 |
| 2 | 0 | 2 | 0 |
| 0 | 4 | 0 | 0 |
| 1 | 2 | 1 | 0 |
c) \(E =8\epsilon\), \(\Omega = 4\), \(S\) = 1.9140e-23
| \(0 \epsilon\) | \(1 \epsilon\) | \(2 \epsilon\) | \(4 \epsilon\) |
|---|---|---|---|
| 1 | 0 | 2 | 1 |
| 0 | 0 | 4 | 0 |
| 2 | 0 | 0 | 2 |
| 0 | 2 | 1 | 1 |
d) \(E = 12 \epsilon\), \(\Omega =2 \), \(S\) = 9.5699e-24
| \(0 \epsilon\) | \(1 \epsilon\) | \(2 \epsilon\) | \(4 \epsilon\) |
|---|---|---|---|
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 3 |
e) \(E =16 \epsilon\), \(\Omega = 1\), \(S\) = 0
| \(0 \epsilon\) | \(1 \epsilon\) | \(2 \epsilon\) | \(4 \epsilon\) |
|---|---|---|---|
| 0 | 0 | 0 | 4 |
Q2
Look up the definitions of A, U, S etc. and look at how they depend on the derivatives of the partition function. Perform the necessary derivatives on \(Q\), but do so keeping it in the \(\Sigma\) form (rather than expanding the sum).
\(Z = \Sigma \exp(- E_i/kT) = \Sigma \exp(-\beta E_i) \)
\(\dfrac{dZ}{dT} = \dfrac{d(\Sigma \exp(- E_i/kT) )}{dT} = \dfrac{1}{kT^2} \Sigma E_i \exp(E_i/kT) \)
\(\dfrac{dZ}{d\beta} = \dfrac{d(\Sigma \exp(-\beta E_i)}{d \beta} = -\Sigma E_i exp(-\beta E_i) \)
\(U = -\dfrac{d(\ln Z)}{d \beta} = -(\dfrac{d(\ln Z)}{dZ}) (\dfrac{dZ}{d \beta}) = \dfrac{1}{Z} \Sigma E_i exp(-E_i/kT) \)
\(C_v = \dfrac{dU}{dT} = \dfrac{1}{Z} \dfrac{\Sigma E_i exp(-E_i/kT)}{dT} = \dfrac{1}{kT^2} \dfrac{\Sigma E_i^2 exp(-E_i/kT)}{Z} \)
\(A = - kT \ln(Z) \)
\(S = -\dfrac{dA}{dT} = -\dfrac{d(-kT \ln(Z))}{dT} = k \ln(Z) - kT ( \frac{1}{kT^2}) \Sigma E_i \exp(-E_i/kT) = k \ln(Z) - ( \frac{1}{T}) \Sigma E_i \exp(-E_i/kT) \)
Q3
Look up the particle in a box equation and set it equal to the \(\dfrac{3k_BT}{2} \) term to solve for the quantum number \(n\).
Q4
We know the form of energy of this harmonic oscillator system. We can then use these energies to find the ratio of the Boltzmann terms (the ratio of probabilities) and set that equal to the population of the states and solve for the temperature.
If the system is fully equilibrated the rotational temperature will be the same as vibrational and translation (as long as those degrees are populated). In not equilibrated systems like molecular beams it is possible for the temperatures to differ, but that is not an equilibrated thermal distribution.
If the system is fully equilibrated the rotational temperature will be the same as vibrational and translation (as long as those degrees are populated). In not equilibrated systems like molecular beams it is possible for the temperatures to differ, but that is not an equilibrated thermal distribution.
Q5
We cannot find the absolute entropy but we can settle for solving for volume and having the microstates proportionate to volume: \(\Omega = C V\)
Q6
You'll need to look up the rotational constant for \(N_2\) and use it to calculate the energies of different states for this question.
a) Look at the general form of the terms in the rotational partition function and differentiate with respect to the quantum number \(J\):
\(q_{rot} = \Sigma (2J+1) \exp(-\frac{\Theta_{rot}}{T}J(J+1)) \)
\( \dfrac{d(q_{rot} )}{dJ} = 2 \exp(-\frac{\Theta_{rot}}{T} J(J+1)) + (2J+1) \exp(-\frac{\Theta_{rot}}{T}J(J+1)) (-\dfrac{\Theta_{rot}}{T}(2J+1)) \)
=\( \exp(-\dfrac{ \Theta_{rot}}{T} J(J+1)) \left[2 - (2J+1)^2 \dfrac{\Theta_{rot}}{T} \right] = 0 \)
This implies: \( 2 = (2J+1)^2 \dfrac{\Theta_{rot}}{T}\) so \( (2J+1) \) =
\((\dfrac{2T}{\Theta_{rot}})^{1/2} \) and \( J = (\dfrac{T}{2 \Theta_{rot}})^{1/2} -\dfrac{1}{2} \)
b). Calculate the energies of each state. Find the partition function for the system. Use this to calculate the probabilities of each state and therefore, the fraction of particles in each state..
c). If there were no degeneracy, then the most probably quantum number is that which is associated with the lowest energy, or \(\nu = 0\). However, the distribution is skewed by this. Plug in the parameters: \( J = (\dfrac{T}{2 \Theta_{rot}})^{1/2} -\dfrac{1}{2} \)