This lecture introduces the equipartition theorem.
Write an explicit expression for the total classical Newtonian energy of the material in terms of the independent free variables.
For the ideal monatomic gas, we have \[ E = \frac12mv_1^2 + \frac12mv_2^2 + \frac12mv_3^2 + \cdots \text{(total of $N$ atoms)} \] where if we write things in terms of Cartesian coordinates gives us \[ \frac12mv_1^2 = \frac12mv_{1x}^2 + \frac12mv_{1y}^2 + \frac12mv_{1z}^2 \] so you end up with \(3N\) terms that depend on independent free variables.
Count the number of indepdendent free variables that are squared in the expression for the energy (“quadratic terms”). We call this the number of degrees of freedom \(f\).
For the ideal monatomic gas \(f=3N\).
The equipartition theorem predicts that \begin{align} U_{\text{classical}}(T) &= \frac{f}{2}k_BT \end{align}
For the ideal monatomic gas \(U_{\text{classical}}(T) = \frac{3}{2}Nk_BT\).
After this we do something else.
group Small Group Activity
30 min.
assignment Homework
group Small Group Activity
30 min.
assignment Homework
Find the chemical potential of an ideal monatomic gas in two dimensions, with \(N\) atoms confined to a square of area \(A=L^2\). The spin is zero.
Find an expression for the energy \(U\) of the gas.
Find an expression for the entropy \(\sigma\). The temperature is \(kT\).
assignment Homework
A one-dimensional harmonic oscillator has an infinite series of equally spaced energy states, with \(\varepsilon_n = n\hbar\omega\), where \(n\) is an integer \(\ge 0\), and \(\omega\) is the classical frequency of the oscillator. We have chosen the zero of energy at the state \(n=0\) which we can get away with here, but is not actually the zero of energy! To find the true energy we would have to add a \(\frac12\hbar\omega\) for each oscillator.
Show that for a harmonic oscillator the free energy is \begin{equation} F = k_BT\log\left(1 - e^{-\frac{\hbar\omega}{k_BT}}\right) \end{equation} Note that at high temperatures such that \(k_BT\gg\hbar\omega\) we may expand the argument of the logarithm to obtain \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).
From the free energy above, show that the entropy is \begin{equation} \frac{S}{k_B} = \frac{\frac{\hbar\omega}{kT}}{e^{\frac{\hbar\omega}{kT}}-1} - \log\left(1-e^{-\frac{\hbar\omega}{kT}}\right) \end{equation}
This entropy is shown in the nearby figure, as well as the heat capacity.assignment Homework
Consider one mole of an ideal monatomic gas at 300K and 1 atm. First, let the gas expand isothermally and reversibly to twice the initial volume; second, let this be followed by an isentropic expansion from twice to four times the original volume.
How much heat (in joules) is added to the gas in each of these two processes?
What is the temperature at the end of the second process?
Suppose the first process is replaced by an irreversible expansion into a vacuum, to a total volume twice the initial volume. What is the increase of entropy in the irreversible expansion, in J/K?
assignment Homework
Suppose \(g(U) = CU^{3N/2}\), where \(C\) is a constant and \(N\) is the number of particles.
Show that \(U=\frac32 N k_BT\).
Show that \(\left(\frac{\partial^2S}{\partial U^2}\right)_N\) is negative. This form of \(g(U)\) actually applies to a monatomic ideal gas.
group Small Group Activity
30 min.
face Lecture
120 min.
Planck distribution blackbody radiation photon statistical mechanics
These notes from the fourth week of Thermal and Statistical Physics cover blackbody radiation and the Planck distribution. They include a number of small group activities.assignment Homework
In this entire problem, keep results to first order in the van der Waals correction terms \(a\) and $b.
Show that the entropy of the van der Waals gas is \begin{align} S &= Nk\left\{\ln\left(\frac{n_Q(V-Nb)}{N}\right)+\frac52\right\} \end{align}
Show that the energy is \begin{align} U &= \frac32 NkT - \frac{N^2a}{V} \end{align}
Show that the enthalpy \(H\equiv U+pV\) is \begin{align} H(T,V) &= \frac52NkT + \frac{N^2bkT}{V} - 2\frac{N^2a}{V} \\ H(T,p) &= \frac52NkT + Nbp - \frac{2Nap}{kT} \end{align}