assignment Homework
Show that a Fermi electron gas in the ground state exerts a pressure \begin{align} p = \frac{\left(3\pi^2\right)^{\frac23}}{5} \frac{\hbar^2}{m}\left(\frac{N}{V}\right)^{\frac53} \end{align} In a uniform decrease of the volume of a cube every orbital has its energy raised: The energy of each orbital is proportional to \(\frac1{L^2}\) or to \(\frac1{V^{\frac23}}\).
Find an expression for the entropy of a Fermi electron gas in the region \(kT\ll \varepsilon_F\). Notice that \(S\rightarrow 0\) as \(T\rightarrow 0\).
face Lecture
120 min.
Fermi level fermion boson Bose gas Bose-Einstein condensate ideal gas statistical mechanics phase transition
These lecture notes from week 7 of Thermal and Statistical Physics apply the grand canonical ensemble to fermion and bosons ideal gasses. They include a few small group activities.face Lecture
120 min.
ideal gas particle in a box grand canonical ensemble chemical potential statistical mechanics
These notes from week 6 of Thermal and Statistical Physics cover the ideal gas from a grand canonical standpoint starting with the solutions to a particle in a three-dimensional box. They include a number of small group activities.assignment Homework
The concentration of potassium \(\text{K}^+\) ions in the internal sap of a plant cell (for example, a fresh water alga) may exceed by a factor of \(10^4\) the concentration of \(\text{K}^+\) ions in the pond water in which the cell is growing. The chemical potential of the \(\text{K}^+\) ions is higher in the sap because their concentration \(n\) is higher there. Estimate the difference in chemical potential at \(300\text{K}\) and show that it is equivalent to a voltage of \(0.24\text{V}\) across the cell wall. Take \(\mu\) as for an ideal gas. Because the values of the chemical potential are different, the ions in the cell and in the pond are not in diffusive equilibrium. The plant cell membrane is highly impermeable to the passive leakage of ions through it. Important questions in cell physics include these: How is the high concentration of ions built up within the cell? How is metabolic energy applied to energize the active ion transport?
assignment Homework
face Lecture
120 min.
chemical potential Gibbs distribution grand canonical ensemble statistical mechanics
These notes from the fifth week of Thermal and Statistical Physics cover the grand canonical ensemble. They include several small group activities.assignment Homework
Let us imagine a new mechanics in which the allowed occupancies of an orbital are 0, 1, and 2. The values of the energy associated with these occupancies are assumed to be \(0\), \(\varepsilon\), and \(2\varepsilon\), respectively.
Derive an expression for the ensemble average occupancy \(\langle N\rangle\), when the system composed of this orbital is in thermal and diffusive contact with a resevoir at temperature \(T\) and chemical potential \(\mu\).
Return now to the usual quantum mechanics, and derive an expression for the ensemble average occupancy of an energy level which is doubly degenerate; that is, two orbitals have the identical energy \(\varepsilon\). If both orbitals are occupied the toal energy is \(2\varepsilon\). How does this differ from part (a)?
face Lecture
30 min.
thermodynamics statistical mechanics
These are notes, essentially the equation sheet, from the final review session for Thermal and Statistical Physics.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
The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.
The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).