assignment Homework
Make sure that you have memorized the following identities and can use them in simple algebra problems: \begin{align} e^{u+v}&=e^u \, e^v\\ \ln{uv}&=\ln{u}+\ln{v}\\ u^v&=e^{v\ln{u}} \end{align}
assignment Homework
You are given the following Gibbs free energy: \begin{equation*} G=-k T N \ln \left(\frac{a T^{5 / 2}}{p}\right) \end{equation*} where \(a\) is a constant (whose dimensions make the argument of the logarithm dimensionless).
Compute the entropy.
Work out the heat capacity at constant pressure \(C_p\).
Find the connection among \(V\), \(p\), \(N\), and \(T\), which is called the equation of state (Hint: find the volume as a partial derivative of the Gibbs free energy).
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.face Lecture
30 min.
face Lecture
120 min.
paramagnet entropy temperature statistical mechanics
These lecture notes for the second week of Thermal and Statistical Physics involve relating entropy and temperature in the microcanonical ensemble, using a paramagnet as an example. These notes include a few small group activities.assignment Homework
Nuclei of a particular isotope species contained in a crystal have spin \(I=1\), and thus, \(m = \{+1,0,-1\}\). The interaction between the nuclear quadrupole moment and the gradient of the crystalline electric field produces a situation where the nucleus has the same energy, \(E=\varepsilon\), in the state \(m=+1\) and the state \(m=-1\), compared with an energy \(E=0\) in the state \(m=0\), i.e. each nucleus can be in one of 3 states, two of which have energy \(E=\varepsilon\) and one has energy \(E=0\).
Find the Helmholtz free energy \(F = U-TS\) for a crystal containing \(N\) nuclei which do not interact with each other.
Find an expression for the entropy as a function of temperature for this system. (Hint: use results of part a.)
face Lecture
120 min.
ideal gas entropy canonical ensemble Boltzmann probability Helmholtz free energy statistical mechanics
These notes, from the third week of Thermal and Statistical Physics cover the canonical ensemble and Helmholtz free energy. They include a number of small group activities.assignment Homework
Consider two noninteracting systems \(A\) and \(B\). We can either treat these systems as separate, or as a single combined system \(AB\). We can enumerate all states of the combined by enumerating all states of each separate system. The probability of the combined state \((i_A,j_B)\) is given by \(P_{ij}^{AB} = P_i^AP_j^B\). In other words, the probabilities combine in the same way as two dice rolls would, or the probabilities of any other uncorrelated events.
assignment Homework
assignment Homework
Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.
Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.
If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.
Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.
This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).
\(\ln{x}+\ln{y}\)
\(\ln{a}-\ln{b}\)
\(2\ln{f}+3\ln{f}\)
\(e^{m}e^{k}\)
\(\frac{e^{c}}{e^{d}}\)
\(e^{(3h-j)}\)
\(e^{2(c+w)}\)
\(\ln{h/g}\)
\(\ln(kT)\)
\(\ln{\sqrt{\frac{q}{r}}}\)