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}
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).
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}
Entropy of a simple harmonic oscillatorHeat capacity of a simple harmonic oscillator
This entropy is shown in the nearby figure, as well
as the heat capacity.
This lecture introduces the idea of entropy, including the relationship between entropy and multiplicity as well as the relationship between changes in entropy and heat.
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.
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.)
Indicate what your results predict for the entropy at the
extremes of very high temperature and very low temperature.
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.
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.
Show that the entropy of the combined system \(S_{AB}\) is the
sum of entropies of the two separate systems considered
individually, i.e. \(S_{AB} = S_A+S_B\). This means that entropy is
extensive. Use the Gibbs entropy for this computation. You need
make no approximation in solving this problem.
Show that if you have \(N\) identical non-interacting systems,
their total entropy is \(NS_1\) where \(S_1\) is the entropy of a
single system.
Note
In real materials, we treat properties as being extensive even
when there are interactions in the system. In this case,
extensivity is a property of large systems, in which surface
effects may be neglected.
Static Fields 2023 (4 years)
Determine the following derivatives and evaluate the following integrals, all by hand. You should also learn how to check these answers on Wolfram Alpha.
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.
Note
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).
