Students will determine the change in entropy (positive, negative, or none) for both the system and surroundings in three different cases. This is followed by an active whole-class discussion about where the entropy comes from during an irreversible process.

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.

The internal energy is of any ideal gas can be written as
\begin{align}
U &= U(T,N)
\end{align}
meaning that the internal energy depends only on the number of
particles and the temperature, but not the volume.^{*}

The ideal gas law
\begin{align}
pV &= Nk_BT
\end{align}
defines the relationship between \(p\), \(V\) and \(T\). You may take the
number of molecules \(N\) to be constant. Consider the free adiabatic
expansion of an ideal gas to twice its volume. “Free expansion”
means that no work is done, but also that the process is also
neither quasistatic nor reversible.

What is the change in entropy of the gas? How do you know
this?