Ideal gasEntropyTempuratureThermal and Statistical Physics 2020
Consider an ideal gas of
\(N\) particles, each of mass \(M\), confined to a one-dimensional
line of length \(L\). The particles have spin zero (so you can ignore
spin) and do not interact with one another. Find the entropy at
temperature \(T\). You may assume that the temperature is high enough
that \(k_B T\) is much greater than the ground state energy of one
particle.
Find the entropy of a set of \(N\) oscillators of frequency
\(\omega\) as a function of the total quantum number \(n\). Use the
multiplicity function: \begin{equation}
g(N,n) = \frac{(N+n-1)!}{n!(N-1)!}
\end{equation} and assume that \(N\gg 1\). This means you can
make the Sitrling approximation that
\(\log N! \approx N\log N - N\). It also means that
\(N-1 \approx N\).
Let \(U\) denote the total energy \(n\hbar\omega\) of the
oscillators. Express the entropy as \(S(U,N)\). Show that the total
energy at temperature \(T\) is \begin{equation}
U = \frac{N\hbar\omega}{e^{\frac{\hbar\omega}{kT}}-1}
\end{equation} This is the Planck result found the hard
way. We will get to the easy way soon, and you will never again need
to work with a multiplicity function like this.
EnergyTemperatureParamagnetismThermal and Statistical Physics 2020
Find the equilibrium value at temperature \(T\)
of the fractional magnetization \begin{equation}
\frac{\mu_{tot}}{Nm} \equiv \frac{2\langle s\rangle}{N}
\end{equation} of a system of \(N\) spins each of magnetic moment
\(m\) in a magnetic field \(B\). The spin excess is \(2s\). The energy
of this system is given by \begin{align}
U &= -\mu_{tot}B
\end{align} where \(\mu_{tot}\) is the total magnetization. Take the
entropy as the logarithm of the multiplicity \(g(N,s)\) as given in
(1.35 in the text): \begin{equation}
S(s) \approx k_B\log g(N,0) - k_B\frac{2s^2}{N}
\end{equation} for \(|s|\ll N\), where \(s\) is the spin excess, which
is related to the magnetization by \(\mu_{tot} = 2sm\). Hint:
Show that in this approximation \begin{equation}
S(U) = S_0 - k_B\frac{U^2}{2m^2B^2N},
\end{equation} with \(S_0=k_B\log g(N,0)\). Further, show that
\(\frac1{kT} = -\frac{U}{m^2B^2N}\), where \(U\) denotes
\(\langle U\rangle\), the thermal average 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.
Task:Draw a right triangle. Put a circle around the right angle, that is, the angle that is \(\frac\pi2\) radians.
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Consider the bottle in a bottle problem in a previous problem set, summarized here.
A small bottle of
helium is placed inside a large bottle, which otherwise contains
vacuum. The inner bottle contains a slow leak, so that the helium
leaks into the outer bottle. The inner bottle contains one tenth
the volume of the outer bottle, which is insulated.
The volume of the small bottle is 0.001 m23 and the volume of the big bottle is 0.01 m3. The initial state of the gas in the small bottle was \(p=106\) Pa and its temperature \(T=300\) K. Approximate the helium gas as an ideal gas of equations of state \(pV=Nk_BT\) and \(U=\frac32 Nk_BT\).
How many molecules of gas does the large bottle contain? What is the final temperature of the gas?
Compute the integral \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\) and the change of entropy \(\Delta S\) between the initial state (gas in the small bottle) and the final state (gas leaked in the big bottle).
Discuss your results.
Thermal and Statistical Physics 2020
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.
