Students sketch the temperature-dependent heat capacity of molecular nitrogen. They apply the equipartition theorem and compute the temperatures at which degrees of freedom “freeze out.”
Students calculate probabilities for a particle on a ring using three different notations: Dirac bra-ket, matrix, and wave function. After calculating the angular momentum and energy measurement probabilities, students compare their calculation methods for notation.
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.
Students calculate probabilities for energy, angular momentum, and position as a function of time for an initial state that is a linear combination of energy/angular momentum eigenstates for a particle confined to a ring written in bra-ket notation. This activity helps students build an understanding of when they can expect a quantity to depend on time and to give them more practice moving between representations.
Consider a system of
fixed volume in thermal contact with a resevoir. Show that the mean
square fluctuations in the energy of the system is \begin{equation}
\left<\left(\varepsilon-\langle\varepsilon\rangle\right)^2\right>
= k_BT^2\left(\frac{\partial U}{\partial T}\right)_{V}
\end{equation} Here \(U\) is the conventional symbol for
\(\langle\varepsilon\rangle\). Hint: Use the partition function
\(Z\) to relate \(\left(\frac{\partial U}{\partial T}\right)_V\) to
the mean square fluctuation. Also, multiply out the term
\((\cdots)^2\).
A student is invited to “act out” motion corresponding to a plot of effective potential vs. distance. The student plays the role of the “Earth” while the instructor plays the “Sun”.
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 \(\beta = \frac1{kT}\).
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.
Students examine a plastic “surface” graph of the gravitational potential energy of an Earth-satellite system to explore the properties of gravitational potential energy for a spherically symmetric system.
Students consider the change in internal energy during three different processes involving a container of water vapor on a stove. Using the 1st Law of Thermodynamics, students reason about how the internal energy would change and then compare this prediction with data from NIST presented as a contour plot.
Consider a column of atoms each of mass \(M\) at temperature \(T\) in
a uniform gravitational field \(g\). Find the thermal average
potential energy per atom. The thermal average kinetic energy is
independent of height. Find the total heat capacity per atom. The
total heat capacity is the sum of contributions from the kinetic
energy and from the potential energy. Take the zero of the
gravitational energy at the bottom \(h=0\) of the column. Integrate
from \(h=0\) to \(h=\infty\). You may assume the gas is ideal.
A valuable model for figuring out how we're going to save the Earth
Let's start by visualizing the energy flow associated with driving a gasoline-powered car. We will use a box and arrow diagram, where boxes represent where energy can accumulate, and arrows show energy flow.
The energy clearly starts in the form of gasoline in the tank. Where does it go?
Actually ask this of students.
Visualize the energy as an indestructable, incompressible liquid.
“Energy is conserved”
The heat can look like
Hot exhaust gas
The radiator (its job is to dissipate heat)
Friction heating in the drive train
The work contribute to
Rubber tires heated by deformation
Wind, which ultimately ends up as heating the atmosphere
The most important factors for a coarse-grain model of highway driving:
The 75:25 split between “heat” and “work”
The trail of wind behind a car
What might we have missed? Where else might energy have gone?
We ignored the kinetic energy of the car, and the energy dissipated as heat in the brakes. On the interstate this is appropriate, but for city driving the dominant “work” may be in accelerating the car to 30 mph, and with that energy then converted into heat by the brakes.
Transitioning from the PDM back to thermodynamic systems
Heating
In the partial derivative machine, the change in internal energy corresponds to the work done on the left string and the right string:
\begin{align}
dU &= F_L dx_L + F_R dx_R
\end{align}
The ”thing we changed” was \(dx_L\) or \(dx_R\). From that we could determine the change in internal energy.
When we transfer energy to something by heating, it's hard to measure the “thing we changed,” which was entropy. It is, however, possible in some cases to measure the amount of energy transfered by heating, and from that we can work backwards to find out how much the entropy changed.
An infinitesimal amount of energy transfered by heating is called \({\mathit{\unicode{273}}} Q\). The symbol \({\mathit{\unicode{273}}} \) indicates an inexact differential, which you can think of as a “small chunk” that is not the change of something. \({\mathit{\unicode{273}}} Q\) is nota small change in the amount of energy transfered by heating, but rather is a small amount of energy transfered by heating.
When playing with the partial derivative machine, we can say the work done on the left string, \(F_Ldx_L\), is analogous to heat entering a thermodynamic system.
Latent heat
A phase transition is when a material changes state of matter, as in melting or
boiling. At most phase transitions (technically, abrupt phase transitions
as you will learnin the Capstone), the temperature remains constant while the
material is changing from one state to the other. So you know that as long as
you have ice and water coexisting in equilibrium at one atmosphere of pressure,
the temperature must be \(0^\circ\)C. Similarly, as long as water is boiling at
one atmosphere of pressure, the temperature must be \(100^\circ\)C. In both of
these cases, you can transfer energy to the system (as we will) by heating
without changing the temperature! This relates to why I keep awkwardly
saying
“transfer energy to a system by heating” rather than just “heating a system”
which means the same thing. We have deeply ingrained the idea that “heating”
is synonymous with “raising the temperature,” which does not align with the
physics meaning.
So now let me define the latent heat. The latent heat is the amount
of energy that must be transfered to a material by heating in order to change
it from one phase to another. The latent heat of fusion is the amount
of energy required to melt a solid, and the latent heat of vaporization
is the amount of energy required to turn a liquid into a gas. We will be
measuring both of these for water.
A question you may ask is whether the latent heat is extensive or intensive.
Technically the latent heat is extensive, since if you have more material
then more energy is required to melt/boil it. However, when you hear latent heat
quoted, it is almost always the specific latent heat,
which is the energy
transfer by heating required per unit of mass. It can be confusing that people
use the same words to refer to both quantities. Fortunately, dimensional checking
can always give you a way to verify which is being referred to. If \(L\) is an
energy per mass, then it must be the specific latent heat, while if it is an
energy, then it must be the latent heat.
Heat capacity and specific heat
The heat capacity is the amount of energy transfer required per
temperature to raise the temperature of a system. If we hold the pressure fixed
(as in our experiment) we can write this as:
\begin{align}
{\mathit{\unicode{273}}} Q &= C_p dT
\end{align}
where \(C_p\) is the heat capacity at fixed pressure.
You might think to rewrite this expression as a derivative, but we can't
do that since the energy transfered by heating is not a state function.
Note that the heat capacity, like the latent heat, is an extensive quantity.
The specific heat is the the heat capacity per unit mass, which is an
intensive quantity that we can consider a property of a material independently
of the quantity of that material.
I'll just mention as an aside that the term “heat capacity” is another one of
those unfortunate phrases that reflect the inaccurate idea that heat is a
property of a system.
Entropy
Finally, we can get to entropy. Entropy is the “thing that changes” when you
transfer energy by heating. I'll just give this away:
\begin{align}
{\mathit{\unicode{273}}} Q &= TdS
\end{align}
where this equation is only true if you make the change quasistatically
(see another lecture). This allows us to find the change in entropy if we know
how much energy was transfered by heating, and the temperature in the process.
\begin{align}
\Delta S &= \int \frac1T {\mathit{\unicode{273}}} Q
\end{align}
where again, we need to know the temperature as we add heat.
Students implement a finite-difference approximation for the kinetic energy operator as a matrix, and then use numpy to solve for eigenvalues and eigenstates, which they visualize.
A beam of spin-\(\frac{1}{2}\) particles is prepared in the initial state \[ \left\vert \psi\right\rangle = \sqrt{\frac{2}{5}}\; |+\rangle_x - \sqrt{\frac{3}{5}}\; |-\rangle_x \](Note: this state is written in the \(S_x\) basis!)
What are the possible results of a measurement of \(S_x\), with what probabilities?
Repeat part a for measurements of \(S_z\).
Suppose you start with a particle in the state given above, measure \(S_x\), and happen to get \(+\hbar /2\). You then take that same particle and measure \(S_z\). What are the possible results and with what probability would you measure each possible result?
With the Spins simulation set for a spin 1/2 system, measure the probabilities of all the possible spin components for each of the unknown initial states \(\left|{\psi_3}\right\rangle \) and \(\left|{\psi_4}\right\rangle \). (Since \(\left|{\psi_3}\right\rangle \) has already been covered in class, please only do \(\left|{\psi_4}\right\rangle \) )
Use your measured probabilities to find each of the unknown states as a linear superposition of the \(S_z\)-basis states \(\left|{+}\right\rangle \) and \(\left|{-}\right\rangle \).
Articulate a Process: Write a set of general instructions that would allow another student in next year's class to find an unknown state from measured probabilities.
Compare Theory with Experiment: Design an experiment that will allow you to test whether your prediction for each of the unknown states is correct. Describe your experiment here, clearly but succinctly, as if you were writing it up for a paper. Do the experiment and discuss your results.
Make a Conceptual Connection: In general, can you determine a quantum state with spin-component probability measurements in only two spin-component-directions? Why or why not?
Break the links between the analyzer and the counters by clicking on the boxes with up and down arrow labels on the analyzer.
Click and drag a new connection from the analyzer to empty space to create a new element. A new analyzer is one of the options.
Measure \(S_z\) twice in succession.
What is the probability that a particle leaving the first analyzer with \(S_z=\frac{+\hbar}{2}\) will be measured by the second analyzer to have \(S_z=\frac{-\hbar}{2}\)?
Try all four possible combinations of input/outputs for the second analyzer.
What have you learned from these experiments?
Try All Combinations of Sequential Measurements
In the table, enter the probability of a particle exiting the 2nd analyzer with the spin indicated in row if the particle enters the 2nd analyzer with the spin indicated in each column.
You can rotate the Stern-Gerlach analyzers to any direction you want (using spherical coordinates).
Choose an arbitrary direction (not along one of the coordinate axes) for the 1st analyzer and measure the spin along the coordinate directions for the 2nd analyzer.
Students see probability density for eigenstates and linear combinations of eigenstates for a particle on a ring. The three visual representations: standard position vs probability density plot, a ring with colormapping, and cylindrical plot with height and colormapping, are also animated to visualize time-evolution.
This activity allows students to puzzle through indexing, the from of operators in quantum mechanics, and working with the new quantum numbers on the sphere in an applied context.
Students, working in pairs, use their left arms to represent each component in a two-state quantum spin 1/2 system. Reinforces the idea that quantum states are complex valued vectors. Students make connections between Dirac, matrix, and Arms representation.
Students calculate the expectation value of energy and angular momentum as a function of time for an initial state for a particle on a ring. This state is a linear combination of energy/angular momentum eigenstates written in bra-ket notation.
Students use Tinker Toys to represent each component in a two-state quantum spin system in all three standard bases (\(x\), \(y\), and \(z\)). Through a short series of instructor-led prompts, students explore the difference between overall phase (which does NOT change the state of the system) and relative phase (which does change the state of the system). This activity is optional in the Arms Sequence Arms Sequence for Complex Numbers and Quantum States.
Students become acquainted with the Spins Simulations of Stern-Gerlach Experiments and record measurement probabilities of spin components for a spin-1/2 system. Students start developing intuitions for the results of quantum measurements for this system.
Students review using the Arms representation to represent states for discrete quantum systems and connecting the Arms representation to histogram and matrix representation. The student then extend the Arms representation to begin exploring the continuous position basis.
Students, working in pairs, use the Arms representations to represent states of spin 1/2 system. Through a short series of instructor-led prompts, students explore the difference between overall phase (which does NOT distinguish quantum states) and relative phase (which does distinguish quantum states).
Students calculate probabilities for a particle on a ring whose wavefunction is not easily separated into eigenstates by inspection. To find the energy, angular momentum, and position probabilities, students perform integrations with the wavefunction or decompose the wavefunction into a superposition of eigenfunctions.
The Gibbs free energy,
\(G\), is given by
\begin{align*}
G = U + pV - TS.
\end{align*}
Find the total differential of \(G\). As always, show your work.
Interpret the coefficients of the total differential \(dG\) in
order to find a derivative expression for the entropy \(S\).
From the total differential \(dG\), obtain a different
thermodynamic derivative that is equal to
\[ \left(\frac{\partial {S}}{\partial {p}}\right)_{T} \]
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 \(\beta = \frac1{kT}\).
Found in: Thermal and Statistical Physics course(s)
(Messy algebra) Convince yourself that the expressions for kinetic energy in original and center of mass coordinates are equivalent. The same for angular momentum.
Consider a system of two particles of mass \(m_1\) and \(m_2\).
Show that the total kinetic energy of the system is the same as that of two
“fictitious” particles: one of mass \(M=m_1+m_2\) moving with the velocity of the
center of mass and one of mass \(\mu\) (the reduced mass) moving with the
velocity of the relative position.
Show that the total angular momentum of the system can similarly be decomposed
into the angular momenta of these two fictitious particles.
Consider a system which has an internal energy \(U\) defined by:
\begin{align}
U &= \gamma V^\alpha S^\beta
\end{align}
where \(\alpha\), \(\beta\) and \(\gamma\) are constants. The internal
energy is an extensive quantity. What constraint does this place on
the values \(\alpha\) and \(\beta\) may have?
Groups are asked to analyze the following standard problem:
Two identical lumps of clay of (rest) mass m collide head on, with each
moving at 3/5 the speed of light. What is the mass of the resulting lump of
clay?
As discussed in
class, we can consider a black body as a large box with a small hole
in it. If we treat the large box a metal cube with side length \(L\)
and metal walls, the frequency of each normal mode will be given by:
\begin{align}
\omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2}
\end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have
positive integer values. This simply comes from the fact that a half
wavelength must fit in the box. There is an additional quantum number
for polarization, which has two possible values, but does not affect
the frequency. Note that in this problem I'm using different
boundary conditions from what I use in class. It is worth learning to
work with either set of quantum numbers. Each normal mode is a
harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\)
where we will not include the zero-point energy
\(\frac12\hbar\omega\), since that energy cannot be extracted from the
box. (See the
Casimir effect
for an example where the zero point energy of photon modes does have
an effect.)
Note
This is a slight approximation, as the boundary conditions for light
are a bit more complicated. However, for large \(n\) values this gives
the correct result.
Show that the free energy is given by \begin{align}
F &= 8\pi \frac{V(kT)^4}{h^3c^3}
\int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi
\\
&= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3}
\\
&= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3}
\end{align} provided the box is big enough that
\(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a
slightly different dimensionless integral that numerically evaluates
to the same result, which would be fine. I also do not expect you to
solve this definite integral analytically, a numerical confirmation
is fine. However, you must manipulate your integral until it
is dimensionless and has all the dimensionful quantities removed
from it!
Show that the entropy of this box full of photons at temperature
\(T\) is \begin{align}
S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3
\\
&= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3
\end{align}
Show that the internal energy of this box full of photons at
temperature \(T\) is \begin{align}
\frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3}
\\
&= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3}
\end{align}
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.
Consider a system that may be unoccupied with energy zero, or
occupied by one particle in either of two states, one of energy zero
and one of energy \(\varepsilon\). Find the Gibbs sum for this
system is in terms of the activity \(\lambda\equiv e^{\beta\mu}\).
Note that the system can hold a maximum of one particle.
Solve for the thermal average occupancy of the system in terms of
\(\lambda\).
Show that the thermal average occupancy of the state at energy
\(\varepsilon\) is \begin{align}
\langle N(\varepsilon)\rangle =
\frac{\lambda e^{-\frac{\varepsilon}{kT}}}{\mathcal{Z}}
\end{align}
Find an expression for the thermal average energy of the system.
Allow the possibility that the orbitals at \(0\) and at
\(\varepsilon\) may each be occupied each by one particle at the
same time; Show that \begin{align}
\mathcal{Z} &= 1 + \lambda + \lambda e^{-\frac{\varepsilon}{kT}} +
\lambda^2 e^{-\frac{\varepsilon}{kT}}
\\
&= (1+\lambda)\left(1+e^{-\frac{\varepsilon}{kT}}\right)
\end{align} Because \(\mathcal{Z}\) can be factored as shown, we
have in effect two independent systems.
For electrons with an energy \(\varepsilon\gg mc^2\), where
\(m\) is the mass of the electron, the energy is given by
\(\varepsilon\approx pc\) where \(p\) is the momentum. For electrons
in a cube of volume \(V=L^3\) the momentum takes the same values as
for a non-relativistic particle in a box.
Show that in this extreme relativistic limit the Fermi energy of a
gas of \(N\) electrons is given by \begin{align}
\varepsilon_F &= \hbar\pi c\left(\frac{3n}{\pi}\right)^{\frac13}
\end{align} where \(n\equiv \frac{N}{V}\) is the number density.
Show that the total energy of the ground state of the gas is
\begin{align}
U_0 &= \frac34 N\varepsilon_F
\end{align}
Consider a white dwarf of mass \(M\) and radius \(R\). The dwarf
consists of ionized hydrogen, thus a bunch of free electrons and
protons, each of which are fermions. Let the electrons be degenerate
but nonrelativistic; the protons are nondegenerate.
Show that the order of magnitude of the gravitational self-energy is
\(-\frac{GM^2}{R}\), where \(G\) is the gravitational constant. (If
the mass density is constant within the sphere of radius \(R\), the
exact potential energy is \(-\frac53\frac{GM^2}{R}\)).
Show that the order of magnitude of the kinetic energy of the
electrons in the ground state is \begin{align}
\frac{\hbar^2N^{\frac53}}{mR^2}
\approx \frac{\hbar^2M^{\frac53}}{mM_H^{\frac53}R^2}
\end{align} where \(m\) is the mass of an electron and \(M_H\) is
the mas of a proton.
Show that if the gravitational and kinetic energies are of the same
order of magnitude (as required by the virial theorem of mechanics),
\(M^{\frac13}R \approx 10^{20} \text{g}^{\frac13}\text{cm}\).
If the mass is equal to that of the Sun (\(2\times 10^{33}g\)), what
is the density of the white dwarf?
It is believed that pulsars are stars composed of a cold degenerate
gas of neutrons (i.e. neutron stars). Show that for a neutron star
\(M^{\frac13}R \approx 10^{17}\text{g}^{\frac13}\text{cm}\). What is
the value of the radius for a neutron star with a mass equal to that
of the Sun? Express the result in \(\text{km}\).
In our week on radiation, we saw that the Helmholtz free energy of a
box of radiation at temperature \(T\) is \begin{align}
F &= -8\pi \frac{V(kT)^4}{h^3c^3}\frac{\pi^4}{45}
\end{align} From this we also found the internal energy and entropy
\begin{align}
U &= 24\pi \frac{(kT)^4}{h^3c^3}\frac{\pi^4}{45} V \\
S &= 32\pi kV\left(\frac{kT}{hc}\right)^3 \frac{\pi^4}{45}
\end{align} Given these results, let us consider a Carnot engine that
uses an empty metalic piston (i.e. a photon gas).
Given \(T_H\) and \(T_C\), as well as \(V_1\) and \(V_2\) (the two
volumes at \(T_H\)), determine \(V_3\) and \(V_4\) (the two volumes
at \(T_C\)).
What is the heat \(Q_H\) taken up and the work done by the gas
during the first isothermal expansion? Are they equal to each other,
as for the ideal gas?
Does the work done on the two isentropic stages cancel each other,
as for the ideal gas?
Calculate the total work done by the gas during one cycle. Compare
it with the heat taken up at \(T_H\) and show that the energy
conversion efficiency is the Carnot efficiency.
