You’ll find here to sections. The first section are notes taken from the book, while the second section is my written solutions to the problem-set at the end of chapters.

📝 Notes

Chapter 1 - The Statistical Basis of Thermodynamics

1.1 The macroscopic and the microscopic states

We make a distinction between the macroscopic and microscopic states of a system. The macroscopic states are defined by the properties of the collective of particles. The volume $V$, the number of particles $N$ and the total energy $E$. In macroscopic systems we work in the thermodynamic limit, $N \rightarrow \infty, V \rightarrow \infty$ but the ratio $n=\frac{N}{V}$ (the particle density) remains constant.

The macroscopic state is made out of many particles in microscopic states. We write that there are $n_i$ particles in state $i$, each with energy $\epsilon_i$. It’s obvious then that

Assuming there is no interaction between the particles we can also write for the total energy $E$

Since there are many ways to realize a macrostate with particle microstates, we define the number of microstates that realize a specific macrostate $(N, V, E)$ by the function $\Omega(N, V, E)$. Amazingly, from this value all the thermodynamic properties can be derived!

1.2 Contact between statistics and thermodynamics: physical significance of the number $\Omega(N, V, E)$

We consider two physical systems, $A_1$ and $A_2$, which are separately in equilibrium. The two system are in thermal contact so that energy exchange is possible but the amount of particles and the volume of each system stays constant. Also the total energy of the system is constant

We can easily show that $\Omega^{(0)}$ varies only with $E_1$ (while other values such as $E^{(0)}$, $V_{1/2}$ etc… are parameters). We can separate $\Omega^{(0)}$ using separation of variables because of the multiplicity of the the number of microstates

At equilibrium we maximize $\Omega^{(0)}$ so

And since $\partial E_2 / \partial E_1 = -1$ and some algebra we get

This is our condition for equilibrium. We then define

So the condition for equilibrium between two system reduces to $\beta_1 = \beta_2$.

1.3 Further contact between statistics and thermodynamics

1.4 The classical ideal gas

1.5 The entropy of mixing and the Gibbs paradox

1.6 The “correct” enumeration of the micro-states

✍ Solutions

Chapter 1

1.1

1.1.a

Problem: Show that, for two large systems in thermal contact, the number $\Omega^{(0)} (E^{(0)}, E_1)$ of Sec. 1.2 can be expressed as a Gaussian in the variable $E_1$. Determine the root-mean-square deviation of $E_1$ from the value $\bar{E}_1$ in terms of other quantities pertaining to the problem.

Solution: Let’s expand the quantity $\ln{\Omega^{(0)}(E_1)}$ in terms of $E_1 - \bar{E}_1$.

The first terms is just some constant, the second term vanishes (see Eq. 1.2.1a), we’ll now evaluate the third term, and the higher terms are neglected for large systems

Let’s assign ($T_1 = T_2$ at equilibrium)

Hence we get that

and therefore

As expected, the deviation of this distribution is

1.1.b

Problem: Make an explicit evaluation of the root-mean-square deviation of $E_1$ in the special case when the systems $A_1$ and $A_2$ are ideal classical gases

Solution: In this case $(C_v){1/2} = \frac{3}{2}N{1/2}k$, so we get that

so

1.2

Problem: Assuming that the entropy $S$ and the statistical number $\Omega$ of a physical system are related through an arbitrary functional form

show that the additive character of $S$ and the multiplicative character of $\Omega$ necessarily require that the function $f(\sigma)$ be of the form (1.2.6) $S = f(\Omega) = k \ln{\Omega}$

Solution: Consider a system composed of two sub-systems, A and B, we shall call the total system AB. The additive nature of the entropy tells us tht $S_{AB} = S_A + S_B$ and the multiplicative nature of the statistical number tells us that $\Omega_{AB} = \Omega_A\Omega_B$. Therefore we’ll write on the one hand

and on the other hand

Combining these two equation we get

Let us now also define $g(x) = f(e^x)$, we’ll notice that

from this we can easily derive that

We can calculate the derivative of $g(x)$ at a point $x=x_0$ directly from definition of the derivative

We can expand $g(\Delta x)$ as a Taylor series of $\Delta x$ around $0$ like so (also it’s easy to see that $g(0) = 0$)

let’s define $g’(0)$ as $g’(0)=k$, we get that $g’(x)=k$ and therefore

from here we can go back to $f(x)$ by

and using the original physical symbols,

as expected. $\blacksquare$

1.3

Problem: Two systems A and B, of identical composition, are brought together and allowed to exchange both energy and particles, keeping volumes $V_A$ and $V_B$ constant. Show that the minimum value of the quantity $(\frac{dE_A}{dN_A})$ is given by

where $\mu$ and the $T$ are the respective chemical potentials and temperatures.

Solution: We start with equation (1.3.4)

Applying for each system separately we get

We may use the relation $E=E_A+E_B$, $N=N_A+N_B$ and $S=S_A+S_B$ on B (where E, N and S are the energy and entropy of the total composite system). Also we shall use the fact that the voumes remain constant to write down $dV_{A/B}=0$. We thus get

Since the total system is isolated from the environment there is no exchange of energy or particles between the system and the environment, therefore, $dN=dE=0$ so we can write

Solving for $dS_A$ we get

on the other hand, for the original equation of A, solving for $dS_A$, we get that

comparing the two expressions we have the following equation

solving for $dE_A$

From the second law we know that $dS \ge 0$, so if $T_B > T_A$ then the right-most element in the expression is positive and equals $0$ at the minimum (from the third law), so we can write, as expected

1.4

Problem: In a classical gas of hard spheres (of diameter $\sigma$), the spatial distribution of the particles is no longer uncorrelated. Roughly speaking, the presence of $n$ particles in the system leaves only a volume $(V-nv_0)$ available for the $(n+1)$th particle; clearly, $v_0$ would be proportional to $\sigma^3$. Assuming that $Nv_0\ll V$, determine the dependence of $\Omega(N,V,E)$ on V {cf. (1.4.1)} and show that, as a result of this, V in the gas law (1.4.3) gets replaced by $(V-b)$, where $b$ is four times the actual space occupied by the particles.

Solution: We start by calculating the dependence of the number $\Omega$ on V. We consider each additional particle takes volume $v_0$ from the possible volume for the next particle. Since the particles can’t overlap, $v_0$ would be the volume of for which if we added a particle in it, the added particle would overlap with existing particle. Since both particles have diameter $\sigma$, that volume is of a sphere with radius $\sigma$, so $v_0 = \frac{4}{3}\pi\sigma^3=4 \cdot v_p$ where $v_p$ is the actual volume of a single particle. We’ll stick with $v_0$ notation from now on.

The first particle have volume $V$ to be found in, the next has volume $V-v_0$, the next has volume $V-2v_0$ and so on and so forth… We can calculate similarly to (1.3.1) that the number $\Omega(N,V,E)$ is proportional to

We usually want to deal with $\log{\Omega}$ and not $\Omega$ itself, so let’s calculate it

From (1.3.9) and (1.3.10) we get that

since $nv_0 \ll V$ we can expand $1/x$ in a taylor series around $V$ as

Applying to the earlier equation we get

Where in the last approximation we used the taylor expansion in the other direction and that $N-1 \approx N$. Putting it all together we get that

We got what we expected, and the $b$ that the question refers to is $b=\frac{N}{2} v_0$. The actual space occupied by the particles is $N \cdot v_p$ (where, again, $v_p$ is the volume of a single particle),

or, to put in the phrasing of the question

1.5

Problem: Read Appendix A and establish formulae (1.4.15) and (1.4.16). Estimate the importance of the linear term in these formulae, relative to the main term $(π/6){ε^∗}^{3/2}$, for an oxygen molecule confined to a cube of side 10 cm; take $ε = 0.05$ eV.

Solution: I don’t really get the question… It is basically solved in the appendix. From (A.12) we immediately get

Plugging that into (A.11) we get

Now, since we’re considering a particle in a box we may write $V=L^3$ and $S = 6L^2$, so the equation becomes

which is exactly (1.4.15/16). For an oxygen molecule we can write $m=2.98\cdot 10^{10} \frac{ev}{c^2}$, $\varepsilon=0.05ev$ and $L=0.1m$, plugging into (1.4.6) we get

so the significance of the linear term compared to the main term is

Which is, as you might have noticed, pretty fucking small :). $\blacksquare$

1.6

Problem: A cylindrical vessel $1$m long and $0.1$m in diameter is filled with a monatomic gas at $P=1$atm and $T=300$ K. The gas is heated by an electrical discharge, along the axis of the vessel, which releases an energy of $10^4$ joules. What will the temperature of the gas be immediately after the discharge?

Solution: We assume that the total energy of the system is conseved, as well as the volume and the number of particles. We may write that the energy immidiatly after the electrical discharge the total energy of the system is the (thermal) energy of the gas before the discharge, plus the energy released from the axis

Where (assuming ideal gas) from (1.4.23)

Since we also know the pressure we may use (1.4.26)

therefore

Plugin back into $E_i$ we get $E_i\approx3.15 \cdot 10^3J$, so the total energy after the discharge is $E=1.32 \cdot 10^4J$, using (1.4.23) again we get

1.7

Problem: Study the statistical mechanics of an extreme relativistic gas characterized by the single-particle energy states

instead of (1.4.5), along the lines followed in Section 1.4. Show that the ratio $C_P/C_V$ in this case is $4/3$, instead of $5/3$.

Solution: The number of distinct eigenvalues (i.e. microstates), which is the number $\Omega$ that we seek to calculate, is equal to the number of independent, positive-integral solutions of the equation $\varepsilon=\varepsilon(n_x,n_y,n_z)$ where $\epsilon$ is the energy of the particle. Since we’re dealing with a gas of many particles, the actual equation should be the number of solutions of $E=E({n_r})$ where ${n_r}$ is a set of the single-particle state numbers of all the particles and $E$ is the total energy of the system. We can write this equation explicitly

An immediate results from this relation is that $E$ and $V^{1/3}$ only come together, thus $\Omega( N,E,V)$ is really $\Omega(N, V^{1/3}E)$, so for a reversible adiabatic process ($N$ and $S$ constant) we get that

From the thermodynamic relation we get

From applying both we get

From thermodynamics we know that this implies that $\gamma=C_P/C_V=4/3$ as expected. This might not be a fully satisfactory answer since we didn’t calculate them explicitly, let us continue then. Now we encounter our first issue, most of the thermodynamic properties of the system arise from $\Omega$ and it’s derivatives, which just happened to be extremely discontinues and not differentiable, $\Omega$ is some integer for integer values of $E^$ and is $0$ for all other values of $E^$, we’ll come back to that issue later. For now, instead of looking at $\Omega(E^)$ which is the number of states with energy $E^$, we’ll look at $\Sigma(E^)$ which we define as the number of microstates with energy $E^$ or less.

This function fluctuates much less with $E^*$, and we’ll see later how we would get thermodynamic properties out of it. From (C.7) we’ll write (for Dirichlet boundary conditions)

and substituting for $E^*$ we get that

Using Stirling’s approximation we get

OK, we got the expression for $\Sigma$ now what? Well, we’ll use now some ad-hoc solution to the problem that $\Omega$ was not smooth. A system cannot be completely isolated from it’s environment, so it’s energy is never exactly some specific $E$, we’ll thus define $\Delta$ to be the size of the range of the possible energies of the system, i.e. the energy is between $E-\frac{1}{2}\Delta$ and $E+\frac{1}{2}\Delta$, we want ($\Delta \ll E$). To get back $\Omega$ from $\Sigma$ we note that at the limit of large energies, the sum used to define $\Sigma(N,V,E)$ becomes and integral over $\Omega$, thus, from the fundamental theorem of calculus, $\Omega \approx \frac{\partial \Sigma} {\partial E}$. Let’s define the number of states in the energy range defined earlier by $\Gamma$, we can then write

whence

Since $\Delta/E = O(1/\sqrt{N})$ we can ignore the right term in curly brackets, and since $N \gg 1$ we can ignore the left term in the curly brackets (since $\frac{3}{2}N \gg \ln{\frac{3N}{2}}$), thus

Amazingly, this expression does not depend on the size of the range $\Delta$ as long as it’s sufficiently small! We can easily get the expression for the entropy from Boltzmann’s relation

Solving for E we get

We can easily get from (1.3.10), $\partial S/\partial E = 1/T$, the energy-temperature relation

The same as for an ideal gas. So from (1.3.17)

We can obtain the equation of state by

And we can obtain $C_P$ from (1.3.18)

So we can write

As expected. We can easily get now any thermodynamic quantity since we have an expression for the entropy $S(N,V,E)$.

1.8

Problem: Consider a system of quasi-particles whose energy eigenvalues are given by

Obtain an asymptotic expression for the number $\Omega$ of this system for a given number $N$ of the quasi-particles and a given total energy $E$. Determine the temperature $T$ of the system as a function of $E/N$ and $hν$, and examine the situation for which $E/(Nhν) \gg 1$.

Solution: We’ll solve this problem in two, very different looking ways. The first is more similar to problem (1.8) and the second uses combinatorics expressions rather than volume approximations. Note that the first method is less accurate and uses the condition $E/(Nhν) \gg 1$ by default.

Method 1 - Volume approximations

We’ll follow steps similar to problem (1.8), so I won’t get too much into the details. The single particle energy is given by $\varepsilon(n)=hvn$, so assuming no interactions, the total energy of the system is given by

For convenience we define $E^* = \frac{E}{hv}$. Thus we want to calculate the number of positive integral solutions ${n_r}$ of the equation

As before, it’s easier to calculate the number of solutions with energy $E^$ or less, denoted $\Sigma(E^)$

This is roughly equal to the “volume” created between the curve $\sum_{r=0}^N n_r = E^$ and the axes $n_1, n_2, \dots, n_N$. This is a simplex (multidimensional triangle) in an $N$ dimensional space, with all edges coming out of the origin has length $E^$. This volume is given by

We’ll also want to know $\ln{\Sigma(E^*)}$

Now we shall define $\Gamma_\Delta (E)$ to be the number of microstates between $E-\frac{1}{2}\Delta$ and $E+\frac{1}{2}\Delta$, it is approximately equal to

The number we’re really interested in is the the logarithm of $\Gamma_\Delta$

The value of the curly brackets is much smaller than $\ln{\Sigma}$ thus we can ignore it. Using Boltzmann’s equation for the entropy

Now, from (1.3.10)

or

This solution, while true, assumes $E/N \gg hv$ by default (since the volume approximation is only true under this approximation), the question also asks about when this limit is not true, so we’ll need to take a different approach (method 2).

Method 2 - Combinatorics

Instead of approximating the number of solution to $\sum_{r=0}^N n_r = E^$ for a given $E^$ with the volume of an $N$ dimensional shape, we can use combinatorics. We have $N$ particles, each particle has some specific energy given by $\varepsilon(n) = hvn$. We can think of this as if there are $N$ boxes (particles), each one can take some integer number of quanta $hv$ from a total of $E$ i.e. we have $E^*=E/hv$ balls and we need to allocate them into $N$ different boxes. This is a simple combinatorics problem and the solution is

From this we can get the entropy by Boltzmann’s relation (also Stirling’s approximation)

We’ll get the temperature from (1.3.10)

or

At the limit $E/N \gg hv$ ($hv N/E \ll 1$) we may expand the denominator as a taylor series $\ln{(1+hvN/E)} \approx hvN/E$ and so the equation becomes

just as before :)

1.9

Problem: Making use of the fact that the entropy $S(N,V,E)$ of a thermodynamic system is an extensive quantity, show that

Not that this result implies: $(-N\mu+PV+E)/T=S$, i.e. $N\mu=E+PV-TS$

Solution: TODO

1.10

Problem: A mole of argon and a mole of helium are contained in vessels of equal volume. If argon is at $300 K$, what should the temperature of helium be so that the two have the same entropy?

Solution: Using equation (1.4.21) we get that the expression for the entropy of the gasses is given by

Making the temperature the subject of the equation we get that $mT$ is some function of $S, N$ and $V$, therefore, if we keep these three variables constant, $mT$ will also stay constant. For our gasses that means that

We can plug in $T_{\text{He}}=300K$ and that the mass ratio between Helium and Argon is $\frac{1}{40}$ we get that

1.11

Problem: Four moles of nitrogen and one mole of oxygen at $P = 1$ atm and $T = 300K$ are mixed together to form air at the same pressure and temperature. Calculate the entropy of mixing per mole of the air formed.

Solution: TODO

1.12

Problem: Show that the various expressions for the entropy of mixing, derived in Sec. 1.5, satisfy the following relations:

• For all $N_1$, $V_1$, $N_2$, and $V_2$, \((\Delta S)_{1=2} = \{(\Delta S) - (\Delta S)^*\} \ge 0\)

• For a given value of $(N_1 + N_2)$, \((\Delta S)^* \le (N_1+N_2)k\ln{2}\)

Solution: TODO

1.13

Problem: If the two gases considered in the mixing process of Section 1.5 were initially at different temperatures, say $T_1$ and $T_2$, what would the entropy of mixing be in that case? Would the contribution arising from this cause depend on whether the two gases were different or identical?

Solution: TODO

1.14

Problem: Show that for an ideal gas composed of monatomic molecules the entropy change, between any two temperatures, when the pressure is kept constant is $5/3$ times the corresponding entropy change when the volume is kept constant. Verify this result numerically by calculating the actual values of $(\Delta S)_P$ and $(\Delta S)_V$ per mole of an ideal gas whose temperature is raised from $300 K$ to $400 K$.

Solution: TODO

1.15

Problem: We have seen that the (P, V)-relationship during a reversible adiabatic process in an ideal gas is governed by the exponent $\gamma$, such that

Consider a mixture of two ideal gases, with mole fractions $f_1$ and $f_2$ and respective exponents $\gamma_1$ and $\gamma_2$. Show that the effective exponent $\gamma$ for the mixture is given by

Solution: TODO

1.16

Problem: Establish thermodynamic formulae

Express the pressure $P$ of an ideal classical gas in terms of the variable $\mu$ and $T$, and verify the above formulae.

Solution: TODO