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This blog consists of my notes and solutions to problems in various areas of mainstream physics. An index to the topics covered is contained in the links in the sidebar on the right, or in the menu at the top of the page.

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# Ocean water as a heat engine

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.4.

A different kind of proposed power plant uses the thermal gradient of the ocean drive a heat engine. If the temperature of the water at the surface of the ocean in some (presumably fairly tropical; it never reaches that temperature here in Scotland) location is ${T_{h}=22^{\circ}\mbox{ C}=295\mbox{ K}}$ and at the bottom is ${T_{c}=4^{\circ}\mbox{ C}=277\mbox{ K}}$, then the maximum efficiency of an engine operating between these two reservoirs is

$\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}}=0.061 \ \ \ \ \ (1)$

This is a pretty dire efficiency. If we want to produce 1 GW of electricity, so the plant is comparable to land-based power plants, how much surface water would we need to process to extract the required heat ${Q_{h}}$? We have

$\displaystyle e_{max}=\frac{W}{Q_{h}} \ \ \ \ \ (2)$

The amount of heat to be extracted per second is therefore

$\displaystyle Q_{h}=\frac{10^{9}}{0.061}=1.64\times10^{10}\mbox{ J s}^{-1} \ \ \ \ \ (3)$

The heat expelled to the cold reservoir is

$\displaystyle Q_{c}=Q_{h}-W=1.54\times10^{10}\mbox{ J s}^{-1} \ \ \ \ \ (4)$

To find the minimum volume of water that must be processed per second, we can try the following argument. [Disclaimer: I’m not 100% convinced this argument is correct, but it seems to make some sense. Comments welcome.]

The amount of water needed clearly depends on how much of a temperature drop the extraction of heat ${Q_{h}}$ produces; the smaller the volume of water, the larger the temperature drop produced. The same is true (in reverse) when expelling heat ${Q_{c}}$ into the cold layer; the smaller the volume of water, the larger the temperature increase.

Now, we can’t decrease the temperature of the upper layer to a final temperature ${T_{f}}$ that is lower than the final temperature of the cold layer, so the maximum temperature drop of the upper layer is to a value of ${T_{f}}$ that is equal to the final temperature of the lower layer after ${Q_{c}}$ is dumped into it. If we take the masses (I’ll use mass instead of volume, since heat capacities are expressed in terms of mass) of water processed to be the same in both layers, then we have

 $\displaystyle T_{h}-T_{f}$ $\displaystyle =$ $\displaystyle \frac{Q_{h}}{mc}\ \ \ \ \ (5)$ $\displaystyle T_{f}-T_{c}$ $\displaystyle =$ $\displaystyle \frac{Q_{c}}{mc} \ \ \ \ \ (6)$

where ${m}$ is the mass of water processed in each layer and ${c}$ is the specific heat capacity. We can add these two equations to get

 $\displaystyle T_{h}-T_{c}$ $\displaystyle =$ $\displaystyle \frac{Q_{h}+Q_{c}}{mc}\ \ \ \ \ (7)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{Q_{h}+Q_{c}}{c\left(T_{h}-T_{c}\right)} \ \ \ \ \ (8)$

Plugging in the numbers we get

$\displaystyle m=\frac{\left(1.64+1.54\right)\times10^{10}}{4181\times18}=4.23\times10^{5}\mbox{ kg} \ \ \ \ \ (9)$

This corresponds to a volume (since ${1\mbox{ m}^{3}}$ of water has a mass of 1000 kg) of

$\displaystyle V=423\mbox{ m}^{3} \ \ \ \ \ (10)$

The total volume of water processed (hot + cold) is twice this, or

$\displaystyle V_{total}=846\mbox{ m}^{3} \ \ \ \ \ (11)$

# Power plants as heat engines

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 4.2 – 4.3.

Most power generating plants are versions of heat engines, in that some fuel (coal, gas or nuclear fission) is used to heat up a reservoir to a temperature ${T_{h}}$ from which the generators extract an amount of heat ${Q_{h}}$, converting an amount ${W}$ into electric power and expelling the remainder ${Q_{c}=Q_{h}-W}$ as waste heat into a cold reservoir at temperature ${T_{c}}$.

The efficiency ${e}$ of the heat engine is the fraction of ${Q_{h}}$ that is converted into work, so

$\displaystyle e=\frac{W}{Q_{h}}=\frac{Q_{h}-Q_{c}}{Q_{h}}=1-\frac{Q_{c}}{Q_{h}} \ \ \ \ \ (1)$

Example 1 Suppose a plant produces 1 GW (${10^{9}\mbox{ W})}$ of electricity by taking in steam at a temperature of ${T_{h}=500^{\circ}\mbox{ C}=773\mbox{ K}}$ and expelling waste heat into the surrounding environment at a temperature of ${T_{c}=20^{\circ}\mbox{ C}=293\mbox{ K}}$. The maximum possible efficiency for an engine operating between these two temperatures is

$\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}}=0.620 \ \ \ \ \ (2)$

If ${T_{h}}$ could be raised to ${600^{\circ}\mbox{ C}=873\mbox{ K}}$, the maximum efficiency increases to

$\displaystyle e_{max}=0.664 \ \ \ \ \ (3)$

To work out how much extra electricity this allows the plant to generate, we need to know the actual efficiency of the plant, which isn’t given in the question, so presumably we are meant to assume that the plant is operating at maximum efficiency (which isn’t realistic, since that would require a Carnot cycle, which would be far too slow to generate electricity). At any rate, we’ll assume that this is the case, and as we’re told that the amount of fuel consumed by the plant doesn’t change, we can assume that the amount of heat absorbed is unchanged as well.

When the plant is operating at ${T_{h}=500^{\circ}\mbox{ C}}$, the amount of heat absorbed in a year is

$\displaystyle Q_{h}=\frac{W}{e}=\frac{10^{9}\times\left(3600\times24\times365\right)}{0.62}=5.086\times10^{16}\mbox{ J} \ \ \ \ \ (4)$

With the same ${Q_{h}}$ at the higher efficiency, the amount of electricity generated is

$\displaystyle W=eQ_{h}=0.664\times5.086\times10^{16}=3.377\times10^{16}\mbox{ J} \ \ \ \ \ (5)$

The power generating is therefore

$\displaystyle P=\frac{3.377\times10^{16}}{3600\times24\times365}=1.071\times10^{9}\mbox{ W} \ \ \ \ \ (6)$

The extra power is therefore 71000 kilowatts. To see how much more money can be made per year from this, we can take a typical electricity rate here in the UK of around 10p per kwh (around 15 cents in the US; Schroeder’s price of 5 cents has been overtaken by inflation).

$\displaystyle 71\times10^{3}\times24\times365\times\left(0.10\right)=62\times10^{6} \ \ \ \ \ (7)$

The extra income is around £62 million.

Example 2 Now we have a more realistic 1 GW power plant that operates at an efficiency of 40%. It absorbs heat at a rate of (I’ll use the same symbols as above to represent the rates of heat and work (with units of watts) rather than the actual amounts of heat and work, to avoid introducing a bunch of new symbols):

$\displaystyle Q_{h}=\frac{W}{e}=\frac{10^{9}}{0.4}=2.5\times10^{9}\mbox{ W} \ \ \ \ \ (8)$

The rate at which it expels heat into the cold reservoir is

$\displaystyle Q_{c}=Q_{h}-W=1.5\times10^{9}\mbox{ W} \ \ \ \ \ (9)$

If this waste heat is expelled into a river with a flow rate of ${100\mbox{ m}^{3}\mbox{s}^{-1}}$, we can work out how much the temperature of the river increases as a result. The specific heat capacity of water is ${c=4181\mbox{ J kg}^{-1}\mbox{K}^{-1}}$ and in one second, ${100\mbox{ m}^{3}}$ of water flows past the plant, absorbing ${1.5\times10^{9}\mbox{ J}}$ of energy. Each kilogram of water therefore absorbs

$\displaystyle Q=\frac{1.5\times10^{9}}{100\mbox{ m}^{3}\times1000\mbox{ kg m}^{-3}}=1.5\times10^{4}\mbox{ J kg}^{-1} \ \ \ \ \ (10)$

The temperature of the river increases by

$\displaystyle \Delta T=\frac{Q}{c}=3.6\mbox{ K} \ \ \ \ \ (11)$

An alternative method of cooling the plant is by evaporation of the water. Assuming that the river is initially at ${293\mbox{ K}}$, we need a quantity of water so that when it is heated to ${100^{\circ}\mbox{ C}=373\mbox{ K}}$ and then vaporized, this takes ${1.5\times10^{9}\mbox{ J s}^{-1}}$. For one kilogram of water, the latent heat of vaporization is ${2.258\times10^{6}\mbox{ J}}$, so if we vaporize a mass ${m}$ of water starting at 293 K, this takes an amount of heat given by

 $\displaystyle Q$ $\displaystyle =$ $\displaystyle 4181m\left(373-293\right)+2.258\times10^{6}m\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.592\times10^{6}m\mbox{ J kg}^{-1} \ \ \ \ \ (13)$

The first term is the heat required to heat the water to its boiling point and the second term is the heat required to vaporize it at the boiling point.

In one second, the mass of water required is therefore

$\displaystyle m=\frac{1.5\times10^{9}}{2.592\times10^{6}}=579\mbox{ kg s}^{-1} \ \ \ \ \ (14)$

This represents a fraction of the total flow of the river of:

$\displaystyle \frac{579}{100\mbox{ m}^{3}\times1000\mbox{ kg m}^{-3}}=0.00579 \ \ \ \ \ (15)$

# Carnot cycle

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.5.

An ideal heat engine absorbs an amount ${Q_{h}}$ from a hot reservoir at temperature ${T_{h}}$, converts an amount ${W}$ of this heat to work and expels the remaining heat ${Q_{c}}$ to a cold reservoir at tempearture ${T_{c}}$. From the first and second laws of thermodynamics, we can show that the maximum efficiency of such a heat engine is

$\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (1)$

In practice, real engines achieve considerably less than this efficiency, but is it possible to construct an engine that does achieve the maximum? It is, in principle, possible, and the path that such an engine must follow on a ${PV}$ diagram is called the Carnot cycle.

The idea is to minimize the entropy generated at each stage. If the gas absorbs an amount ${Q_{h}}$ from the hot reservoir, the entropy of the reservoir decreases by ${Q_{h}/T_{h}}$ and the entropy of the gas increases by ${Q_{h}/T_{gas}}$. If ${T_{gas}, then the net entropy generated is greater than zero. To minimize this, we must therefore make ${T_{gas}}$ as close to ${T_{h}}$ as we can. We can’t make them equal, since then no heat would flow from the reservoir to the gas, so we’ll make ${T_{gas}}$ infinitesimally less than ${T_{h}}$. Of course, this also means that it will take a very long time for the gas to absorb the heat, since such a small temperature difference means a very slow rate of heat flow.

Similarly, at the cold reservoir, to avoid extra entropy we make the temperature of the gas infinitesimally greater than ${T_{c}}$. In between the hot and cold reservoirs, we would like the heat flow to be zero, since this means no entropy flows at all in these stages.

Thus the two stages where the gas is in contact with either the hot or cold reservoir take place at a constant temperature (the temperature of the respective reservoir), so they are isothermal processes. The two stages of the cycle where the gas cools off and then warms up again take place with no heat transfer (so the temperature changes are achieved by changing the pressure and volume). Thus these two stages are adiabatic processes.

On a ${PV}$ diagram, and isothermal curve obeys the relation

$\displaystyle P=\frac{NkT}{V} \ \ \ \ \ (2)$

$\displaystyle P=\frac{K}{V^{\gamma}} \ \ \ \ \ (3)$

where ${K}$ is a constant and ${\gamma=\left(f+2\right)/f}$ where ${f}$ is the number of degrees of freedom of each molecule. For a monatomic ideal gas, ${\gamma=\frac{5}{3}}$. A Carnot cycle looks like this:

[The units are arbitrary. It’s the shape of the curve that’s important.] The gas traverses the cycle in a clockwise direction, with the red ${A\rightarrow B}$ and blue ${C\rightarrow D}$ legs being the isothermals, and the green ${B\rightarrow C}$ and yellow ${D\rightarrow A}$ legs the adiabats.

To prove that the Carnot cycle does in fact achieve the maximum efficiency, we can calculate the heats ${Q_{h}}$ and ${Q_{c}}$. In the leg ${A\rightarrow B}$, the gas is in contact with the hot reservoir and undergoes an isothermal expansion. Since the temperature doesn’t change, the energy of the gas remains constant, so the heat added goes entirely into the work of expansion of the gas. That is

 $\displaystyle Q_{h}$ $\displaystyle =$ $\displaystyle \int_{V_{A}}^{V_{B}}P\;dV\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle NkT_{h}\int_{V_{A}}^{V_{B}}\frac{dV}{V}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle NkT_{h}\ln\frac{V_{B}}{V_{A}} \ \ \ \ \ (6)$

Similarly, the heat expelled to the cold reservoir along the leg ${C\rightarrow D}$ is

$\displaystyle Q_{c}=NkT_{c}\ln\frac{V_{C}}{V_{D}} \ \ \ \ \ (7)$

[Note that ${Q_{h}}$ and ${Q_{c}}$ flow in opposite directions relative to the gas, although they both have the same sign.]

From 3 we have for the ${B\rightarrow C}$ leg

$\displaystyle P_{B}V_{B}^{\gamma}=P_{C}V_{C}^{\gamma} \ \ \ \ \ (8)$

However, at point ${B}$, the adiabat joins the isothermal curve, so

$\displaystyle P_{B}=\frac{NkT_{B}}{V_{B}} \ \ \ \ \ (9)$

and similarly for point ${C}$, so

 $\displaystyle NkT_{B}V_{B}^{\gamma-1}$ $\displaystyle =$ $\displaystyle NkT_{C}V_{C}^{\gamma-1}\ \ \ \ \ (10)$ $\displaystyle \frac{T_{B}}{T_{C}}$ $\displaystyle =$ $\displaystyle \left(\frac{V_{C}}{V_{B}}\right)^{\gamma-1} \ \ \ \ \ (11)$

Applying the same logic to the ${D\rightarrow A}$ leg, and using ${T_{A}=T_{B}}$ and ${T_{C}=T_{D}}$ we get

$\displaystyle \frac{T_{A}}{T_{D}}=\frac{T_{B}}{T_{C}}=\left(\frac{V_{D}}{V_{A}}\right)^{\gamma-1} \ \ \ \ \ (12)$

Comparing the last two equations we find that

$\displaystyle \frac{V_{D}}{V_{A}}=\frac{V_{C}}{V_{B}} \ \ \ \ \ (13)$

or, rearranging:

$\displaystyle \frac{V_{B}}{V_{A}}=\frac{V_{C}}{V_{D}} \ \ \ \ \ (14)$

Inserting this into 6 and 7 and taking the ratio we get

$\displaystyle \frac{Q_{c}}{Q_{h}}=\frac{NkT_{c}\ln\frac{V_{C}}{V_{D}}}{NkT_{h}\ln\frac{V_{B}}{V_{A}}}=\frac{T_{c}}{T_{h}} \ \ \ \ \ (15)$

Thus the efficiency of a Carnot cycle is

$\displaystyle e=1-\frac{Q_{c}}{Q_{h}}=1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (16)$

QED.

# Heat engines

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.1.

A heat engine is a system that absorbs heat and converts part of that heat into work, which can be extracted to perform other tasks. In practice, all heat engines absorb a certain amount of heat and can convert only part of that heat into work, with the remainder being expelled back into the environment. An idealized heat engine absorbs an amount of heat ${Q_{h}}$ from a thermal reservoir at constant temperature ${T_{h}}$, converts an amount ${W}$ of this heat into work, and expels the remainder ${Q_{c}}$ into another thermal reservoir at a constant temperature ${T_{c}}$. The efficiency ${e}$ of the heat engine is the fraction of ${Q_{h}}$ that is converted into work, so

$\displaystyle e=\frac{W}{Q_{h}}=\frac{Q_{h}-Q_{c}}{Q_{h}}=1-\frac{Q_{c}}{Q_{h}} \ \ \ \ \ (1)$

Since the absorbed heat ${Q_{h}}$ is equal to the sum of the work done and the heat expelled, the heat engine itself returns back to its original state at the end of the process. By driving the engine in a cycle of such processes, a continuously operating engine can be obtained.

At this point, it might seem that by merely arranging for no heat to be expelled so that ${Q_{c}=0}$, we could develop a perfectly efficient engine that converts all its input heat into work. However, the second law of thermodynamics throws a spanner into the works, since it requires that entropy increases in any process that involves heat transfer. From the macroscopic definition of entropy, the amount of entropy absorbed from the hot reservoir is

$\displaystyle S_{h}=-\frac{Q_{h}}{T_{h}} \ \ \ \ \ (2)$

(I’ve written it with a minus sign, since the entropy of the hot reservoir decreases.) The amount expelled into the cold reservoir is

$\displaystyle S_{c}=\frac{Q_{c}}{T_{c}} \ \ \ \ \ (3)$

The overall change in entropy must be positive, so

$\displaystyle S_{h}+S_{c}=\frac{Q_{c}}{T_{c}}-\frac{Q_{h}}{T_{h}}>0 \ \ \ \ \ (4)$

In other words

$\displaystyle \frac{Q_{c}}{Q_{h}}>\frac{T_{c}}{T_{h}} \ \ \ \ \ (5)$

Thus the only way ${Q_{c}}$ can be zero (to get 100% efficiency) is if the cold reservoir is at absolute zero. In fact, the best we can do is to get an efficiency of

$\displaystyle e\le1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (6)$

As an example, suppose we look again at the monatomic ideal gas that is driven around a rectangular cycle on a ${PV}$ diagram. To review, the system is as follows. The path is a rectangle starting at ${P_{1}}$ and ${V_{1}}$. On side A, the pressure is increased to ${P_{2}}$ while the volume is held constant. Then on side B, the volume is increased to ${V_{2}}$ while the pressure is constant at ${P_{2}}$. On side C, the pressure is decreased back to ${P_{1}}$ with the volume constant at ${V_{2}}$. Finally, on side D, the volume is decreased back to ${V_{1}}$ with the pressure constant at ${P_{1}}$. From the table in the earlier post, we see that heat is added to the gas along sides A and B, and removed from the gas along sides C and D. The net work done by the gas is given by

$\displaystyle W=Q_{A}+Q_{B}+Q_{C}+Q_{D} \ \ \ \ \ (7)$

where ${Q_{C}<0}$ and ${Q_{D}<0}$ to indicate that heat is leaving the gas. We can calculate the efficiency of this cycle if it is used as a heat engine:

 $\displaystyle e$ $\displaystyle =$ $\displaystyle 1-\frac{Q_{expelled}}{Q_{absorbed}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\frac{-Q_{C}-Q_{D}}{Q_{A}+Q_{B}} \ \ \ \ \ (9)$

For the particular case ${P_{2}=2P_{1}}$ and ${V_{2}=3V_{1}}$ given in the problem, and reading the formulas for the heats off the table in the previous post, we get

 $\displaystyle Q_{A}$ $\displaystyle =$ $\displaystyle \frac{5}{2}V_{1}\left(P_{2}-P_{1}\right)=2.5P_{1}V_{1}\ \ \ \ \ (10)$ $\displaystyle Q_{B}$ $\displaystyle =$ $\displaystyle \frac{7}{2}P_{2}\left(V_{2}-V_{1}\right)=14P_{1}V_{1}\ \ \ \ \ (11)$ $\displaystyle -Q_{C}$ $\displaystyle =$ $\displaystyle \frac{5}{2}V_{2}\left(P_{2}-P_{1}\right)=7.5P_{1}V_{1}\ \ \ \ \ (12)$ $\displaystyle -Q_{D}$ $\displaystyle =$ $\displaystyle \frac{7}{2}P_{1}\left(V_{2}-V_{1}\right)=7P_{1}V_{1} \ \ \ \ \ (13)$

[The sign conventions can be a bit confusing, since Schroeder refers to all heat and work quantities as positive, whereas in the earlier chapters a distinction was made between heat absorbed by the gas and emitted by the gas.]

The efficiency is therefore

$\displaystyle e=1-\frac{14.5}{16.5}=0.12 \ \ \ \ \ (14)$

This isn’t a very efficient cycle for a heat engine.

In the course of the cycle, the gas varies between a low temperature of ${T_{c}=P_{1}V_{1}/Nk}$ and a high temperature of ${T_{h}=P_{2}V_{2}/Nk=6T_{c}}$. Thus the theoretical maximum efficiency is

$\displaystyle e_{max}=1-\frac{1}{6}=0.83 \ \ \ \ \ (15)$

# Thermodynamic properties of a 2-dim ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.39.

We now revisit the 2-d ideal gas for which the Sackur-Tetrode equation is

$\displaystyle S=Nk\left[\ln\frac{2\pi mAU}{\left(hN\right)^{2}}+2\right] \ \ \ \ \ (1)$

where ${A}$ is the area occupied by the gas, ${N}$ is the number of molecules, each of mass ${m}$, and ${U}$ is the total energy. We can work out the temperature, pressure and chemical potential by applying the thermodynamic identity adapted for 2 dimensions (by replacing the volume ${V}$ by the area ${A}$):

$\displaystyle dU=TdS-PdA+\mu dN \ \ \ \ \ (2)$

The temperature is determined from the entropy as

 $\displaystyle \frac{1}{T}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial S}{\partial U}\right)_{A,N}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Nk\frac{\left(hN\right)^{2}}{2\pi mAU}\frac{2\pi mA}{\left(hN\right)^{2}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Nk}{U} \ \ \ \ \ (5)$

This just gives us the formula from the equipartition theorem for a system with 2 degrees of freedom:

$\displaystyle U=\frac{2}{2}NkT=NkT \ \ \ \ \ (6)$

The pressure can be obtained from

 $\displaystyle P$ $\displaystyle =$ $\displaystyle T\left(\frac{\partial S}{\partial A}\right)_{U,N}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Nk\frac{\left(hN\right)^{2}}{2\pi mAU}\frac{2\pi mU}{\left(hN\right)^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{NkT}{A} \ \ \ \ \ (9)$

This is just the 2-dim analogue of the ideal gas law:

$\displaystyle PA=NkT \ \ \ \ \ (10)$

Finally, chemical potential is defined in terms of the entropy as

 $\displaystyle \mu$ $\displaystyle \equiv$ $\displaystyle -T\left(\frac{\partial S}{\partial N}\right)_{U,A}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\left[\ln\frac{2\pi mAU}{\left(hN\right)^{2}}+2\right]-NkT\left(-\frac{2}{N}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\ln\frac{2\pi mAU}{\left(hN\right)^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\ln\left(\frac{A}{N}\frac{2\pi mkT}{h^{2}}\right) \ \ \ \ \ (14)$

We can compare this to the chemical potential for a 3-d ideal gas

$\displaystyle \mu=-kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right] \ \ \ \ \ (15)$

The only differences are the replacement of ${V}$ by ${A}$ and the change in the exponent inside the logarithm from ${\frac{3}{2}}$ to 1. The latter arises from the derivation of the multiplicity, where the exponent depends on the number of degrees of freedom in the system. For a 3-d gas, there are ${3N}$ degrees of freedom, while for a 2-d gas, there are ${2N}$. Thus the exponent in the 2-d case is ${\frac{2}{3}}$ that in the 3-d case. [You’d need to follow through the derivation in detail to see the difference, but basically that’s where it comes from.]

# Chemical potential of a mixture of ideal gases

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.38.

The chemical potential is defined in terms of the entropy as

$\displaystyle \mu\equiv-T\left(\frac{\partial S}{\partial N}\right)_{U,V} \ \ \ \ \ (1)$

This definition leads to a general thermodynamic identity

$\displaystyle dU=TdS-PdV+\mu dN \ \ \ \ \ (2)$

For a mixture of ideal gases, each species ${i}$ constitutes a molar fraction ${x_{i}}$ of the total number ${N_{total}}$ of molecules, so each species has its own chemical potential defined as

$\displaystyle \mu_{i}=-T\left(\frac{\partial S}{\partial N_{i}}\right)_{U,V,N_{j\ne i}} \ \ \ \ \ (3)$

where all ${N_{j}}$ with ${j\ne i}$ are held constant in the derivative.

Also, for an ideal gas, each species contributes its own portion of the overall entropy, independently of the other species. We can see this by noting that if we have a mixture of, say, 2 gases, then for each configuration of the gas ${A}$ molecules there is a multiplicity of ${\Omega_{B}}$ of the gas ${B}$ molecules and since for an ideal gas, the molecules don’t interact, the total multiplicity of the mixture is ${\Omega_{total}=\Omega_{A}\Omega_{B}}$, so the entropy is the sum of the entropies for the separate species: ${S_{total}=S_{A}+S_{B}}$.

Since ideal gas molecules don’t interact, species ${i}$ contributes a fraction ${x_{i}}$ of the total pressure, or in other words, its partial pressure is

$\displaystyle P_{i}=x_{i}P \ \ \ \ \ (4)$

We can therefore write the thermodynamic identity for a mixture of ideal gases as

$\displaystyle dU=T\sum_{i}dS_{i}-\left(\sum_{i}P_{i}\right)dV+\sum_{i}\mu_{i}dN_{i} \ \ \ \ \ (5)$

Since ${dS_{j\ne i}=0}$ in 3 (because only the number ${N_{i}}$ of species ${i}$ is changing, and no properties of any of the other species are changing), we can write the chemical potential of species ${i}$ as

$\displaystyle \mu_{i}=-T\left(\frac{\partial S_{i}}{\partial N_{i}}\right)_{U,V,N_{j\ne i}} \ \ \ \ \ (6)$

But this is the definition of chemical potential in a system containing only species ${i}$ at partial pressure ${P_{i}}$ in volume ${V}$. Thus, for a mixture of ideal gases, the chemical potential of each species is independent of the other species. In a mixture of real gases, however, this is probably not the case, since interactions between the species means the total entropy isn’t a simple sum of the entropies of the individual species.

# Chemical potential of an ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.37.

The chemical potential is defined in terms of the entropy as

$\displaystyle \mu\equiv-T\left(\frac{\partial S}{\partial N}\right)_{U,V} \ \ \ \ \ (1)$

This definition leads to a general thermodynamic identity

$\displaystyle dU=TdS-PdV+\mu dN \ \ \ \ \ (2)$

We can get another formula for ${\mu}$ from this:

$\displaystyle \mu=\left(\frac{\partial U}{\partial N}\right)_{S,V} \ \ \ \ \ (3)$

We can apply these formulas to get the chemical potential of an ideal gas from the Sackur-Tetrode equation

$\displaystyle S=Nk\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right] \ \ \ \ \ (4)$

Schroeder applies 1 to this equation and gets the chemical potential as

$\displaystyle \mu=-kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right] \ \ \ \ \ (5)$

Now consider a volume of gas a distance ${z}$ above the Earth’s surface. In this case, the energy of the gas is composed of both the kinetic energy of the moving molecules and the potential energy of the molecules, the latter of which is ${mgz}$ per molecule. In order to find the chemical potential of this volume of gas, we need to modify 4 to write ${U}$ in terms of the potential and kinetic energy.

To see how to do this, we need to review the derivation of the multiplicity of an ideal gas (Schroeder’s equation 2.40). This derivation relied on arguments from quantum mechanics that gave the number of position and momentum states that are possible in a volume of gas. The energy ${U}$ that appeared in this derivation is entirely kinetic energy. Merely raising the volume of gas a distance above the Earth’s surface doesn’t change the number of possible states that gas can occupy; it merely shifts the total energy by a fixed amount ${Nmgz}$. Thus the ${U}$ in 4 should really be written as ${U_{K}}$, the kinetic energy:

$\displaystyle S=Nk\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU_{K}}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right] \ \ \ \ \ (6)$

However, the ${U}$ being held constant in 1 is the total energy, which is

$\displaystyle U=U_{K}+Nmgz=\frac{3}{2}NkT+Nmgz \ \ \ \ \ (7)$

so to write 4 in terms of this constant energy and the potential energy, we have

 $\displaystyle S$ $\displaystyle =$ $\displaystyle Nk\left[\ln\left(\frac{V}{N}\left(\frac{4\pi m\left(U-Nmgz\right)}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Nk\left[\ln\left(V\left(\frac{4\pi m\left(U-Nmgz\right)}{3h}\right)^{3/2}\right)-\ln N^{5/2}+\frac{5}{2}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Nk\left[\ln\left(\alpha\left(U-Nmgz\right)^{3/2}\right)-\ln N^{5/2}+\frac{5}{2}\right] \ \ \ \ \ (10)$

where

$\displaystyle \alpha\equiv V\left(\frac{4\pi m}{3h}\right)^{3/2} \ \ \ \ \ (11)$

We can now find the chemical potential by taking the derivative 1:

 $\displaystyle \mu$ $\displaystyle =$ $\displaystyle -kT\left[\ln\left(\alpha\left(U-Nmgz\right)^{3/2}\right)-\ln N^{5/2}+\frac{5}{2}\right]-\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle NkT\left[\frac{\alpha\frac{3}{2}\left(U-Nmgz\right)^{1/2}\left(-mgz\right)}{\alpha\left(U-Nmgz\right)^{3/2}}-\frac{1}{N^{5/2}}\left(\frac{5}{2}N^{3/2}\right)\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\left[\ln\left(\alpha\left(U-Nmgz\right)^{3/2}\right)-\ln N^{5/2}\right]+NkT\left[\frac{3mgz}{2\left(U-Nmgz\right)}\right] \ \ \ \ \ (14)$

We can now substitute ${\alpha}$ back in and use 7 to get

 $\displaystyle \mu$ $\displaystyle =$ $\displaystyle -kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]+\frac{\frac{3}{2}NkT}{\frac{3}{2}NkT}mgz\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]+mgz \ \ \ \ \ (16)$

Thus the chemical potential is changed by exactly the potential energy of a single molecule.

Now suppose that we have two samples of the same ideal gas with equal volumes and temperatures, but with one at height ${z}$ and the other at the Earth’s surface so ${z=0}$. If the two volumes are in diffusive equilibrium, then their chemical potentials are equal, so we have

 $\displaystyle \mu\left(z\right)$ $\displaystyle =$ $\displaystyle \mu\left(0\right)\ \ \ \ \ (17)$ $\displaystyle -kT\ln\left[\frac{V}{N\left(z\right)}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]+mgz$ $\displaystyle =$ $\displaystyle -kT\ln\left[\frac{V}{N\left(0\right)}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right] \ \ \ \ \ (18)$

Dividing both sides by ${-kT}$ and exponentiating, we get

 $\displaystyle \frac{V}{N\left(z\right)}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}e^{-mgz/kT}$ $\displaystyle =$ $\displaystyle \frac{V}{N\left(0\right)}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\ \ \ \ \ (19)$ $\displaystyle N\left(z\right)$ $\displaystyle =$ $\displaystyle N\left(0\right)e^{-mgz/kT} \ \ \ \ \ (20)$

Since the volumes and temperatures are equal, from the ideal gas law ${PV=NkT}$ this equation implies the same relation applies for the pressure as a function of height

$\displaystyle P\left(z\right)=P\left(0\right)e^{-mgz/kT} \ \ \ \ \ (21)$

which is the barometric equation we obtained earlier.

# Chemical potential; application to the Einstein solid

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.35 – 3.36.

Thermodynamic systems can be in equilibrium in various ways: thermal equilibrium results from systems being able to exchange energy, resulting in them being at the same temperature and mechanical equilibrium results from being able to exchange volume, resulting in the pressures being equal. The final type of equilibrium is diffusive equilibrium, when systems can exchange actual matter (numbers of particles) with each other. Now the entropy is taken to be a function of energy ${U}$, volume ${V}$ and particle number ${N}$ and using the same logic as in deriving temperature and pressure from derivatives of entropy, we find that at diffusive equilibrium between two systems ${A}$ and ${B}$ with a constant total particle number ${N=N_{A}+N_{B}}$, the condition that entropy achieve its maximum value results in

$\displaystyle \frac{\partial S_{A}}{\partial N_{A}}=\frac{\partial S_{B}}{\partial N_{B}} \ \ \ \ \ (1)$

This condition is used to define the chemical potential ${\mu}$ as

$\displaystyle \mu\equiv-T\left(\frac{\partial S}{\partial N}\right)_{U,V} \ \ \ \ \ (2)$

If the two systems are also in thermal equilibrium, the temperatures are equal, so at equilibrium

$\displaystyle \mu_{A}=\mu_{B} \ \ \ \ \ (3)$

If the systems are not in equilibrium, then the tendency is for the overall entropy of the combined system to increase as it tends towards equilibrium. If ${\frac{\partial S_{A}}{\partial N_{A}}>\frac{\partial S_{B}}{\partial N_{B}}}$, then an increase in ${N_{A}}$ results in a greater increase in entropy than an increase in ${N_{B}}$, so the diffusion will tend to transfer particles from ${B}$ to ${A}$. From the definition of ${\mu}$, a larger ${\frac{\partial S}{\partial N}}$ means a lower value of ${\mu}$ (due to the minus sign), so diffusion tends to transfer particles from the system with a higher chemical potential to the system with a lower chemical potential.

If a system is allowed to vary ${U}$, ${V}$ and ${N}$, the overall change in entropy is the sum of the contributions from all three processes, so the generalized form of the thermodynamic identity is

$\displaystyle dS=\left(\frac{\partial S}{\partial U}\right)_{N,V}dU+\left(\frac{\partial S}{\partial V}\right)_{U,N}dV+\left(\frac{\partial S}{\partial N}\right)_{U,V}dN \ \ \ \ \ (4)$

or, in its more usual form

$\displaystyle dU=TdS-PdV+\mu dN \ \ \ \ \ (5)$

Example 1 Schroeder does an example of a very small Einstein solid containing ${N=3}$ oscillators and ${q=3}$ energy quanta. Although true derivatives aren’t valid in such a small system, we can get an idea of how chemical potential works by considering what happens if we add another oscillator to the system in such a way that ${S}$ and ${V}$ don’t change. The entropy before the addition is

 $\displaystyle S$ $\displaystyle =$ $\displaystyle k\ln\Omega\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\ln\binom{3+3-1}{3}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\ln10 \ \ \ \ \ (8)$

If we change ${N}$ to 4, then to keep ${S}$ constant, we need to decrease ${q}$. This example is contrived so that we can actually do this and get the same value for ${S}$, since with ${N=4}$ and ${q=2}$, we find ${S=k\ln10}$. Thus in this case ${\Delta U=-\epsilon}$ where ${\epsilon}$ is the energy of a single quantum and so the chemical potential is (approximately)

$\displaystyle \mu=\frac{\Delta U}{\Delta N}=\frac{-\epsilon}{1}=-\epsilon \ \ \ \ \ (9)$

Now suppose we started with ${N=3}$ and ${q=4}$, and then try to add another oscillator while keeping ${S}$ constant. The entropy before the addition is

$\displaystyle S=k\ln\binom{3+4-1}{4}=k\ln15 \ \ \ \ \ (10)$

Reducing ${q}$ to 3 after increasing ${N}$ to 4 results in

$\displaystyle S=k\ln\binom{4+3-1}{3}=k\ln20 \ \ \ \ \ (11)$

so we’re still not down to the original entropy. However, if we reduce ${q}$ to 2, we get

$\displaystyle S=k\ln\binom{4+2-1}{2}=k\ln10 \ \ \ \ \ (12)$

so now we’ve dropped below the original entropy. To keep ${S}$ constant, we’d need to remove somewhere around 1.5 quanta, so ${\mu<-\epsilon}$ and the chemical potential is lower (more negative) than in the first case.

Example 2 Still with an Einstein solid, but now at the other extreme where both ${q}$ and ${N}$ are large numbers. In this case, the multiplicity is approximately

 $\displaystyle \Omega$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{N}{2\pi q\left(q+N\right)}}\left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N} \ \ \ \ \ (14)$

where we’ve dropped the square root as it is merely ‘large’ compared to the other two factors being ‘very large’.

The entropy is therefore

 $\displaystyle S$ $\displaystyle =$ $\displaystyle k\ln\Omega\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \left(q+N\right)\ln\left(q+N\right)-q\ln q-N\ln N \ \ \ \ \ (16)$

Using 2, this gives a chemical potential of

 $\displaystyle \mu$ $\displaystyle =$ $\displaystyle -kT\left[\ln\left(q+N\right)+1-\ln N-1\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -kT\ln\frac{q+N}{N} \ \ \ \ \ (18)$

For ${N\gg q}$, this reduces to

$\displaystyle \mu\rightarrow-kT\ln\left(1+\frac{q}{N}\right)\approx-kT\frac{q}{N} \ \ \ \ \ (19)$

At the other extreme, ${N\ll q}$ and

$\displaystyle \mu\rightarrow-kT\ln\left(\frac{q}{N}\right)\rightarrow-\infty \ \ \ \ \ (20)$

In the ${N\gg q}$ case, there are many more oscillators than energy quanta to put in them, so adding an extra oscillator won’t make much difference to the multiplicity. Think of a simple case where you’ve got lots of bins and only one ball to put in them. In that case, adding an extra bin creates only one extra possible state. Thus we’d expect ${\partial S/\partial N}$ to be fairly small in this case.

In the ${N\ll q}$ case, there are many more quanta than oscillators to put them in, so adding an extra oscillator creates many more possible microstates, since we can place any number of quanta from 0 right up to ${q}$ in the new oscillator. Thus the multiplicity, and hence the entropy, increases more rapidly in this case.

# Rubber bands and entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.34.

We can model a rubber band as a one-dimensional chain of ${N}$ polymer links each of length ${\ell}$, where each link can point either to the left or right. The total length ${L}$ of the rubber band is therefore

$\displaystyle L=\left(N_{R}-N_{L}\right)\ell=\left(2N_{R}-N\right)\ell \ \ \ \ \ (1)$

where ${N_{R,L}}$ is the number of links pointing to the right or left, so that ${N=N_{R}+N_{L}}$. We’ll assume that ${N_{R}>N_{L}}$, although the converse is just the mirror image and gives the same behaviour.

The entropy can be found by noting that the system is equivalent to a coin-flipping experiment, so for a given ${N_{R}}$, the multiplicity is

$\displaystyle \Omega=\binom{N}{N_{R}}=\frac{N!}{N_{R}!\left(N-N_{R}\right)!} \ \ \ \ \ (2)$

Using Stirling’s approximation, the entropy is

$\displaystyle S=k\left[N\ln N-N_{R}\ln N_{R}-\left(N-N_{R}\right)\ln\left(N-N_{R}\right)\right] \ \ \ \ \ (3)$

Since this is a one-dimensional system, the role of the pressure in a 3-d system is taken here by the tension force ${F}$ generated by stretching the rubber band. If ${F>0}$ when the band is pulling inward, then work ${F\cdot dL}$ is done on the band when it is stretched through a distance ${dL>0}$.

The thermodynamic identity for this system is therefore

$\displaystyle dU=TdS+FdL \ \ \ \ \ (4)$

If the band is stretched in such a way that its energy remains constant (e.g. by losing heat), then ${dU=0}$ and the force is given by

$\displaystyle F=-T\left(\frac{\partial S}{\partial L}\right)_{U} \ \ \ \ \ (5)$

From 1

$\displaystyle dL=2\ell dN_{R} \ \ \ \ \ (6)$

so from 3

 $\displaystyle \frac{\partial S}{\partial L}$ $\displaystyle =$ $\displaystyle \frac{1}{2\ell}\frac{\partial S}{\partial N_{R}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\left[-\ln N_{R}-1+\ln\left(N-N_{R}\right)+1\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\ln\frac{N-N_{R}}{N_{R}} \ \ \ \ \ (9)$

From 1

$\displaystyle N_{R}=\frac{1}{2}\left(\frac{L}{\ell}+N\right) \ \ \ \ \ (10)$

so

 $\displaystyle \frac{\partial S}{\partial L}$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\ln\frac{\frac{N}{2}-\frac{L}{2\ell}}{\frac{N}{2}+\frac{L}{2\ell}}=\frac{k}{2\ell}\ln\frac{N\ell-L}{N\ell+L}\ \ \ \ \ (11)$ $\displaystyle F$ $\displaystyle =$ $\displaystyle -\frac{kT}{2\ell}\ln\frac{N\ell-L}{N\ell+L} \ \ \ \ \ (12)$

If ${L\ll N\ell}$, the band is almost fully contracted since its length is much less than the maximum length. In this case we can get an estimate of the force:

 $\displaystyle F$ $\displaystyle =$ $\displaystyle -\frac{kT}{2\ell}\ln\frac{1-L/N\ell}{1+L/N\ell}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{kT}{2\ell}\ln\left[\left(1-L/N\ell\right)^{2}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{kT}{\ell}\ln\left[1-L/N\ell\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{kT}{N\ell^{2}}L \ \ \ \ \ (16)$

That is, ${F\propto L}$ which is Hooke’s law for a spring, with spring constant ${kT/N\ell^{2}}$.

The formula 12 for ${F}$ says that ${F}$ should increase as the temperature is increased, that is, a rubber band should contract when heated (up to a point, obviously; after a while it will just melt). Although this relation was derived in the special case of constant energy ${U}$, it does seem to make sense. A higher temperature would increase the entropy, meaning that ${N_{R}}$ gets closer to its maximum entropy value of ${\frac{N}{2}}$, which means that ${L}$ gets smaller. By the same token, we’d expect a rubber band to get warmer if it is suddenly stretched. I did try the experiment suggested, although I couldn’t find a heavy rubber band, and it did seem to warm up a bit when it was stretched.

# Heat capacities in terms of entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.33.

The thermodynamic identity for an infinitesimal process is

$\displaystyle dU=TdS-PdV \ \ \ \ \ (1)$

For constant volume processes ${dV=0}$, so we can derive the expression for the heat capacity by dividing both sides by ${dT}$:

$\displaystyle C_{V}=\left(\frac{\partial U}{\partial T}\right)_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V} \ \ \ \ \ (2)$

For constant pressure processes, the heat capacity is defined in terms of the enthalpy. The enthalpy is defined as

$\displaystyle H=U+PV \ \ \ \ \ (3)$

It is the energy required to create the system, which is a combination of the internal energy ${U}$ of the system itself, plus the work required to clear the volume ${V}$ that the system occupies. If this work is done at constant pressure, the work is ${PV}$. In a system in which the only work done is from expansion, ${H=Q}$ so the heat capacity at constant pressure is

$\displaystyle C_{P}=\left(\frac{\partial Q}{\partial T}\right)_{P}=\left(\frac{\partial H}{\partial T}\right)_{P} \ \ \ \ \ (4)$

The change in ${H}$ is

 $\displaystyle dH$ $\displaystyle =$ $\displaystyle dU+PdV+VdP\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle TdS-PdV+PdV+VdP\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle TdS+VdP \ \ \ \ \ (7)$

At constant pressure ${dP=0}$ so dividing both sides by ${dT}$ we get

$\displaystyle C_{P}=T\left(\frac{\partial S}{\partial T}\right)_{P} \ \ \ \ \ (8)$