Extensive and intensive quantities

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.21.

The various thermodynamic properties can be classified according to whether they are intensive or extensive. Basically, if we take a system and duplicate it exactly, an extensive quantity will also double, while an intensive quantity will remain unchanged. The intensive quantities include temperature, pressure and any form of density, such as mass density or energy density. Most other quantities are extensive. If we duplicate a system, clearly its volume doubles as does the number of particles. All forms of energy (${U}$, ${H}$, ${F}$ and ${G}$, for example) will double, as will the system’s mass.

The ratio of two extensive quantities produces an intensive quantity, since the common factor that appears when a system is duplicated cancels out in the division. Multiplying two intensive quantities gives another intensive quantity, since the absolute size of the system doesn’t appear in the product. Multiplying an intensive by an extensive quantity produces another extensive quantity, since the absolute size occurs once in the product (in the extensive quantity). You might think that multiplying two extensive quantities gives a quantity that increases according to the product of the sizes of the systems, but in fact such products don’t appear in thermal physics.

For some examples, we’ll look at the following.

The entropy is an extensive quantity. It is defined as

$\displaystyle S=k\ln\Omega \ \ \ \ \ (1)$

where ${\Omega}$ is the number of microstates available to the system. If we duplicate a system then ${\Omega\rightarrow\Omega^{2}}$ since for each microstate in the original system, any of the microstates is available to the duplicate system. Therefore ${S\rightarrow2S}$, so entropy is extensive.

The chemical potential is defined as

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

The derivative is intensive, since it is the ratio of two extensive quantities ${S}$ and ${N}$. ${T}$ is also intensive, so ${\mu}$ is the product of two intensive quantities, making it intensive.

The total heat capacity ${C}$ is extensive, since it is the amount of heat required to raise the entire system by one kelvin. If you double the system, you’ll need twice as much heat. The specific heat capacity ${c}$, however, is intensive, since it gives the amount of heat required to raise a fixed amount of a substance by one kelvin, so it’s independent of the size of the system.

Heat capacities using Maxwell relations

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.13 – 5.15.

We can use the Maxwell relations to derive some formulas relating to heat capacities. First, recall the thermal expansion coefficient

$\displaystyle \beta\equiv\frac{\Delta V/V}{\Delta T} \ \ \ \ \ (1)$

which is the fractional change in volume per Kelvin, assumed to be a constant pressure. For small changes, we can write this as a partial derivative:

$\displaystyle \beta=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (2)$

From the Maxwell relation derived from the Gibbs energy:

$\displaystyle \left(\frac{\partial V}{\partial T}\right)_{P}=-\left(\frac{\partial S}{\partial P}\right)_{T} \ \ \ \ \ (3)$

so

$\displaystyle \beta=-\frac{1}{V}\left(\frac{\partial S}{\partial P}\right)_{T} \ \ \ \ \ (4)$

The third law of thermodynamics says that as ${T\rightarrow0}$, ${S\rightarrow0}$. I’m not quite sure how we can use this to show that ${\beta\rightarrow0}$ as ${T\rightarrow0}$ since it’s not clear how entropy depends on pressure at low temperatures. However, if ${S\rightarrow0}$ for any system as ${T\rightarrow0}$, then presumably it must be true no matter what the pressure is, so in that sense, ${S}$ is independent of pressure and then ${\left(\frac{\partial S}{\partial P}\right)_{T}=0}$. I’m not sure that constitutes a ‘proof’ as requested in Schroeder’s problem, though.

We’re on firmer ground when we wish to derive a relation between the heat capacities ${C_{V}}$ (constant volume) and ${C_{P}}$ (constant pressure). In terms of entropy, they are

 $\displaystyle C_{V}$ $\displaystyle =$ $\displaystyle T\left(\frac{\partial S}{\partial T}\right)_{V}\ \ \ \ \ (5)$ $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle T\left(\frac{\partial S}{\partial T}\right)_{P} \ \ \ \ \ (6)$

If we write ${S=S\left(V,T\right)}$ then

$\displaystyle dS=\left(\frac{\partial S}{\partial V}\right)_{T}dV+\left(\frac{\partial S}{\partial T}\right)_{V}dT \ \ \ \ \ (7)$

Also, starting with ${V=V\left(P,T\right)}$ we have

$\displaystyle dV=\left(\frac{\partial V}{\partial P}\right)_{T}dP+\left(\frac{\partial V}{\partial T}\right)_{P}dT \ \ \ \ \ (8)$

Inserting this into 7 and setting ${dP=0}$ (constant pressure) gives

 $\displaystyle dS$ $\displaystyle =$ $\displaystyle \left[\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial S}{\partial T}\right)_{V}\right]dT\ \ \ \ \ (9)$ $\displaystyle \left(\frac{\partial S}{\partial T}\right)_{P}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial S}{\partial T}\right)_{V}\ \ \ \ \ (10)$ $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (11)$

where we’ve used 5 and 6 in the last line.

From the Helmholtz energy Maxwell relation

$\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T}=\left(\frac{\partial P}{\partial T}\right)_{V} \ \ \ \ \ (12)$

we have

$\displaystyle C_{P}=T\left(\frac{\partial P}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (13)$

Using 2 with its relation to the isothermal compressibility:

$\displaystyle \frac{\beta}{\kappa_{T}}=\left(\frac{\partial P}{\partial T}\right)_{V} \ \ \ \ \ (14)$

we have

$\displaystyle C_{P}=C_{V}+\frac{TV\beta^{2}}{\kappa_{T}} \ \ \ \ \ (15)$

For an ideal gas, ${PV=NkT}$, so 13 becomes

 $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle T\frac{Nk}{P}\frac{Nk}{V}+C_{V}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{NkT}{PV}Nk+C_{V}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle C_{V}+Nk \ \ \ \ \ (18)$

which agrees with Schroeder’s equation 1.48.

From 15, we can see that ${C_{P}>C_{V}}$ provided that the term ${\frac{TV\beta^{2}}{\kappa_{T}}}$ is always positive. The numerator is certainly positive, and from the definition of the isothermal compressibility

$\displaystyle \kappa_{T}\equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T} \ \ \ \ \ (19)$

we see that it is positive provided that an increase in pressure causes a decrease in volume, which is pretty well always true for any realistic material. Since ${\beta\rightarrow0}$ for low temperatures and as we wouldn’t expect ${V}$ or ${\kappa_{T}}$ to change much for very low temperatures (where we’d expect most substances to be solid, or possibly liquid, such as helium), then we’d expect ${C_{P}\approx C_{V}}$ for low temperatures, with ${C_{P}>C_{V}}$ as temperature increases. This agrees with Schroeder’s Figure 1.14.

To put in some numbers, we can use the data from the earlier problem. For water, the values given by Schroeder are (at ${25^{\circ}\mbox{ C}=298\mbox{ K}}$):

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle 2.57\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (20)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle 4.52\times10^{-10}\mbox{ Pa}^{-1} \ \ \ \ \ (21)$

For one mole of water, the volume is ${18.068\times10^{-6}\mbox{ m}^{3}}$, so

 $\displaystyle C_{P}-C_{V}$ $\displaystyle =$ $\displaystyle \frac{\left(298\right)\left(18.068\times10^{-6}\right)\left(2.57\times10^{-4}\right)^{2}}{4.52\times10^{-10}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.787\mbox{ J K}^{-1} \ \ \ \ \ (23)$

${C_{P}=75.29\mbox{ J K}^{-1}}$ for one mole of water so the difference between the heat capacities is around 1% of ${C_{P}}$.

For mercury we have

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle 1.81\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (24)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle 4.04\times10^{-11}\mbox{ Pa}^{-1} \ \ \ \ \ (25)$

One mole of mercury has a volume of ${14.81\times10^{-6}\mbox{ m}^{3}}$ so we get

 $\displaystyle C_{P}-C_{V}$ $\displaystyle =$ $\displaystyle \frac{\left(298\right)\left(14.81\times10^{-6}\right)\left(1.81\times10^{-4}\right)^{2}}{4.04\times10^{-11}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.58\mbox{ J K}^{-1} \ \ \ \ \ (27)$

The heat capacity for one mole of mercury (from the appendix to Schroeder’s book) is ${27.98\mbox{ J K}^{-1}}$ so the difference is around 12.8% of ${C_{P}}$.

Finally, we can derive 15 by starting with ${U}$ and ${H}$ instead of ${S}$ and ${V}$. The heat capacities are defined in terms of internal energy and enthalpy as

 $\displaystyle C_{V}$ $\displaystyle \equiv$ $\displaystyle \left(\frac{\partial U}{\partial T}\right)_{V}\ \ \ \ \ (28)$ $\displaystyle C_{P}$ $\displaystyle \equiv$ $\displaystyle \left(\frac{\partial H}{\partial T}\right)_{P} \ \ \ \ \ (29)$

Writing ${U=U\left(V,T\right)}$ we have

$\displaystyle dU=\left(\frac{\partial U}{\partial V}\right)_{T}dV+\left(\frac{\partial U}{\partial T}\right)_{V}dT \ \ \ \ \ (30)$

The enthalpy is defined as ${H=U+PV}$, so at constant pressure

 $\displaystyle dH$ $\displaystyle =$ $\displaystyle dU+P\;dV\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]dV+\left(\frac{\partial U}{\partial T}\right)_{V}dT \ \ \ \ \ (32)$

Dividing through by ${dT}$ at constant pressure gives

 $\displaystyle \left(\frac{\partial H}{\partial T}\right)_{P}$ $\displaystyle =$ $\displaystyle \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial U}{\partial T}\right)_{V}\ \ \ \ \ (33)$ $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (34)$

The Helmholtz free energy is defined as ${F=U-TS}$ from which we can derive the relation

$\displaystyle P=-\left(\frac{\partial F}{\partial V}\right)_{T} \ \ \ \ \ (35)$

Using this, we have

 $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle \left[\left(\frac{\partial U}{\partial V}\right)_{T}-\left(\frac{\partial F}{\partial V}\right)_{T}\right]\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\partial\left(U-F\right)}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (38)$

This is the same as 11, so from here on the derivation is the same as before, and we get 15 again.

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

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

Entropy of diamond and graphite

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.30 – 3.31.

In a quasistatic process, the relation between entropy, temperature and the heat flow is

$\displaystyle dS=\frac{Q}{T} \ \ \ \ \ (1)$

where ${Q}$ is the (infinitesimal) amount of heat flowing into or out of the system at temperature ${T}$. For a process at constant pressure but changing temperature, ${Q}$ can be written in terms of the heat capacity ${C_{P}}$ at constant pressure, since the amount of heat required to change the temperature by ${dT}$ is ${C_{P}dT}$. In that case, the entropy change between temperatures ${T_{i}}$ and ${T_{f}}$ is

$\displaystyle \Delta S=\int_{T_{i}}^{T_{f}}\frac{C_{P}\left(T\right)}{T}dT \ \ \ \ \ (2)$

Example 1 From Schroeder’s Figure 1.14, we can estimate a linear relation for ${C_{P}}$ for a mole of diamond between ${T=300\mbox{ K}}$ and ${T=400\mbox{ K}}$. Reading off the graph we get

 $\displaystyle C_{P}\left(300\right)$ $\displaystyle =$ $\displaystyle 6.5\mbox{ J K}^{-1}\ \ \ \ \ (3)$ $\displaystyle C_{P}\left(400\right)$ $\displaystyle =$ $\displaystyle 11\mbox{ J K}^{-1} \ \ \ \ \ (4)$

Between these temperatures, a formula for ${C_{P}\left(T\right)}$ is therefore a straight line:

 $\displaystyle \frac{C_{P}\left(T\right)-6.5}{T-300}$ $\displaystyle =$ $\displaystyle \frac{11-6.5}{400-300}=0.045\ \ \ \ \ (5)$ $\displaystyle C_{P}\left(T\right)$ $\displaystyle =$ $\displaystyle 0.045T-7 \ \ \ \ \ (6)$

If we assume this is valid over the range of temperature from 298 K up to 500 K, we can get the entropy change over that range for a mole of diamond (incidentally, if you want to try this experiment, you’ll need a very big diamond. A mole of diamond (carbon) is around 12 grams, and there are 5 carats per gram, so you’re looking for a 60 carat diamond). The entropy change is

 $\displaystyle \Delta S$ $\displaystyle =$ $\displaystyle \int_{298}^{500}\frac{0.045T-7}{T}dT\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[0.045T-7\ln T\right]_{298}^{500}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.47\mbox{ J K}^{-1} \ \ \ \ \ (9)$

The entropy of a mole of diamond at ${T=298\mbox{ K}}$ is given in Schroeder’s appendix as ${2.38\mbox{ J K}^{-1}}$ so the total entropy at ${T=500\mbox{ K}}$ is

$\displaystyle S\left(500\right)=2.38+5.47=7.85\mbox{ J K}^{-1} \ \ \ \ \ (10)$

Example 2 An empirical formula obtained by fitting to measured data for ${C_{P}}$ for one mole of graphite is

$\displaystyle C_{P}=a+bT-\frac{c}{T^{2}} \ \ \ \ \ (11)$

where the constants are

 $\displaystyle a$ $\displaystyle =$ $\displaystyle 16.86\mbox{ J K}^{-1}\ \ \ \ \ (12)$ $\displaystyle b$ $\displaystyle =$ $\displaystyle 4.77\times10^{-3}\mbox{ J K}^{-2}\ \ \ \ \ (13)$ $\displaystyle c$ $\displaystyle =$ $\displaystyle 8.54\times10^{5}\mbox{ J K} \ \ \ \ \ (14)$

The entropy change of a mole of graphite over the range of temperature from 298 K up to 500 K is therefore

 $\displaystyle \Delta S$ $\displaystyle =$ $\displaystyle \int_{298}^{500}\frac{a+bT-\frac{c}{T^{2}}}{T}dT\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[a\ln T+bT+\frac{c}{2T^{2}}\right]_{298}^{500}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.59\mbox{ J K}^{-1} \ \ \ \ \ (17)$

Adding on the tabulated value for ${T=298\mbox{ K}}$ we get

$\displaystyle S\left(500\right)=5.74+6.59=12.33\mbox{ J K}^{-1} \ \ \ \ \ (18)$

The entropy of graphite is larger than that of diamond which we’d expect since diamond’s crystal structure is more ordered than that of graphite.

Einstein solid: analytic solution for heat capacity

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.25.

The properties of an Einstein solid can be investigated analytically. The multiplicity of a solid with ${N}$ oscillators and ${q}$ energy quanta is

$\displaystyle \Omega=\binom{q+N-1}{q}=\frac{\left(q+N-1\right)!}{q!\left(N-1\right)!} \ \ \ \ \ (1)$

For large ${q}$ and ${N}$, we can approximate this by

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

For very large systems, the square root factor can be neglected compared to the two power terms as it is merely “large” compared to the “very large” power terms. We thus get

$\displaystyle \Omega\approx\left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N} \ \ \ \ \ (3)$

From this we can get the entropy

 $\displaystyle S$ $\displaystyle =$ $\displaystyle k\ln\Omega\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\left(q+N\right)\ln\left(q+N\right)-kq\ln q-kN\ln N \ \ \ \ \ (5)$

The total energy of the system is

$\displaystyle U=q\epsilon \ \ \ \ \ (6)$

where ${\epsilon}$ is the energy of a single quantum, so we can get the temperature as defined from the entropy

 $\displaystyle \frac{1}{T}$ $\displaystyle =$ $\displaystyle \frac{\partial S}{\partial U}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\frac{\partial S}{\partial q}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{\epsilon}\left[\ln\left(q+N\right)+1-\ln q-1\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{\epsilon}\ln\frac{q+N}{q}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{\epsilon}\ln\frac{U+N\epsilon}{U}\ \ \ \ \ (11)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\epsilon}{k\ln\frac{U+N\epsilon}{U}} \ \ \ \ \ (12)$

We can invert this to obtain ${U\left(T\right)}$:

 $\displaystyle e^{\epsilon/kT}$ $\displaystyle =$ $\displaystyle \frac{U+\epsilon N}{U}\ \ \ \ \ (13)$ $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{\epsilon N}{e^{\epsilon/kT}-1} \ \ \ \ \ (14)$

Finally, the heat capacity is obtained from

 $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{\partial U}{\partial T}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\epsilon Ne^{\epsilon/kT}}{\left(e^{\epsilon/kT}-1\right)^{2}}\left(-\frac{\epsilon}{kT^{2}}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\epsilon^{2}Ne^{\epsilon/kT}}{kT^{2}\left(e^{\epsilon/kT}-1\right)^{2}} \ \ \ \ \ (17)$

A plot of the dimensionless quantities ${C/Nk}$ versus ${kT/\epsilon}$ looks like this:

To compare this with Schroeder’s Figure 1.14, we need to insert some actual numbers so we can plot ${C}$ versus ${T}$. Figure 1.14 is for one mole of atoms, but an Einstein solid treats ${N}$ oscillators, where each oscillator is a one-dimensional simple harmonic oscillator. Each atom in a real solid can vibrate in three directions, so one mole of atoms in a real solid contains three moles of oscillators in an Einstein solid. Thus to compare the plots for one mole of “real” atoms, we should use ${N=3\times6.02\times10^{23}}$.

Using ${k=8.62\times10^{-5}\mbox{ eV K}^{-1}}$ and multiplying the result of ${C}$ by ${1.602\times10^{-19}}$ to convert the result from ${\mbox{eV K}^{-1}}$ to ${\mbox{J K}^{-1}}$ (which are the units used in Figure 1.14), we then need to choose some values for ${\epsilon}$ to get the curves for lead, aluminum and diamond.

As a starting point, I used the values from the numerical solution done earlier. The values for lead and aluminum turned out (surprisingly, considering the numerical solution was worked out for very small systems) to work quite well, but the value for diamond needed to be adjusted a bit. The final values were

 $\displaystyle \epsilon_{Pb}$ $\displaystyle =$ $\displaystyle 0.00862\mbox{ eV}\ \ \ \ \ (18)$ $\displaystyle \epsilon_{Al}$ $\displaystyle =$ $\displaystyle 0.0259\mbox{ eV}\ \ \ \ \ (19)$ $\displaystyle \epsilon_{Dia}$ $\displaystyle =$ $\displaystyle 0.12\mbox{ eV} \ \ \ \ \ (20)$

The resulting plots are:

The curves are lead (red), aluminum (green) and diamond (blue). They are a good match to the curves shown in Figure 1.14.

We can get an expression for the heat capacity in the limit of high temperatures by expanding the exponential. As a first approximation we’ll use

$\displaystyle e^{\epsilon/kT}\approx1+\frac{\epsilon}{kT} \ \ \ \ \ (21)$

Then from 17

$\displaystyle C\approx\frac{\epsilon^{2}N}{k}\frac{\left(1+\epsilon/kT\right)}{T^{2}\left(\epsilon^{2}/k^{2}T^{2}\right)}=Nk\left(1+\frac{\epsilon}{kT}\right)\approx Nk \ \ \ \ \ (22)$

This is the expected result from the equipartition theorem, since the heat capacity for a system whose thermal energy is entirely in the form of quadratic degrees of freedom is

$\displaystyle C=\frac{1}{2}Nfk \ \ \ \ \ (23)$

where ${f}$ is the number of degrees of freedom. For a harmonic oscillator, ${f=2}$ (one from the kinetic energy term and one from the potential energy), so we’d expect ${C=Nk}$.

To get a more accurate approximation, we can expand the exponential out to a few more terms using the Taylor expansion for small ${x\equiv\epsilon/kT}$:

$\displaystyle e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots \ \ \ \ \ (24)$

To get the next term in the expansion beyond 22, we need to expand out to ${x^{3}}$, so we get

 $\displaystyle \frac{C}{Nk}$ $\displaystyle =$ $\displaystyle x^{2}\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle x^{2}\frac{1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}}{\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{6}\right)^{2}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}}{\left(1+\frac{x}{2}+\frac{x^{2}}{6}\right)^{2}} \ \ \ \ \ (27)$

The denominator can now be expanded again using a Taylor series. This gets quite tedious so I used Maple to do the expansion (you’re welcome to try it by hand by calculating the derivatives) with the result

 $\displaystyle \frac{C}{Nk}$ $\displaystyle \approx$ $\displaystyle \left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}\right)\left(1-x+\frac{5}{12}x^{2}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\frac{x^{2}}{12}+\mathcal{O}\left(x^{3}\right)\ \ \ \ \ (29)$ $\displaystyle C$ $\displaystyle \approx$ $\displaystyle Nk\left(1-\frac{1}{12}\left(\frac{\epsilon}{kT}\right)^{2}\right) \ \ \ \ \ (30)$

Einstein solid – numerical solution

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.24.

The properties of an Einstein solid can be investigated numerically for relatively small systems. The multiplicity of a solid with ${N}$ oscillators and ${q}$ energy quanta is

$\displaystyle \Omega=\binom{q+N-1}{q}=\frac{\left(q+N-1\right)!}{q!\left(N-1\right)!} \ \ \ \ \ (1)$

From this we can get the entropy

$\displaystyle S=k\ln\Omega \ \ \ \ \ (2)$

which for small systems can be worked out without using Stirling’s approximation.

The total energy of the system is

$\displaystyle U=q\epsilon \ \ \ \ \ (3)$

where ${\epsilon}$ is the energy of a single quantum, so we can get the temperature as defined from the entropy

$\displaystyle \frac{1}{T}=\frac{\partial S}{\partial U} \ \ \ \ \ (4)$

Finally, the heat capacity per oscillator is obtained from

$\displaystyle \frac{C}{N}=\frac{1}{N}\frac{\partial U}{\partial T} \ \ \ \ \ (5)$

If we produce a table of these quantities for each value of ${q}$ from 0 upwards, we can approximate the derivatives to estimate values for ${T}$ and ${C}$ using the same techniques that we used for the two-state paramagnet.

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\Delta U}{\Delta S}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\epsilon}{k}\frac{\Delta q}{\Delta\left(\ln\Omega\right)}\ \ \ \ \ (7)$ $\displaystyle \frac{C}{N}$ $\displaystyle =$ $\displaystyle \frac{1}{N}\frac{\Delta U}{\Delta T}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\epsilon\Delta q}{N\frac{\epsilon}{k}\Delta\left(\frac{\Delta q}{\Delta\left(\ln\Omega\right)}\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k\Delta q}{N\Delta\left(\frac{\Delta q}{\Delta\left(\ln\Omega\right)}\right)} \ \ \ \ \ (10)$

For example, we can calculate ${T\left(q\right)}$ as

$\displaystyle T\left(q\right)=\frac{U\left(q+1\right)-U\left(q-1\right)}{S\left(q+1\right)-S\left(q-1\right)} \ \ \ \ \ (11)$

and then ${\frac{C\left(q\right)}{N}}$ as

$\displaystyle \frac{C\left(q\right)}{N}=\frac{1}{N}\frac{U\left(q+1\right)-U\left(q-1\right)}{T\left(q+1\right)-T\left(q-1\right)} \ \ \ \ \ (12)$

In practice, we are actually calculating ${S/k}$, ${kT/\epsilon}$ and ${C/Nk}$.

For the case ${N=50}$ and ${q=0,\ldots,100}$, the entropy as a function of energy (actually ${U/\epsilon=q}$) is as shown

Schroeder also asks us to calculate the values for ${N=5000}$ with ${q}$ still in the same range. This dilutes the energy among a larger number of oscillators, which effectively lowers the temperature of the system since the amount of energy per oscillator is reduced. He then asks us to compare these results to the heat capacity plots in his Figure 1.14, which shows the heat capacity for one mole of lead, aluminum and diamond as functions of temperature. It’s not quite clear to me how we can compare the results of this problem with that plot, since we’re working out the heat capacity for very small numbers of oscillators (certainly much less than a mole), so the total heat capacity of our small samples is much less than the values shown in Figure 1.14. However, if the goal is to get curves that have similar shapes to those in Schroeder, we can get a crude comparison.

To do this, we need to plot the actual heat capacity and temperature rather than the dimensionless quantities we’ve used in generating the numbers. That is, we need to calculate

 $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{k\Delta q}{\Delta\left(\frac{\Delta q}{\Delta\left(\ln\Omega\right)}\right)}\ \ \ \ \ (13)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\epsilon}{k}\frac{\Delta q}{\Delta\left(\ln\Omega\right)} \ \ \ \ \ (14)$

We can vary the shape of the curve by changing the size ${\epsilon}$ of the energy quantum. For lead and aluminum, we can use the ${N=50}$ calculation to generate the following plot

The red curve uses ${\epsilon=\left(100\mbox{ K}\right)k\left(\mbox{eV K}^{-1}\right)=8.62\times10^{-3}\mbox{ eV}}$ (and extends ${q}$ out to 200), and the green curve uses ${\epsilon=\left(300\mbox{ K}\right)k\left(\mbox{eV K}^{-1}\right)=0.0259\mbox{ eV}}$ (with ${q}$ out to 100). The two curves are roughly the same shape as Schroeder’s curves for lead (red) and aluminum (green). The vertical scale for the heat capacity is, of course, much smaller than the values in Figure 1.14 since we’re dealing with only 50 particles rather than a mole.

For diamond, the heat capacity doesn’t increase much until we get to higher temperatures, so we can use the ${N=5000}$ results for that. For ${\epsilon=\left(2000\mbox{ K}\right)k\left(\mbox{eV K}^{-1}\right)=0.1724\mbox{ eV}}$ we get the following curve

which is roughly the same shape as the diamond curve in Schroeder. (The kink in the graph is due to the fact that we’re using discrete values of ${q}$.) We had to extend ${q}$ out to 200 to get this curve. Note that the vertical scale is a couple of orders of magnitude larger than the lead and aluminum plot, since we’re dealing with a hundred times as many oscillators.

I don’t know how much confidence we can place in these values for ${\epsilon}$ (probably not much), except to say that increasing ${\epsilon}$ at any given temperature tends to decrease the heat capacity. This may seem odd, since equation 13 appears to be independent of ${\epsilon}$ so it might appear that ${C}$ should not depend on ${\epsilon}$. However, if we trace the quantities in 13 back to their definitions, we see that they depend only on ${N}$ and ${q}$, so it is true that if we hold ${N}$ and ${q}$ fixed, ${C}$ does not change even if we change ${\epsilon}$. However, if we’re viewing ${C}$ as a function of temperature and not ${q}$, we note from 14 that for fixed values of ${N}$ and ${q}$, the temperature does increase as we increase ${\epsilon}$ (which makes sense, since if we have a fixed number ${q}$ of quanta and increase the energy contained in each quantum, the energy of the system increases and so does its temperature). In terms of the red/green plot above, suppose we pick a point on the red curve with heat capacity ${C_{0}}$. This point corresponds to some temperature ${T_{0}}$ which in turn is determined by values ${q_{0}}$ and ${\epsilon_{0}}$ (with everything else held fixed). If we now hold ${q_{0}}$ fixed but increase ${\epsilon}$ to a new value ${\epsilon_{1}}$, the heat capacity ${C_{0}}$ remains unchanged but the temperature increases to a new value ${T_{1}}$ which is further to the right. Thus the point on the graph moves horizontally to the right and (if we pick ${\epsilon_{1}}$ correctly) could move onto the green curve. This point at temperature ${T_{1}}$ is below the point at temperature ${T_{1}}$ on the red curve. In other words, if we want to hold the temperature constant and increase ${\epsilon}$, we must decrease ${q}$ which in turn decreases ${C}$.

Entropy of a star

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.15.

The heat capacity of a star was estimated using the virial theorem as

$\displaystyle C_{V}=-\frac{3}{2}Nk \ \ \ \ \ (1)$

where ${N}$ is the number of particles (typically dissociated protons and electrons) in the star. A negative heat capacity is typical of gravitationally bound systems.

We can use this to work out the entropy from the formula

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \int\frac{C_{V}\left(T\right)}{T}dT\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3}{2}Nk\ln T+f\left(N,V\right) \ \ \ \ \ (3)$

where ${f}$ is some function that depends on ${N}$ and the volume ${V}$, but not on ${T}$.

The total energy of a gravitationally bound system is negative and, from the virial theorem, we have

$\displaystyle U=-K=-\frac{3}{2}NkT \ \ \ \ \ (4)$

where ${K}$ is the kinetic energy and the formula is obtained from the equipartition theorem. Thus the entropy can be written in terms of the energy as

$\displaystyle S=-\frac{3}{2}Nk\ln\left|\frac{2U}{3Nk}\right|+f\left(N,V\right) \ \ \ \ \ (5)$

We can incorporate everything inside the logarithm except for ${U}$ into the function ${f}$ (call the new function ${g}$, say), so that

$\displaystyle S=-\frac{3}{2}Nk\ln\left|U\right|+g\left(N,V\right) \ \ \ \ \ (6)$

The general shape of this curve is like this (units on the axes are arbitrary as I’m just trying to show the shape of the graph):

The graph is concave upwards, which is typical of systems with negative heat capacity as we discussed earlier. For sufficiently low (large negative) values of ${U}$, the graph would go negative, but I would guess that the temperature at that point would be higher than anything found in real stars so that would never happen. Note that as ${U\rightarrow0}$, ${S}$ effectively becomes infinite. At ${U=0}$, however, the system becomes gravitationally unbound, so the particles would presumably then be able to wander over the entire universe, giving them an infinite number of possible states.

Entropy changes in macroscopic systems

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.10 – 3.11.

The original definition of entropy was in terms of heat flow, rather than the multiplicity of states in a system. If an amount of heat ${Q}$ flows into a substance at a constant temperature ${T}$, the change in entropy is

$\displaystyle \Delta S=\frac{Q}{T} \ \ \ \ \ (1)$

If heat flows out of the substance, then ${Q}$ is negative and the system loses entropy. In order for a system to maintain a constant temperature when it gains or loses heat, it must differ from the types of systems we’ve considered up to now. One possibility is that the amount of heat gained is very small compared to the existing internal energy of the system so that the added heat makes a negligible difference to the temperature. Such a system is called a heat reservoir. Another example is a phase change, as when ice melts into liquid water, as during a phase change, heat is gained or lost and the substance doesn’t change its temperature.

Example 1 Suppose a 30 g ice cube at ${0^{\circ}\mbox{ C}=273\mbox{ K}}$ is placed on a table in a room at ${25^{\circ}\mbox{ C}=298\mbox{ K}}$. The ice will first melt into water, still at 273 K (because it’s a phase change), then the water will warm up to 298 K. All of this heat is transferred from the air in the room, which we can consider to be a heat reservoir at a constant temperature of 298 K. The changes in entropy are then:

The latent heat of fusion of water at 273 K is ${334\mbox{ J g}^{-1}}$, so the amount of heat required to melt the ice is

$\displaystyle Q=334\times30=10020\mbox{ J} \ \ \ \ \ (2)$

Since it occurs at a constant temperature, the entropy change of the water is

$\displaystyle \Delta S_{1}=\frac{Q}{T}=\frac{10020}{273}=36.7\mbox{ J K}^{-1} \ \ \ \ \ (3)$

As the water warms up, it absorbs heat, but the temperature varies. We can then use our relation between entropy and heat capacity, along with the fact that the specific heat capacity of water is roughly constant over the liquid range at ${1\mbox{ cal g}^{-1}\mbox{K}^{-1}=4.181\mbox{ J g}^{-1}\mbox{K}^{-1}}$ to get

 $\displaystyle \Delta S_{2}$ $\displaystyle =$ $\displaystyle C_{V}\int_{T_{i}}^{T_{f}}\frac{1}{T}dT\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle C_{V}\ln\frac{T_{f}}{T_{i}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(4.181\times30\right)\ln\frac{298}{273}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 11.0\mbox{ J K}^{-1} \ \ \ \ \ (7)$

Thus the total entropy increase of the water is

$\displaystyle \Delta S_{H_{2}O}=47.7\mbox{ J K}^{-1} \ \ \ \ \ (8)$

The total amount of heat transferred to the water from the room’s air is

$\displaystyle Q=10020+4.181\times30\times\left(298-273\right)=13155.75\mbox{ J} \ \ \ \ \ (9)$

This happens at a constant room temperature of 298 K so the entropy lost by the room is

$\displaystyle \Delta S_{room}=-\frac{13155.75}{298}=-44.1\mbox{ J K}^{-1} \ \ \ \ \ (10)$

The net entropy change of the universe is therefore

$\displaystyle \Delta S=\Delta S_{H_{2}O}+\Delta S_{room}=+3.6\mbox{ J K}^{-1} \ \ \ \ \ (11)$

which is positive, as required by the second law of thermodynamics.

Example 2 To draw a bath (in the days before hot and cold running water, presumably) we mix 50 litres of hot water at ${55^{\circ}\mbox{ C}=328\mbox{ K}}$ with 25 litres of cold water at ${10^{\circ}\mbox{ C}=283\mbox{ K}}$. The final temperature of the water is the weighted average:

$\displaystyle T_{f}=\frac{50\times328+25\times283}{75}=313\mbox{ K}\left(=40^{\circ}\mbox{ C}\right) \ \ \ \ \ (12)$

To find the entropy change, we can use 5. The hot water cools down and thus loses heat, so its entropy change is

 $\displaystyle \Delta S_{hot}$ $\displaystyle =$ $\displaystyle 4.181\times5\times10^{4}\times\ln\frac{313}{328}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -9786\mbox{ J K}^{-1} \ \ \ \ \ (14)$

The entropy gained by the cold water is

 $\displaystyle \Delta S_{cold}$ $\displaystyle =$ $\displaystyle 4.181\times2.5\times10^{4}\times\ln\frac{313}{283}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle +10532\mbox{ J K}^{-1} \ \ \ \ \ (16)$

Thus the net entropy change is

$\displaystyle \Delta S=+746\mbox{ J K}^{-1} \ \ \ \ \ (17)$

which is again positive.

Third law of thermodynamics; residual entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.9.

The entropy is related to temperature by

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

Using the chain rule, and keeping everything at constant ${N}$ and ${V}$, we can measure the change in entropy due to a change in temperature as

$\displaystyle dS=\frac{dU}{T}=\left(\frac{\partial U}{\partial T}\right)_{N,V}\frac{dT}{T}=C_{V}\frac{dT}{T} \ \ \ \ \ (2)$

where ${C_{V}}$ is the heat capacity at constant volume:

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

If we know ${C_{V}\left(T\right)}$ as a function of temperature, we can therefore find the change in entropy for a finite change in temperature by integration:

$\displaystyle \Delta S=S_{f}-S_{i}=\int_{T_{i}}^{T_{f}}\frac{C_{V}\left(T\right)}{T}dT \ \ \ \ \ (4)$

The total entropy in a system at temperature ${T_{f}}$ could theoretically be found by setting ${T_{i}=0}$ in the integral

$\displaystyle S_{f}-S\left(0\right)=\int_{0}^{T_{f}}\frac{C_{V}\left(T\right)}{T}dT \ \ \ \ \ (5)$

In theory, at absolute zero, any system should be in its (presumably) unique lowest energy state so the multiplicity of the zero state is 1, meaning that ${S\left(0\right)=0}$, and this integral does in fact give the actual entropy in a system at temperature ${T_{f}}$. It’s also obvious that for this integral to be finite (and positive) ${C_{V}\rightarrow0}$ as ${T\rightarrow0}$ at a rate such that the integral doesn’t diverge at its lower limit. Thus we must have ${C_{V}\left(T\right)\propto T^{a}}$ where ${a>0}$ as ${T\rightarrow0}$. Either of these conditions is a statement of the third law of thermodynamics, which basically says that at absolute zero, the entropy of any system is zero.

In practice, as a substance is cooled, its molecular configuration can get frozen into one of several possible ground states, so that there is a residual entropy even when ${T=0\mbox{ K}}$.

Example Carbon monoxide molecules are linear and in the solid form, they can line up in two orientations: OC and CO. Thus at absolute zero, the collection of molecules can be considered as a frozen-in matrix of molecules oriented randomly, so for a sample of ${N}$ molecules, there are ${2^{N}}$ possible structures. For a mole, the residual entropy is therefore

$\displaystyle S_{res}=k\ln2^{6.02\times10^{23}}=\left(1.38\times10^{-23}\right)\left(6.02\times10^{23}\right)\ln2=5.76\mbox{ J K}^{-1} \ \ \ \ \ (6)$

Predicting heat capacity

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.8.

If we know the energy ${U}$ of a system as a function of temperature, we can predict its heat capacity at constant volume using

$\displaystyle C_{V}=\left(\frac{\partial U}{\partial T}\right)_{N,V} \ \ \ \ \ (1)$

For an Einstein solid at low temperature, we have

$\displaystyle U=N\epsilon e^{-\epsilon/kT} \ \ \ \ \ (2)$

where ${\epsilon}$ is the magnitude of one energy quantum. The heat capacity is therefore

$\displaystyle C_{V}=\frac{N\epsilon^{2}}{kT^{2}}e^{-\epsilon/kT} \ \ \ \ \ (3)$

We can plot ${C_{V}/Nk}$ versus ${kT/\epsilon}$ to see the shape of the curve:

The Einstein solid has a peak in its heat capacity at a very low temperature of ${T=\epsilon/2k}$ (differentiate ${C_{V}}$ with respect to ${T}$ and set to zero). However, as Schroeder points out, this isn’t the observed heat capacity of real solids at low temperatures.

This method of predicting heat capacity relies on being able to find ${U\left(T\right)}$, which so far we’ve been able to do only by finding an expression for the multiplicity of states ${\Omega}$ in terms of the energy ${U}$ (and other parameters), then using that to find the entropy ${S=k\ln\Omega}$, then finding ${U\left(T\right)}$ from the expression ${1/T=\partial S/\partial U}$. In most cases, finding an expression for the multiplicity of states is not possible, so we need another method of getting ${U\left(T\right)}$ which we’ll hopefully get to eventually.