# 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.

# Entropy of an ideal gas; Sackur-Tetrode equation

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 2.31 – 2.33.

The entropy of a substance is given as

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

where ${\Omega}$ is the number of microstates accessible to the substance.

For a 3-d ideal gas, this is given by Schroeder’s equation 2.40:

$\displaystyle \Omega\approx\frac{V^{N}\left(2\pi mU\right)^{3N/2}}{h^{3N}N!\left(3N/2\right)!} \ \ \ \ \ (2)$

where ${V}$ is the volume, ${U}$ is the energy, ${N}$ is the number of molecules, ${m}$ is the mass of a single molecule and ${h}$ is Planck’s constant. We can further approximate this formula by using Stirling’s approximation for the factorials:

 $\displaystyle N!$ $\displaystyle \approx$ $\displaystyle \sqrt{2\pi N}N^{N}e^{-N}\ \ \ \ \ (3)$ $\displaystyle \left(3N/2\right)!$ $\displaystyle \approx$ $\displaystyle \sqrt{3\pi N}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2} \ \ \ \ \ (4)$

We get

$\displaystyle \Omega\approx\frac{V^{N}\left(\pi mU\right)^{3N/2}}{h^{3N}}\frac{2^{3N}e^{5N/2}}{\sqrt{6}3^{3N/2}\pi N^{5N/2+1}} \ \ \ \ \ (5)$

When ${N}$ is large, we can throw away a couple of factors and take the logarithm:

 $\displaystyle \Omega$ $\displaystyle \approx$ $\displaystyle \frac{V^{N}\left(\pi mU\right)^{3N/2}}{h^{3N}}\frac{2^{3N}e^{5N/2}}{3^{3N/2}N^{5N/2}}\ \ \ \ \ (6)$ $\displaystyle \ln\Omega$ $\displaystyle =$ $\displaystyle N\ln\left(V\left(\frac{\pi mU}{3}\right)^{3/2}\left(\frac{2}{h}\right)^{3}\frac{1}{N^{5/2}}\right)+\frac{5N}{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle N\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right] \ \ \ \ \ (8)$

This gives the entropy of an ideal gas as

$\displaystyle S=Nk\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^{2}}\right)^{3/2}\right)+\frac{5}{2}\right] \ \ \ \ \ (9)$

which is known as the Sackur-Tetrode equation.

Example 1 A variant of this equation can be derived in a similar way for the 2-d ideal gas considered earlier. In that case we had

$\displaystyle \Omega\approx\frac{\left(\pi A\right)^{N}}{\left(N!\right)^{2}h^{2N}}\left(\sqrt{2mU}\right)^{2N} \ \ \ \ \ (10)$

where ${A}$ is the area of the gas. Using Stirling’s approximation as before, we get

 $\displaystyle \Omega$ $\displaystyle \approx$ $\displaystyle \frac{\left(\pi A\right)^{N}}{2\pi N^{2N+1}e^{-2N}h^{2N}}\left(\sqrt{2mU}\right)^{2N}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{\left(\pi A\right)^{N}}{N^{2N}e^{-2N}h^{2N}}\left(2mU\right)^{N}\ \ \ \ \ (12)$ $\displaystyle S=k\ln\Omega$ $\displaystyle =$ $\displaystyle Nk\left[\ln\frac{2\pi mAU}{\left(hN\right)^{2}}+2\right] \ \ \ \ \ (13)$

Example 2 Schroeder gives the entropy of a mole of helium at room temperature and atmospheric pressure as ${S=126\mbox{ J K}^{-1}}$. For another monatomic gas such as argon, we can work out the same thing. From the ideal gas law, at a pressure of ${1.01\times10^{5}\mbox{ N m}^{-2}}$ and temperature of 300 K, one mole occupies a volume of

$\displaystyle V=\frac{nRT}{P}=\frac{\left(1\right)\left(8.31\right)\left(300\right)}{1.01\times10^{5}}=0.025\mbox{ m}^{3} \ \ \ \ \ (14)$

The internal energy of a monatomic gas is ${\frac{1}{2}kT}$ per molecule per degree of freedom, so for one mole we have

$\displaystyle U=\frac{3}{2}NkT=\frac{3}{2}nRT=3739\mbox{ J} \ \ \ \ \ (15)$

The mass of a mole of argon is ${39.948\times10^{-3}\mbox{ kg}}$, so with ${N=6.02\times10^{23}}$ we have

$\displaystyle m=\frac{39.948\times10^{-3}}{6.02\times10^{23}}=6.64\times10^{-26}\mbox{ kg} \ \ \ \ \ (16)$

Plugging these values into 9, the entropy comes out to

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \left(6.02\times10^{23}\right)\left(1.38\times10^{-23}\right)\left(\ln1.02\times10^{7}+2.5\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 155\mbox{ J K}^{-1} \ \ \ \ \ (18)$

This is a bit higher than the value for helium because of argon’s higher mass.

# Multiplicity of interacting ideal gases

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 2.27.

We’ve seen that the number of microstates available to an ${N}$-molecule 2-d ideal gas with energy ${U}$ contained in area ${A}$ is approximately

$\displaystyle \Omega\approx\frac{\left(\pi A\right)^{N}}{\left(N!\right)^{2}h^{2N}}\left(\sqrt{2mU}\right)^{2N} \ \ \ \ \ (1)$

Schroeder goes through a very similar calculation for the more realistic 3-d gas in a volume ${V}$, with the result

$\displaystyle \Omega\approx\frac{V^{N}}{N!h^{3N}}\frac{\pi^{3N/2}}{\left(3N/2\right)!}\left(2mU\right)^{3N/2} \ \ \ \ \ (2)$

Separating out the factors that depend only on ${N}$, we can write this as

$\displaystyle \Omega\left(U,V,N\right)=f\left(N\right)V^{N}U^{3N/2} \ \ \ \ \ (3)$

where

$\displaystyle f\left(N\right)=\frac{\left(2\pi m\right)^{3N/2}}{N!\left(3N/2\right)!h^{3N}} \ \ \ \ \ (4)$

Suppose we have two such gases ${A}$ and ${B}$, confined in volumes ${V_{A}}$ and ${V_{B}}$ and with energies ${U_{A}}$ and ${U_{B}}$. We’ll assume that both gases have the same number ${N}$ of molecules. If the volumes are separated by a partition that allows energy (but not molecules) to be exchanged, then the total multiplicity of the combined system is just the product of the individual multiplicities:

$\displaystyle \Omega_{total}=\Omega_{A}\Omega_{B}=\left[f\left(N\right)\right]^{2}\left(V_{A}V_{B}\right)^{N}\left(U_{A}U_{B}\right)^{3N/2} \ \ \ \ \ (5)$

The total energy is ${U=U_{A}+U_{B}}$ and is held constant. This is similar to the multiplicity equation for two interacting Einstein solids which had the form

$\displaystyle \Omega=\left(\frac{e}{N}\right)^{2N}\left(q_{A}q_{B}\right)^{N}=\left(\frac{e}{N}\right)^{2N}\left(q_{A}\left(q-q_{A}\right)\right)^{N} \ \ \ \ \ (6)$

Here, the energy quanta in the two solids obey the relation ${q_{A}+q_{B}=q=\mbox{constant}}$, and the calculation in Schroeder’s section 2.4 showed that if we plot ${\Omega}$ as a function of ${q_{A}}$, the curve peaks very sharply about ${q_{A}=q/2}$ and can be approximated by a Gaussian of form

$\displaystyle \Omega\approx\left(\frac{eq}{2N}\right)^{2N}e^{-N\left(2x/q\right)^{2}} \ \ \ \ \ (7)$

where ${x=q_{A}-q/2}$ is the deviation from the most probable value of ${q_{A}=\frac{q}{2}}$. The width of the peak, defined as the distance between the points where ${\Omega=\Omega_{max}/e}$ is

$\displaystyle w=\frac{q}{\sqrt{N}} \ \ \ \ \ (8)$

For our interacting gases that can exchange energy, the ${\left(q_{A}\left(q-q_{A}\right)\right)^{N}}$ factor is replaced by ${\left(U_{A}U_{B}\right)^{3N/2}=\left(U_{A}\left(U_{total}-U_{A}\right)\right)^{3N/2}}$ and if we follow through the same steps to approximate the curve by a Gaussian around its peak, we find that the formula for the width of the peak just replaces the exponent ${N}$ in the ${\left(q_{A}\left(q-q_{A}\right)\right)^{N}}$ factor in 6 by the exponent ${3N/2}$ in the ${\left(U_{A}U_{B}\right)^{3N/2}}$ factor for the interacting gases, and the width is

$\displaystyle w=\frac{U_{total}}{\sqrt{3N/2}} \ \ \ \ \ (9)$

For any macroscopic sample of gas, ${N}$ is large (of the order of ${10^{23}}$) so the width of the peak is very, very small compared to the overall scale of the graph.

Since one of the assumptions of statistical mechanics is that all microstates are equally probable, it should be possible, just by chance, to find that even if a gas has a total volume ${V}$ available to it, sometimes all the molecules will clump up in some smaller portion of the volume, leaving the remaining space empty (a vacuum). How likely is this to happen?

Effectively, what we’re asking is how likely is it that the volume occupied by the gas will spontaneously reduce from ${V}$ to ${aV}$, where ${0. Since the volume is all that changes (both ${N}$ and ${U}$ are unchanged), we can look at 3 and find that reducing the volume reduces the multiplicity to

$\displaystyle \Omega\left(a\right)=f\left(N\right)\left(aV\right)^{N}U^{3N/2} \ \ \ \ \ (10)$

Thus the probability that this will happen spontaeously is

$\displaystyle p\left(a\right)=\frac{\Omega\left(a\right)}{\Omega}=a^{N} \ \ \ \ \ (11)$

Since ${N}$ is a large number, even a value of ${a}$ close to 1 is still very unlikely. For example, if ${a=0.99}$ we find (it’s easier to work with logarithms here):

 ${N}$ ${p\left(0.99\right)}$ 100 0.366 ${10^{4}}$ ${2.25\times10^{-44}}$ ${10^{23}}$ ${\approx10^{-4\times10^{20}}}$

With 100 molecules, we can reasonably expect to see it happen. With 10,000 molecules, we’d probably never see it within the age of the universe. With ${10^{23}}$, that’s about as close to a definition of ‘impossible’ as you’re likely to get.

In fact, even for only 100 molecules, the chance of the gas crowding into a smaller volume is virtually zero for ${a<0.95}$ as we can see from a plot (note that the horizontal axis runs from 0.9 to 1.0):

The take home message is that these sorts of events just never happen in nature.

# Multiplicity of a 2-dim ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 2.26.

Having looked at counting the number of microstates in systems like coin flipping and the Einstein solid, we can now look at counting the microstates in an ideal gas.

At first glance, this might seem to be impossible, since in classical physics at least, a gas molecule confined within a volume ${V}$ with a fixed energy ${U}$ can be in an infinite number of states, since it could be at any location within the volume and the components of its momentum could have any values subject to the constraint that the kinetic energy of the molecule is ${p^{2}/2m}$. That’s true, and to be able to count the number of states of a gas molecule, we need to use quantum mechanics. Because of the uncertainty principle, the location and momentum of a molecule can be determined only up to regions ${\Delta x}$ (in position space) and ${\Delta p}$ (in momentum space) such that ${\Delta x\Delta p\ge\hbar/2}$. Schroeder’s derivation of the number of microstates available to an ideal gas begins with this principle, or at least with the approximation

$\displaystyle \Delta x\Delta p\approx h \ \ \ \ \ (1)$

Because the gas is confined within a volume ${V}$, the situation is somewhat similar to the particle in a box or infinite square well. A particle in such a potential well is restricted to discrete energy states, and also restricted to a location between the infinitely high barriers at either end. It might seem that for a state with fixed energy, the momentum would be precisely known, but in fact, because the particle can move either to the right or left, the momentum doesn’t have a fixed value, and a particle in such a fixed energy state does in fact satisfy the uncertainty principle.

Schroeder’s argument is that, in one dimension, a molecule can be localized to a particular region ${\Delta x}$ in position space and a particular momentum interval ${\Delta p}$ in momentum space, so if the particle is confined to a location range ${L}$ and momentum range ${L_{p}}$, the number of available states in position space is ${L/\Delta x}$ and in momentum space is ${L_{p}/\Delta p}$, and, since the momentum and position ranges are independent, the total number of microstates is

$\displaystyle \Omega=\frac{LL_{p}}{\Delta x\Delta p}=\frac{LL_{p}}{h} \ \ \ \ \ (2)$

Schroeder then develops the theory for a 3-d ideal gas in detail so we won’t go through that again here. Rather, we’ll look at an analogous 2-dimensional case. In 2-d, we can imagine the molecule confined to an area ${A}$ in position space and another area ${A_{p}}$ in momentum space. Since the energy of the molecule is fixed, the momentum space is constrained by the condition

$\displaystyle p_{x}^{2}+p_{y}^{2}=2mU \ \ \ \ \ (3)$

which is the equation of a circle. Thus the ‘area’ ${A_{p}}$ is actually the circumference of the circle:

$\displaystyle A_{p}=2\pi\sqrt{2mU} \ \ \ \ \ (4)$

Since there is an uncertainty of ${h}$ for the products of position and momentum in each of the two dimensions, we get

$\displaystyle \Omega=\frac{A}{h^{2}}2\pi\sqrt{2mU} \ \ \ \ \ (5)$

To generalize this to the case with ${N}$ gas molecules, we note that in position space, the locations of all the molecules are independent, so we’ll get a factor of ${A^{N}/h^{2N}}$. The momenta are constrained by the condition

$\displaystyle \sum_{i=1}^{N}\left(p_{i_{x}}^{2}+p_{i_{y}}^{2}\right)=2mU \ \ \ \ \ (6)$

where the sum index ${i}$ extends over the ${N}$ molecules, so the momenta are not independent. The total ‘volume’ of momentum space is actually the ‘area’ of a ${2N}$ dimensional hypersphere, which Schroeder derives in his Appendix B and gives as

$\displaystyle \mbox{area}=\frac{2\pi^{d/2}}{\left(\frac{d}{2}-1\right)!}r^{d-1} \ \ \ \ \ (7)$

where ${r}$ is the radius and ${d}$ is the dimension. Thus we get, using ${d=2N}$ and ${r=\sqrt{2mU}}$:

$\displaystyle \Omega=\frac{A^{N}}{h^{2N}}\frac{2\pi^{N}}{\left(N-1\right)!}\left(\sqrt{2mU}\right)^{2N-1} \ \ \ \ \ (8)$

This would be the formula if the gas molecules were distinguishable. However, one of the principles of quantum mechanics is that elementary particles of the same type are actually identical. In that case, with ${N}$ molecules, interchanging any pair of them will leave the microstate unchanged, so the above formula actually overcounts the number of states by ${N!}$. The actual number of microstates for a 2-d ideal gas of indistinguishable particles is therefore

$\displaystyle \Omega=\frac{A^{N}}{N!h^{2N}}\frac{2\pi^{N}}{\left(N-1\right)!}\left(\sqrt{2mU}\right)^{2N-1} \ \ \ \ \ (9)$

Since ${N}$ is a large number (on the order of ${10^{23}}$), we can approximate this by

$\displaystyle \Omega\approx\frac{\left(\pi A\right)^{N}}{\left(N!\right)^{2}h^{2N}}\left(\sqrt{2mU}\right)^{2N} \ \ \ \ \ (10)$

In SI units, ${h}$ is a small number (around ${10^{-34}}$) and ${N}$ is a large number, so the number of microstates will typically be very large.

# Diffusion in an ideal gas; Fick’s law

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.70.

Diffusion is the process by which the concentration of some substance varies over time by means of the drifting of molecules through a volume. We must distinguish between diffusion and other methods by which molecular concentrations can vary, such as convection (macroscopic motion of large chunks of the material due to such things as temperature differences) and mechanical stirring.

We can treat diffusion in ideal gases in a similar way to that used for thermal conductivity and viscosity. In diffusion, we are interested in the movement of molecular concentration, rather than energy transport (as in thermal conductivity) or momentum transport (as in viscosity). The derivation of a crude formula for diffusion, however, follows the same steps as in the other two cases.

Consider a thin slab of gas of cross-sectional area ${A}$. We divide the box in half so that the number of molecules on one side of the partition is ${N_{1}}$ and on the other side is ${N_{2}}$. We’re assuming that the whole box is at a constant temperature ${T}$, and that Schroeder’s approximations for the mean free path ${\ell}$ and average speed ${\bar{v}}$ are valid:

 $\displaystyle \ell$ $\displaystyle \approx$ $\displaystyle \frac{1}{4\pi r^{2}}\frac{V}{N}\ \ \ \ \ (1)$ $\displaystyle \bar{v}$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{3kT}{m}} \ \ \ \ \ (2)$

where ${r}$ is the molecular radius and ${m}$ the mass of one molecule. Since only half the molecules will, on average, be moving towards the partition, the net number ${\Delta N}$ of molecules that cross the partition in a time ${\Delta t}$, which is the time it takes a molecule to move distance ${\ell}$, is

$\displaystyle \Delta N=\frac{1}{2}\left(N_{1}-N_{2}\right) \ \ \ \ \ (3)$

If the gradient in molecule number is ${dN/dx}$ then we have

$\displaystyle \Delta N=\frac{1}{2}\ell\frac{dN}{dx} \ \ \ \ \ (4)$

and the net rate at which molecules cross the partition per unit area, known as the flux ${J_{x}}$, is

$\displaystyle \left|J_{x}\right|=\frac{\Delta N}{A\Delta t}=\frac{1}{2}\frac{\ell}{A\Delta t}\frac{dN}{dx}=\frac{1}{2}\frac{\ell^{2}}{\ell A\Delta t}\frac{dN}{dx}=\frac{1}{2}\frac{\ell}{V}\bar{v}\frac{dN}{dx}=\frac{1}{2}\ell\bar{v}\frac{dn}{dx} \ \ \ \ \ (5)$

where ${n=N/V}$ is the molecular concentration. The absolute value indicates that this is the magnitude of the flux. As the flux is in the opposite direction to the gradient, we will have

$\displaystyle J_{x}=-\frac{1}{2}\ell\bar{v}\frac{dn}{dx} \ \ \ \ \ (6)$

The quantity ${\frac{1}{2}\ell\bar{v}}$ is an approximation for the diffusion constant ${D}$ for an ideal gas, and has units of ${\mbox{m}^{2}\mbox{s}^{-1}}$. The more general form of this equation

$\displaystyle J_{x}=-D\frac{dn}{dx} \ \ \ \ \ (7)$

is known as Fick’s law.

For air at room temperature, using the values from Schroeder’s book,

$\displaystyle D\approx\frac{1}{2}\left(1.5\times10^{-7}\right)\left(500\right)=3.75\times10^{-5}\mbox{m}^{2}\mbox{s}^{-1} \ \ \ \ \ (8)$

The measured value is around ${2\times10^{-5}}$ so this isn’t too far off for a rough estimate.

From the values for ${\ell}$ and ${\bar{v}}$ above, we have

$\displaystyle D=\frac{1}{2}\ell\bar{v}=\frac{1}{8\pi r^{2}}\frac{V}{N}\sqrt{\frac{3kT}{m}}=\frac{1}{8\pi r^{2}}\frac{\left(kT\right)^{3/2}}{P}\sqrt{\frac{3}{m}} \ \ \ \ \ (9)$

At fixed pressure, ${D\propto T^{3/2}}$.

# Viscosity of an ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.66.

We can treat viscosity in gases in a similar way to that used for thermal conductivity. A common situation involving viscosity is that of two horizontal, parallel flat plates with a gas or liquid sandwiched between them. If one plate moves parallel to the other, the gas between the plates exerts a drag force inhibiting the motion of the plates.

In the reference frame with the lower plate at rest and the upper plate moving at some speed ${u_{x}}$ to the right, we’d expect the fluid between the plates to be moving at a speed that increases from zero next to the lower plate up to ${u_{x}}$ next to the upper plate. This gradient in speed is the result of momentum transfer between adjacent layers in the fluid. Because of Newton’s law of equal action and reaction, the horizontal drag force exerted on each plate is equal and opposite to the force on the fluid layer directly adjacent to the plate.

The guesstimate derivation given by Schroeder assumes that the force on each plate is proportional to the area ${A}$ of the plate and to the relative speed of the upper and lower plates ${u_{x,top}-u_{x,bottom}}$, and inversely proportional to the distance ${\Delta z}$ between the plates. The last two assumptions are equivalent to saying that the force is proportional to the velocity gradient ${du_{x}/dz}$. That is

$\displaystyle \frac{F_{x}}{A}=\eta\frac{du_{x}}{dz} \ \ \ \ \ (1)$

where ${\eta}$ is the coefficient of viscosity or just the viscosity. From its definition, it has units of ${\left[\mbox{force}\right]\left[\mbox{area}\right]^{-1}\left[\mbox{time}\right]}$ or ${\mbox{N m}^{-2}\mbox{s}}$.

By following a similar argument to that for thermal conductivity, we can get an estimate for ${\eta}$ in the case of an ideal gas. We’ll assume that the mean free path and average molecular velocity for the gas are the same as before:

 $\displaystyle \ell$ $\displaystyle \approx$ $\displaystyle \frac{1}{4\pi r^{2}}\frac{V}{N}\ \ \ \ \ (2)$ $\displaystyle \bar{v}$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{3kT}{m}} \ \ \ \ \ (3)$

where ${r}$ is the molecular radius and ${m}$ the mass of one molecule. Then if we consider a thin horizontal slab of the gas between the plates, those molecules within a distance ${\ell}$ of the midpoint of the slab can cross the midpoint if they are travelling towards the midpoint. On average half the molecules in each half are travelling towards the midpoint so if the average horizontal momentum of the molecules on side ${i}$ is ${p_{i}}$ for ${i=1,2}$, then in a time ${\Delta t}$ (that is, the time it takes an average molecule to travel ${\ell}$) the momentum transferred is

$\displaystyle \Delta p=\frac{1}{2}\left(p_{1}-p_{2}\right)=\frac{M}{2}\left(u_{x,1}-u_{x,2}\right)=\frac{M}{2}\ell\frac{du_{x}}{dz} \ \ \ \ \ (4)$

where ${M}$ is the total mass of gas in a slab of area ${A}$ and thickness ${\ell}$.

The average force per unit area of the plates is then

$\displaystyle \frac{F_{x}}{A}=\frac{\Delta p}{A\Delta t}=\frac{M}{2A\Delta t}\ell\frac{du_{x}}{dz}=\frac{M}{2A\ell\Delta t}\ell^{2}\frac{du_{x}}{dz}=\frac{\rho}{2}\ell\bar{v}\frac{du_{x}}{dz} \ \ \ \ \ (5)$

where ${\rho=M/A\ell}$ is the mass density of the gas and ${\bar{v}=\ell/\Delta t}$ is the average speed.

For an ideal gas ${\rho=mN/V}$ so combining this with the above expressions for ${\ell}$ and ${\bar{v}}$ we get

$\displaystyle \eta=\frac{\sqrt{3mkT}}{4\pi r^{2}} \ \ \ \ \ (6)$

Thus the viscosity for an ideal gas is independent of pressure and depends only on the temperature.

For air, the density is around ${1\mbox{ kg m}^{-3}}$, ${\ell\approx1.5\times10^{-7}\mbox{ m}}$ and ${\bar{v}\approx500\mbox{ m s}^{-1}}$ so

$\displaystyle \eta\approx3.75\times10^{-5}\mbox{N m}^{-2}\mbox{s} \ \ \ \ \ (7)$

This is about double the value of ${19\mbox{ }\mu\mbox{Pa s}}$ (1 Pascal is ${1\mbox{ N m}^{-2}}$) given in the book, but it’s not bad for a rough estimate.

# Thermal conductivity of an ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.63.

Schroeder gives a rough derivation of the thermal conductivity of an ideal gas. I won’t repeat the full derivation here; rather I’ll summarize the key points.

The mean free path ${\ell}$ of a gas molecule is derived by freezing all molecules in place except for one, then doubling the radius of the moving molecule while reducing all other molecules to points. The mean free path is then taken to be the length of a cylinder swept out by the moving molecule, when the volume of that cylinder equals the average volume per molecule ${V/N}$ in the gas. The result is

$\displaystyle \ell=\frac{1}{4\pi r^{2}}\frac{V}{N} \ \ \ \ \ (1)$

where ${r}$ is the radius of a gas molecule.

Next, we consider a box of molecules of width ${2\ell}$ with a temperature gradient in the ${x}$ direction. Imagine a partition down the middle of the box and consider the number of molecules that pass across this partition in each direction (that is, in the ${+x}$ and ${-x}$ directions). On average, half the molecules in each half of the box are moving towards the partition, and since the width of each half of the box is ${\ell}$ we expect about half the molecules in each half of the box to cross the partition. This means that half the energy in each half of the box is transported across the partition, so the net heat transfer is

$\displaystyle Q=\frac{1}{2}\left(U_{1}-U_{2}\right)=-\frac{1}{2}\left(U_{2}-U_{1}\right) \ \ \ \ \ (2)$

The energy in the gas is the heat capacity times the temperature so

$\displaystyle Q=-\frac{C_{V}}{2}\left(T_{2}-T_{1}\right) \ \ \ \ \ (3)$

Assuming a linear temperature gradient, the temperature difference between the centres of the two halves of the box is

$\displaystyle T_{2}-T_{1}=\ell\frac{dT}{dx} \ \ \ \ \ (4)$

so we get a formula for the heat transferred in time ${\Delta t}$ by dividing both sides by ${\Delta t}$:

$\displaystyle \frac{Q}{\Delta t}=-\frac{C_{V}\ell}{2}\frac{dT}{dx} \ \ \ \ \ (5)$

Comparing this with the heuristic form of the Fourier heat conduction equation we had earlier, we get an expression for the thermal conductivity:

$\displaystyle k_{t}=\frac{C_{V}\ell}{2A\Delta t} \ \ \ \ \ (6)$

where ${A}$ is the area of the imaginary partition we inserted into the box. If ${\Delta t}$ is taken to be the average time to travel the distance ${\ell}$ then

$\displaystyle \bar{v}=\frac{\ell}{\Delta t} \ \ \ \ \ (7)$

is the average speed of the molecules. Further, if we take the volume of the box to be ${V=A\ell}$ (actually, given that the width of the box in Schroeder’s derivation is ${2\ell}$ I would think the volume should be ${2A\ell}$, but anyway…) we get

$\displaystyle k_{t}=\frac{C_{V}\ell^{2}}{2A\ell\Delta t}=\frac{C_{V}}{2V}\ell\bar{v} \ \ \ \ \ (8)$

Given that the rms speed (which is roughly the same as ${\bar{v}}$) is

$\displaystyle v_{rms}=\sqrt{\frac{3kT}{m}} \ \ \ \ \ (9)$

the heat capacity is

$\displaystyle C_{V}=\frac{1}{2}fNk \ \ \ \ \ (10)$

where ${f}$ is the number of degrees of freedom per molecule, we have, using 1,

$\displaystyle k_{t}\propto\sqrt{T} \ \ \ \ \ (11)$

This relation agrees with measurements over a large temperature range for many gases.

Example The mean free path in the air around us is very small. As given by Schroeder, using 1 this works out to around ${1.5\times10^{-7}\mbox{ m}}$. Using the ideal gas law ${PV=NkT}$, we can find the pressure at which ${\ell=10\mbox{ cm}}$ at room temperature ${T=300\mbox{ K}}$. We get

$\displaystyle \ell=\frac{1}{4\pi r^{2}}\frac{kT}{P} \ \ \ \ \ (12)$

Taking ${r=1.5\times10^{-10}\mbox{ m}}$ we have

 $\displaystyle P$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi r^{2}}\frac{kT}{\ell}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.15\mbox{ N m}^{-2} \ \ \ \ \ (14)$

Standard atmospheric pressure is around ${10^{5}\mbox{ N m}^{-2}}$ so this is a reasonable laboratory vacuum.

# Isothermal versus adiabatic expansion of an ideal gas bubble

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.38.

As a simple example of isothermal versus adiabatic expansion of an ideal gas, suppose that two identical bubbles form at the bottom of a lake. Bubble A rises quickly so that no heat is exchanged with the surrounding water, while bubble B rises slowly (bumping off the leaves of some lakeweed, for example) so that its temperature remains constant (assuming that the lake’s water temperature is the same everywhere).

Bubble A experiences adiabatic expansion, so it obeys the relation

$\displaystyle PV^{\gamma}=A \ \ \ \ \ (1)$

for some constant ${A}$. Bubble B expands isothermally, so

$\displaystyle PV=NkT=\mbox{constant} \ \ \ \ \ (2)$

The initial volumes ${V_{0}}$ and pressures ${P_{0}}$ of the two bubbles are the same so

 $\displaystyle A$ $\displaystyle =$ $\displaystyle P_{0}V_{0}^{\gamma}\ \ \ \ \ (3)$ $\displaystyle NkT$ $\displaystyle =$ $\displaystyle P_{0}V_{0} \ \ \ \ \ (4)$

When the bubbles reach the surface of the lake, the pressure has reduced to ${P_{1}}$ so the volumes of the bubbles are

 $\displaystyle V_{A}$ $\displaystyle =$ $\displaystyle \left(\frac{P_{0}}{P_{1}}\right)^{1/\gamma}V_{0}\ \ \ \ \ (5)$ $\displaystyle V_{B}$ $\displaystyle =$ $\displaystyle \frac{P_{0}}{P_{1}}V_{0} \ \ \ \ \ (6)$

Since ${\gamma=\left(f+2\right)/f>1}$ where ${f}$ is the number of degrees of freedom, and ${P_{0}>P_{1}}$

$\displaystyle \left(\frac{P_{0}}{P_{1}}\right)^{1/\gamma}<\frac{P_{0}}{P_{1}} \ \ \ \ \ (7)$

so ${V_{B}>V_{A}}$. That is, the the bubble that rises slowly will be larger than the bubble that rises quickly when they reach the surface.