# Entropy: a few examples

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 2.34 – 2.36.

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 the Sackur-Tetrode formula:

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

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.

Although this formula looks a bit complicated, we can see that increasing any of ${V}$, ${U}$ or ${N}$ increases the entropy. For an isothermal expansion, the gas expands quasistatically so that its temperature stays constant. This means that ${U=\frac{3}{2}NkT}$ also stays constant, so that only the volume changes. Since the gas is doing work ${W}$ by expanding, the energy for the work must be provided by an amount of heat ${Q}$ input into the gas to maintain the temperature as constant. This heat is given by the formula

$\displaystyle Q=NkT\ln\frac{V_{f}}{V_{i}} \ \ \ \ \ (3)$

where ${V_{i}}$ and ${V_{f}}$ are the initial and final volumes.

However, from 2, the change in entropy in a process where only the volume changes is

$\displaystyle \Delta S=S_{f}-S_{i}=Nk\ln\frac{V_{f}}{V_{i}} \ \ \ \ \ (4)$

Combining these two equations gives

$\displaystyle \Delta S=\frac{Q}{T} \ \ \ \ \ (5)$

This relation is valid for the case where the expanding gas does work, so that heat must be input to provide the energy for the work. In a free expansion, the gas expands into a vacuum so does no work (well, technically, after some of the gas has entered the vacuum area, it’s no longer a vacuum so that some work is done, but we’ll assume the vacuum area is very large so we can neglect this). In this case, the internal energy ${U}$ still doesn’t change, since the gas neither absorbs any heat nor does any work, so ${\Delta U=Q+W=0}$. However, the volume occupied by the gas does increase (and it’s the only thing that changes) so 4 is still valid, although 5 is not.

Another property of 2 is that if the energy ${U}$ drops low enough, the log term can decrease below ${-\frac{5}{2}}$ making ${S}$ negative. This isn’t possible, so the Sackur-Tetrode equation must break down at low energies. For a monatomic ideal gas, ${U=\frac{3}{2}NkT}$, so this implies that things go wrong for low temperatures. For example, suppose we have a mole of helium and cool it (assuming it remains a gas). Then the critical temperature is found from

 $\displaystyle -\frac{5}{2}$ $\displaystyle =$ $\displaystyle \ln\left(\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right)\ \ \ \ \ (6)$ $\displaystyle T_{crit}$ $\displaystyle =$ $\displaystyle \frac{h^{2}}{2\pi mk}\left(\frac{N}{V}e^{-5/2}\right)^{2/3} \ \ \ \ \ (7)$

If we start at room temperature ${T=300\mbox{ K}}$ and atmospheric pressure ${P=1.01\times10^{5}\mbox{ Pa}}$, and can hold the density ${N/V}$ fixed, this will give an actual temperature at which the entropy becomes zero. The density is

$\displaystyle \frac{N}{V}=\frac{P}{kT}=2.44\times10^{25}\mbox{ m}^{-3} \ \ \ \ \ (8)$

The mass of a helium atom is ${4\times10^{-3}\mbox{ kg mol}^{-1}/6.02\times10^{23},}$ so plugging in the other values gives

$\displaystyle T_{crit}=0.012\mbox{ K} \ \ \ \ \ (9)$

In fact, helium liquefies at around 4 K, so it appears that 2 might actually be valid for the region where helium remains a gas.

As a final example, we can observe that the entropy of an ideal gas is ${Nk}$ multiplied by a logarithm, and of an Einstein solid is also ${Nk}$ multiplied by a logarithm (because ${\Omega\approx\left(qe/N\right)^{N}}$ for high-temperature solids). For any macroscopic object, ${N}$ is a large number and the logarithm is much smaller, so for a rough order-of-magnitude estimate of the entropy, we can neglect the log term and take ${S\sim Nk}$. A few such estimates are:

For a 1 kg book, we can take it to be 1 kg of carbon, with a molar mass of ${12\times10^{-3}\mbox{ kg mol}^{-1}}$, so the entropy of a book is around

$\displaystyle S\sim\frac{6.02\times10^{23}}{12\times10^{-3}}\left(1\right)\left(1.38\times10^{-23}\right)=692\mbox{ J K}^{-1} \ \ \ \ \ (10)$

For a 400 kg moose, which we can approximate by 400 kg of water with molar mass of around ${18\times10^{-3}\mbox{ kg mol}^{-1}}$, we have

$\displaystyle S\sim\frac{6.02\times10^{23}}{18\times10^{-3}}\left(400\right)\left(1.38\times10^{-23}\right)=1.85\times10^{5}\mbox{ J K}^{-1} \ \ \ \ \ (11)$

For the sun, we can take it to be ${2\times10^{30}}$ of ionized hydrogen (protons) with molar mass of ${10^{-3}\mbox{ kg mol}^{-1}}$. The entropy is around

$\displaystyle S\sim\frac{6.02\times10^{23}}{10^{-3}}\left(2\times10^{30}\right)\left(1.38\times10^{-23}\right)=1.66\times10^{34}\mbox{ J K}^{-1} \ \ \ \ \ (12)$

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