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)

2 thoughts on “Entropy: a few examples

  1. Pingback: Black hole entropy | Physics pages

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