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)

7 thoughts on “Entropy: a few examples

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  3. Nikos Christodoulou

    I obtained the same solutions listed here for pr. 2.36, but Schroeder uses coefficients to multiply these answers for water (he uses 10 that he calls it a “small coefficient for Nk”) and for the sun he uses 20. Any idea what these coefficients are? There is no mention in his textbook.
    Thanks in advance

    Reply
    1. gwrowe Post author

      I can’t find where Schroeder refers to the coefficients of 10 and 20 – where did you find that?
      I don’t know where he gets them.

      Reply
      1. Nikos Christodoulou

        I checked my answers with a professor (friend of mine) at the local university who has Schroeder’s book and solutions as one of the sources in the Thermodynamics course he teaches. Apparently Schroeder uses a ‘small’ coefficient of 10 for water and 20 for the H in the sun due to the high temperature as opposed to the room temperature used in the text. I was unable to find any mention of such a coefficient in the textbook. Similarly my friend was also unable to explain the difference. However, it is not a big deal since we all agree with the solution that is listed in your list. Thanks for the response.

        Reply
        1. gwrowe Post author

          It would be very helpful to have the solution manual, that’s true. However, I believe it’s only available to people who can prove they are on the academic staff of a university so I doubt I could get a copy.
          I’ve noticed that quite a few of Schroeder’s problems involve some hand-waving in the solutions, but I suppose that’s normal for a subject like thermodynamics where a lot of approximations and guesstimates are made.

          Reply
  4. Nikos Christodoulou

    Yes only university instructors can obtain it. This is why I am fortunate to have this friend who is teaching a Thermodynamics course to mechanical engineers who helps me from time to time. I do agree that there seems to be a bit of hand-waving in some of the problems, particularly when the factorials are approximated with Sterling’s formula and sometimes we drop the square root and sometime not. I have to admit though that I am enjoying the subject and after this book I may try a more advanced one like Parthia’s or Huang’s.
    Thank you once more for being so helpful. Let me know if I can be sometime.

    Reply

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