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.

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