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

where is the number of microstates accessible to the substance.

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

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

We get

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

This gives the entropy of an ideal gas as

which is known as the Sackur-Tetrode equation.

Example 1A variant of this equation can be derived in a similar way for the 2-d ideal gas considered earlier. In that case we hadwhere is the area of the gas. Using Stirling’s approximation as before, we get

Example 2Schroeder gives the entropy of a mole of helium at room temperature and atmospheric pressure as . For another monatomic gas such as argon, we can work out the same thing. From the ideal gas law, at a pressure of and temperature of 300 K, one mole occupies a volume ofThe internal energy of a monatomic gas is per molecule per degree of freedom, so for one mole we have

The mass of a mole of argon is , so with we have

Plugging these values into 9, the entropy comes out to

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

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MartineIf you use equation 13, I don’t see how you get the entropy you get.. Do you use another formula or something?

gwrowePost authorYou need to use equation 9 for a 3-d gas. Equation 13 is for a 2-d gas. I’ve amended the post to make this clearer.