Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problem 3.37.

The chemical potential is defined in terms of the entropy as

This definition leads to a general thermodynamic identity

We can get another formula for from this:

We can apply these formulas to get the chemical potential of an ideal gas from the Sackur-Tetrode equation

Schroeder applies 1 to this equation and gets the chemical potential as

Now consider a volume of gas a distance above the Earth’s surface. In this case, the energy of the gas is composed of both the kinetic energy of the moving molecules and the potential energy of the molecules, the latter of which is per molecule. In order to find the chemical potential of this volume of gas, we need to modify 4 to write in terms of the potential and kinetic energy.

To see how to do this, we need to review the derivation of the multiplicity of an ideal gas (Schroeder’s equation 2.40). This derivation relied on arguments from quantum mechanics that gave the number of position and momentum states that are possible in a volume of gas. The energy that appeared in this derivation is entirely kinetic energy. Merely raising the volume of gas a distance above the Earth’s surface doesn’t change the number of possible states that gas can occupy; it merely shifts the total energy by a fixed amount . Thus the in 4 should really be written as , the kinetic energy:

However, the being held constant in 1 is the *total* energy, which is

so to write 4 in terms of this constant energy and the potential energy, we have

where

We can now find the chemical potential by taking the derivative 1:

We can now substitute back in and use 7 to get

Thus the chemical potential is changed by exactly the potential energy of a single molecule.

Now suppose that we have two samples of the same ideal gas with equal volumes and temperatures, but with one at height and the other at the Earth’s surface so . If the two volumes are in diffusive equilibrium, then their chemical potentials are equal, so we have

Dividing both sides by and exponentiating, we get

Since the volumes and temperatures are equal, from the ideal gas law this equation implies the same relation applies for the pressure as a function of height

which is the barometric equation we obtained earlier.

### Like this:

Like Loading...

*Related*

Pingback: Thermodynamic properties of a 2-dim ideal gas | Physics pages

Pingback: Gibbs free energy and chemical potential | Physics pages

Pingback: Grand free energy | Physics pages