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

where is the number of microstates accessible to the substance.

For a 3-d ideal gas, this is given by the Sackur-Tetrode formula:

where is the volume, is the energy, is the number of molecules, is the mass of a single molecule and is Planck’s constant.

Although this formula looks a bit complicated, we can see that increasing any of , or increases the entropy. For an isothermal expansion, the gas expands quasistatically so that its temperature stays constant. This means that also stays constant, so that only the volume changes. Since the gas is doing work by expanding, the energy for the work must be provided by an amount of heat input into the gas to maintain the temperature as constant. This heat is given by the formula

where and are the initial and final volumes.

However, from 2, the change in entropy in a process where only the volume changes is

Combining these two equations gives

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 still doesn’t change, since the gas neither absorbs any heat nor does any work, so . 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 drops low enough, the log term can decrease below making negative. This isn’t possible, so the Sackur-Tetrode equation must break down at low energies. For a monatomic ideal gas, , 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

If we start at room temperature and atmospheric pressure , and can hold the density fixed, this will give an actual temperature at which the entropy becomes zero. The density is

The mass of a helium atom is so plugging in the other values gives

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 multiplied by a logarithm, and of an Einstein solid is also multiplied by a logarithm (because for high-temperature solids). For any macroscopic object, 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 . 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 , so the entropy of a book is around

For a 400 kg moose, which we can approximate by 400 kg of water with molar mass of around , we have

For the sun, we can take it to be of ionized hydrogen (protons) with molar mass of . The entropy is around