Ideal Gases in Thermal Physics Lecture Notes

Google+ Pinterest LinkedIn Tumblr +

Lets now turn to ideal gases.  To start our investigation, let’s consider a single particle in a box. Recall that the allowed energy solutions of the Schrodinger are

where m is the mass of the particle and L is the length of one side of the box.  The  partition function is then

If the temperature is high enough so that the spacing between adjacent energy values is small in comparison, we may replace the summations with integrals.  We can also factor each integral so that the triple integral becomes a product of three identical integrals


where .  Let x = anx.  Integrating this, we get

                                                                  Z = nqV                                                       (7.2)

where  is called the quantum concentration.

Once we know Z, we can immediately calculate other functions.  For example, the average energy for the particle is


Classical Regime

If we now put N identifiable particles in a box such that the number density of particles, n = N/V satisfies n << nq, then we are in the classical regime.  Assume that the particles do not interact.  Then each particle can be pictured as being in its own box.  In this case, the partition function for the whole system can be written as

The important fact to remember with this result is that the particles are completely identifiable.  Also, the last line of this result is only true if the particles all have the same mass.  If the masses differ for each particle, then the partition function is just


If the particles are identical, we have to count the number of particles in each state.  If the orbital indices are all different, then each entry in the partition function will occur N! times in Z1N, whereas if the particles are identical they should occur only once.   Thus ZN over counts each the number of states by N!, and so the partition function for N identical particles becomes


For an ideal gas, we can treat the gas as a collection of N identical particles.  Then the energy of the ideal gas is


Similarly, the free energy is

Using Stirling’s expansion, this becomes


From the free energy we can find the pressure as


                                                                  pV = Nt                                                       (7.7)

This is the ideal gas law.  Similarly, the entropy can be derived from


This is known as the Sackur‑Tebisch equation.


About Author

Leave A Reply