THE UNIVERSITY OF BRITISH COLUMBIA

PHYSICS 455

Lecture # 9 :

Fri. 24 Jan. 1997

Guest Lecture by Brian Turrell:

Introduction to Ideal Gases

I. PARTICLE IN A BOX:

• State label: | > = |n_x, n_y, n_z> where the integer indices n_x etc. are the numbers of half-wavelengths in L in each direction.

• Momentum components: p_x = n_xh/2L etc. where h is Planck's constant.

• Kinetic energy: thus = ² (n_x² + n_y² + n_z²) where ² h²/8ML² = (h²/8M²) V^-2/3.

• Pressure: therefore an adiabatic (isentropic) change in the volume of the box will leave the state label () unchanged but will change the energy U₁ = . We can thus use the first definition of the pressure to give p₁ = (2/3) (h²/8M²) V^-5/3 = (2/3) U₁/V.

• Partition function: after a little algebra, a little juggling of summation indices, a conversion of a discrete sum to a continuous integral and a lookup of the Gauss integral in your favourite mathematical tables (unless of course you have it memorized), we obtain Z₁ = V/V_Q where V_Q (2² / M)^3/2 is the so-called quantum volume (a function of temperature) which defines (roughly) the smallest volume to which a particle can be confined and still have this treatment (based on a continuous approximation to the density of states) be valid. The quantum volume is (sort of) the volume of a cube one thermal de Broglie wavelength on a side, where the thermal de Broglie wavelength is (within a slightly ambiguous constant factor of order one) the de Broglie wavelength of a nonrelativistic particle with a kinetic energy of . We may alternatively define the single particle partition function in terms of the quantum concentration n_Q 1/V_Q (M/2²)^3/2 : Z₁ = V n_Q.

• Mean Thermal Energy: Now we can use our differential relation U = ² (log Z)/ to get U₁ = (3/2) .

• Putting together our two expressions for U₁ gives p₁V = , the equation of state for a single particle in abox.

Common sense dictates that if we have N truly non-interacting particles together in the same box, each one will contribute equally to the average thermal energy and the pressure, giving U = (3/2) N and p V = N for the ideal gas. This turns out to be correct; in fact, the equipartition theorem states that, in the high temperature limit, a system will contain /2 of energy for each independent degree of freedom - in the case of a monatomic gas, one for each of x, y and z for each gas atom.

However, students of quantum mechanics have learned not to rely too heavily on the dictates of common sense. Moreover, there are some interesting surprises when we try to decide whether the particles of the gas are distinguishable or truly identical. We shall therefore go through the transition from one particle to many with some caution and rigour.

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Last modified: Mon Feb 3 07:11:03 PST