Thermal & Kinetic Lecture 6 Equipartition of energy (….and some problems with classical physics)

1. Thermal & Kinetic Lecture 6 Equipartition of energy (….and some problems with classical physics) LECTURE 6 OVERVIEW Recap…. Equipartition and degrees of freedom Work done by a gas Specific heats

2. Last time…. Boltzmann factors and probabilities Distribution of velocities in an ideal gas Maxwell – Boltzmann distribution (NB see also CW 2)

3. ½ m<v2> = (3/2)kT revisited In our derivation of the ideal gas law, we set the constant of proportionality between mean kinetic energy and temperature as 3k/2 From the Maxwell-Boltzmann distribution function, we can now show why the mean kinetic energy is given by 3k/2 The integral can be found in standard integral tables (see Lecture Notes Set 2b) and leads to the result:

4. E3 E2 U(x) Classically:absolute zero (0 K) corresponds to there being 0 probability of the system having total energy other than zero. Quantum mechanically:0 K corresponds to system in lowest possible energy state (which will have energy > 0). E1 x Absolute temperature Boltzmann’s constant and the preceding analysis provide thelink between mean energy and absolute temperature. What if, instead of having a continuously distributed variable such as velocity, we have discrete, quantised states? How then do we get a ‘definition’ of temperature? We then – as before – mustconsider populations of individual states(statistical mechanics).

5. We’ve found that the mean kinetic energy for the gas molecules is 3kT/2. However, this expression was derived by considering motions of the molecules in the x, y and z directions. We say each molecule has three degrees of freedom Reconsider atom bound in a solid. For small displacements from equilibrium, potential energy curve is that of a simple harmonic oscillator. U(x) Hence, probability of finding system in a given potential energy state is  exp (-mx2/2kT) Using analysis almost identical to that described for kinetic energy (G & P p.443 – 445), we get: ½ m<x2> = ½ kT x Equipartition of Energy and Degrees of Freedom This statement has very important implications for both classical and quantum theory.

6. “What is meant by ‘quadratic term’?” Translational kinetic energy: 3 ‘squared terms’ ½ m<vx2>, ½ m<vy2>, ½ m<vz2>  3(kT/2) ? From the equipartition theorem, what is the average total energy of a simple harmonic oscillator? Rotational energy: ½ I<w2> = ½ kT about an axis Total energy of SHO: 2 ‘squared terms’ ½ m<vx2> + ½ m<x2>  2(kT/2) = kT Equipartition of energy and degrees of freedom Theorem of equipartition of energy “Each quadratic term in the expression for the average total energy of a particle in thermal equilibrium with its surroundings contributes on average ½ kT to the total energy” or “Each degree of freedom contributes an average energy of ½ kT”

7. We’ll first consider the work done by a gas. The gas must expand (or contract) if the pressure is to remain constant when the temperature is changed (PV = nRT) (http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm) Equipartition of energy and specific heats For an ideal gas of monatomic ‘molecules’, equipartition of energy yields the following result for 1 mole of the gas: U = 3NAkT/2 I have used U to represent the total energy of the ideal gas to maintain consistency with the conventional nomenclature in thermodynamics. Note however than I have also previously used U(r) to denote potential energy. Don’t get mixed up between the two. If we heat one mole of gas its temperature rises by a different amount depending on whether the pressure or the volume is fixed. That is: Cp ≠ Cv

8. Work done by a gas It’s clear from the simple cartoon that the piston is pushed out when the gas expands. The question we need to address is: how much work is done by the gas? We assume that the piston is entirely frictionless: all of the work done by the gas goes into pushing the piston back (work done on the surroundings) and not into overcoming frictional forces. That is, the force is conservative. Furthermore, we let the gas expand exceptionally slowly (quasistatically) so that the gas is in thermal equilibrium at all times. The gas expands reversibly.

9. dx Work done by a gas P = F/A  F = PA dW = FdxdW =PAdx  dW = - PdV (reversible) ! NB We adopt the following convention†: Expansion of gas  dW -ve. Work done ONsurroundings. Compression of gas  dW +ve. Work done BYsurroundings. †Note: engineering textbooks tend to adopt the opposite convention.

10. NB This only applies to a reversible process. dQ = dU + PdV CP and CV The molar specific heat at constant volume, Cv, is: (There are a few typographical errors in G&P related to Cv (eqn. 12.46, p.446)) (NB Only for an ideal gas) Conservation of energy: 1st law of thermodynamics ! First law of thermodynamics (just a re-statement of the conservation of energy) dU = W + Q “The total energy of the system is increased either by doing work on the system or by supplying heat to the system.” From the 1st law, when dV = 0 (constant volume process), dQ = dU. Furthermore, for an ideal gas U is a function of T only 

11. We can rewrite the first law (for an ideal gas) as: hence: g : ratio of specific heats CP and CV The specific heat at constant pressure, Cp, is (for 1 mole): Therefore, for an ideal gas: but, from the previous slide:

12. Monatomic gases (Ar, Kr, Xe) have values of g which are (within error) 1.66  However, diatomic gases (N2, O2, CO) have values of g which at RT are ~ ? Can you suggest a reason why g is greater for a diatomic gas than for a monatomic gas? CP, CV and problems with classical theory “Yes, but what has all this got to do with equipartition of energy……?” All the formulae on the previous slide are based on the fact that U = 3NAkT/2 for 1 mole of an ideal gas – this is the fundamental result of the classical equipartition of energy theorem. How well does this theory explain experimental data? ANS: Diatomic molecules have rotational degrees of freedom