Thermal & Kinetic Lecture 7

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Thermal & Kinetic Lecture 7
Maxwell-Boltzmann distribution, Equipartition of energy (….and some problems with classical physics) LECTURE 7 OVERVIEW Deriving the Maxwell-Boltzmann distribution function Equipartition and degrees of freedom

Last time…. Boltzmann factors and probabilities
Distribution of velocities in an ideal gas

Velocity distribution – TAKE NOTES
Consider velocities of molecules first: Gaussian function, <vx> = 0 ) 2 1 exp( ( ÷ ø ö ç è æ - = s m p x g How might we determine what the constant, A, should be? [See Q7(b) of the ’04-’05 Thermal and Kinetic paper for a similar question]. Given: ? So, our velocity distribution in one dimension is:

Distribution of molecular speeds
We need to do a few more steps to get a formula for the distribution of molecular speeds. First, we can combine the expressions for molecular velocities to get: This expression is written in Cartesian co-ordinates (x,y,z). Switch to spherical polar coordinates. Spherical polar coords q vz vy vx f

Molecular speeds: Polar coordinates
q vz vy vx f Surface element, dS, shown – need to consider volume element, dV Considering all directions, the tip of the velocity vector ‘sweeps’ out a spherical volume (only one ‘octant’ shown above). How many velocity states within v and v + dv ? ? What is the volume of this thin shell? Consider thin shell of sphere whose radius changes from v to v + dv.

What is the volume of the thin shell?
4p v 3/3 2p v dv 4p v 2 dv None of these

Taking into account discussion of spherical polar coordinates: Maxwell-Boltzmann distribution of speeds of molecules in a gas at thermal equilibrium.

Maxwell-Boltzmann distribution
To convert from velocity to speed we have carried out two steps: 1. Convert from 1D to 3D probability. 2. Consider all directions. Function no longer a Gaussian! 293 K 600 K

! Maxwell-Boltzmann distribution
Maxwell-Boltzmann distribution for N2 molecules ! Maximum not at v=0; most probable speed is less than mean speed; (CW 2) curve broadens as T increases 293 K 600 K

Maxwell-Boltzmann distribution 293 K 600 K
Shaded part of graph gives fraction of molecules with speeds between 500 and 1000 ms-1. (Integrate under curve with appropriate limits). Coursework Set 2 includes a number of questions on this distribution function.

Typographical errors in Section 2
Typographical errors in Section 2.6: (i) Boltzmann’s (not Planck’s) constant!; (ii) Factor of 4p missing from Eqn Apologies. ½ m<v2> = (3/2)kT 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:

Equipartition of Energy and Degrees of Freedom
This statement has very important implications for both classical and quantum theory. 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 that each molecule has three 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”