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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

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**Last time…. Boltzmann factors and probabilities**

Distribution of velocities in an ideal gas

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**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:

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**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

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**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.

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**What is the volume of the thin shell?**

4p v 3/3 2p v dv 4p v 2 dv None of these

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**Maxwell-Boltzmann distribution We had:**

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

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**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

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**! 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

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**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.

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**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:

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**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”

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