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Thermal & Kinetic Lecture 7 Maxwell-Boltzmann distribution, Equipartition of energy (….and some problems with classical physics) Deriving the Maxwell-Boltzmann distribution function LECTURE 7 OVERVIEW 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: So, our velocity distribution in one dimension is: 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:? Gaussian function, = 0 ) 2 1 exp( 2 1 )( 2 x xg

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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 vzvz vyvy vxvx

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Surface element, dS, shown – need to consider volume element, dV vzvz vyvy vxvx 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 ? Molecular speeds: Polar coordinates Consider thin shell of sphere whose radius changes from v to v + dv. ? What is the volume of this thin shell?

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What is the volume of the thin shell? a)4 v 3 /3 b)2 v dv c)4 v 2 dv d) 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 N 2 molecules 293 K 600 K ! Maximum not at v=0; most probable speed is less than mean speed; (CW 2) curve broadens as T increases

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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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½ m = ( 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: Typographical errors in Section 2.6: (i) Boltzmann’s (not Planck’s) constant!; (ii) Factor of 4 missing from Eqn Apologies.

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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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