Presentation on theme: "Thermal & Kinetic Lecture 7"— Presentation transcript:
1Thermal & Kinetic Lecture 7 Maxwell-Boltzmann distribution, Equipartition of energy(….and some problems withclassical physics)LECTURE 7 OVERVIEWDeriving the Maxwell-Boltzmann distribution functionEquipartition and degrees of freedom
2Last time…. Boltzmann factors and probabilities Distribution of velocities in an ideal gas
3Velocity distribution – TAKE NOTES Consider velocities of molecules first:Gaussian function, <vx> = 0)21exp((÷øöçèæ-=smpxgHow 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:
4Distribution 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.Sphericalpolarcoordsqvzvyvxf
5Molecular speeds: Polar coordinates qvzvyvxfSurface element, dS, shown – need to consider volume element, dVConsidering 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 spherewhose radius changes fromv to v + dv.
6What is the volume of the thin shell? 4p v 3/32p v dv4p v 2 dvNone of these
7Maxwell-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.
8Maxwell-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 K600 K
9! Maxwell-Boltzmann distribution Maxwell-Boltzmann distribution for N2 molecules!Maximum not at v=0;most probable speed is lessthan mean speed; (CW 2)curve broadens as T increases293 K600 K
10Maxwell-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.
11Typographical 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)kTIn our derivation of the ideal gas law, we set the constant of proportionality between mean kinetic energy and temperature as 3k/2From the Maxwell-Boltzmann distribution function, we can now show why the mean kinetic energy is given by 3k/2The integral can be found in standardintegral tables (see Lecture Notes Set 2b) and leads to the result:
12Equipartition of Energy and Degrees of Freedom This statement has very important implications for bothclassical and quantum theory.We’ve found that the mean kinetic energy for the gas molecules is3kT/2. However, this expression was derived by considering motionsof the molecules in the x, y and z directions.We say that each molecule has three degrees of freedomTheorem 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”