# On the size of air molecules It seems more appropriate to think of the atom itself as a more complex object and consider the oxygen or hydrogen atom as.

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On the size of air molecules It seems more appropriate to think of the atom itself as a more complex object and consider the oxygen or hydrogen atom as conglomerate of many smaller particles......to consider the ether shell to explain the processes of live in the finest ramifications of matter. Since its existence manifests itself more and more clearly in the theory of light and electricity it would be hard to understand how it could not play an important role in the smallest elements of organized matter. We leave the decision about the necessity and the value of such extension of the atomistic theory to the facts to be found in future.

 t v xi Kinetic theory of gases Assumptions: 1) large number of particles, elastic collisions 2) large distance compared to size 3) no preferred direction, position Particle i with velocity v i = (v xi,v yi,v zi ) Number of particles in box with v x = v xi : N i Travel distance in x dir. in time  t:  x i =  t v xi Particle reaches wall in  t from vol. V i = A  x i = Av xi  t Number of particles in V i with v x = v xi : N i V i /V Momentum transfer by reflection:  p i = 2mv xi Force: F = dp/dt =  p/  t [momentum transfer/time] Pressure: P = F/A =  p/(A  t)  P i =  p i N i V i /V /(A  t) = 2N i mv xi ² /V (v xi >0) x y z A i Now sum over all particles in box (only half will have positive v x ): P = ½  P i =  (N i mv xi ²/V) = m  (N i v xi ²) /V mean square velocity: (v x ²) av =  (N i v xi ² )/N  P = m (v x ²) av N/V (v x ²) av = (v y ²) av = (v z ²) av and v² = v x ²+ v y ²+ v z ²  (v²) av = 3(v x ²) av  P = 1/3 m(v²) av N/V = 2/3 (½mv²) av N/V = 2/3 (E kin ) av N/V Ideal gas law: PV = N/N A RT  2/3 (E kin ) av = R/N A ·T  (E kin ) av = 3/2 kT k = 1.381×10 -23 J/K = 8.617×10 -5 eV/K Boltzmann constant -Temperature is measure for kinetic energy -T = 300 K  kT = 1 eV/40 -Factor ‘3’ comes from 3 dimensions of space (v x,v y,v z ), i.e. 3 independent possibilities to distribute E: 3 degrees of freedom

Probability distributions – finding an average Example: Midterm Results # of students / (1 point bin) Total number of students: 0 23  (# of students) = 62 Average: (Total # of points)/(Number of students)  (points)(# of students)  (# of students) = = =15.4 952.5 62 # of students / (5 % bin) Probability: P(percentage)/(5 % bin) (# of students)/(5 % bin)  (# of students) = Average:  ( Percentage )xP( percentage )/(5 % bin)] = 67 % …from 2006

Coordinate Transformation

vxvx vyvy [v, v+dv] [v x, v x +dv x ] 2  v dv [v y, v y +dv y ] In 3 dimensions:  v²dv Maxwell-Boltzmann Distribution

F(v) = (m/(2  kT)) 3/2 4  v² exp(-mv²/2kT) N 2 molecules Maxwell-Boltzmann Distribution

vxvx vyvy #/  v P(E x ) = A e -E x /kT P(E y ) = A e -E y /kT Add energy e.g. kinetic energy in x direction convert E x to E y gain high energies This distribution will be found in any system which is in thermal equilibrium Particles in a Box

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