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Re: Re Thermal Energy



In a message dated 3/8/02 5:49:44 PM Eastern Standard Time, jsd@MONMOUTH.COM
writes:

<<
Perhaps another question worth asking is since the temperature of ,say an
ideal gas, is proportional to the average kinetic energy of it's components

T= KE_ave/((3/2)*K where K is Boltzmann's constant
does the ball's increased kinetic energy cause it's temperature to
increase.

But that formula isn't right. That formula wrongly includes
nonthermal energy. Hint: There is such a thing as a
Maxwell-Boltzmann distribution, which differs from the
plain old Boltzmann distribution by a uniform shift in velocity.
>>
As a further clarification and perhaps to clear up any misconceptions I may
have lets look at the above equation again. This equation is based on the
theorem of equipartition of energy which states that the energy of a system
in thermal equilibrium is equally divided among all degrees of freedom.

Therefore

m*v_x^2=K*T m*v^_y^2=*K*T and m*v_z^2=K*T so that we can
say
KE_ave=3/2*K*T We can calculate the RMS velocity using this
equation which is

V_rms=sqrt[3*K*T/m]

In equilibrium I have always thought this was a pretty good description of
the energy distribution using a Boltzmann's distribution curve. However
Maxwell's contribution to the analysis of the distribution of molecular
speeds in a gas provides a far more sophisticated and accurate description. I
presume this is what you are referring to.
The curve is given by the equation

N_v=((4*PI*(m/(2*Pi*K*T)^3/2
)*v^2)*EXP[-m*v^2/2*K*T]

Where N_v is the fraction of molecules at velocity v. This function provides
an accurate picture of the velocity distribution of the molecules of an ideal
gas. However, as far as I know this more accurate description of the energy
distribution of the molecules in a gas has no effect on the basic
relationship between translational kinetic energy and temperature. Am I
wrong?


Bob Zannelli