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I already read about Maxwell's velocity distribution law for gas molecule. And the expression for that distribution is following dnc=4πnA^3e^(-bc^2)c^2dc Now if we assume that the molecules have no intermolecular interaction means no potential energy so total energy is wholly kinetic so E =(1/2)mc^2 So putting the value in the Maxwell's distribution we get the expression for no. Of molecule within the energy range E to E+dE and that is dnE=2n/(√π(KT)^(3/2))√E e^{-E/KT}dE Now when I evaluate average energy it is (3/2)kT but problem is when I evaluate the kinetic energy correspondence to average velocity it is not same with the average energy (we considered the total energy is wholly kinetic) So why get two different values between average energy and kinetic energy correspondence to average velocity????

Similaly the most probable energy is not same with the kinetic energy correspondence to most probable velocity so why we get different values?????

honeste_vivere
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user101134
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Yes, these are three different values. But why do you call it a problem? This is just so.

Take for example three numbers: 1, 1 and 4. The average of the squares (energy) is 6, the square of the average is 4, and the square of the most probable value is 1.

There is no problem :)

Ilja
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  • No I am not saying that.I want to say that the average energy is not same with kinetic energy correspondence to average velocity why it is so? – user101134 Apr 08 '16 at 19:06