Confusion Regarding Most Probable Kinetic Energy of an Ideal gas

AI Thread Summary
The discussion centers on finding the most probable kinetic energy of an ideal gas molecule, with the standard solution indicating it is kT/2 through differentiation of the probability distribution. A participant expresses confusion after calculating kinetic energy using the most probable velocity, resulting in kT instead. The key point clarified is that the most probable value of velocity does not directly translate to the most probable value of kinetic energy due to the nonlinear nature of the kinetic energy equation. This highlights a common misunderstanding in statistical mechanics regarding probability distributions. Understanding this distinction is crucial for correctly applying the principles of kinetic theory.
Sunil Simha
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Homework Statement



Find the most probable kinetic energy of an ideal gas molecule.

Homework Equations


v=sqrt(2RT/M)
where v= most probable velocity
k= Boltzmann's constant
T= temperature of the system
M= molar mass of the gas

Maxwell's formula of probability distribution of kinetic energy

The Attempt at a Solution



The standard solution, as prescribed by many textbooks and reference materials, involves differentiation of the probability distribution equation w.r.t. kinetic energy to find the maximum and hence showing that the most probable kinetic energy is kT/2. This seems logical and I assume it is correct.

But my doubt arises due to a different approach that I employed. Instead of differentiating the probability distribution equation if I simply plug in the value of most probable velocity in the equation for kinetic energy (mv^2/)2 then my answer turns out to be kT. I'm confused as isn't it logical to assume that the molecules with most probable velocity have the most probable kinetic energy. Please help.
 
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The most probable value of v is not the most probable value of v^2. This is an unintuitive mathematical property, and can be seen for other nonlinear functions as well.
 
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