Confusion Regarding Most Probable Kinetic Energy of an Ideal gas

  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
35,134
11,372
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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