A non-interacting gas has speed density $$f(v)dv=4\pi v^2\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/(2k_BT)}dv.$$ In its corresponding particle-energy distribution, find the most probable $E/(k_BT)$, rounded to one decimal place.
Use $E=mv^2/2$ and $dv/dE=1/(mv)$. The energy density is $g(E)=f(v)|dv/dE|\propto E^{1/2}e^{-E/(k_BT)}$. Its logarithmic derivative is $1/(2E)-1/(k_BT)$, which vanishes at $E=k_BT/2$. Hence the required dimensionless value is 0.5. Transforming the density, including its Jacobian, is essential.
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