In Classical gas, why the most probable energy of molecule is not the energy of a molecule moving at the most probable speed?
That is a property of the thermal equilibrium of the ensemble of particles under consideration. For a given temperature, this equilibrium implies a certain distribution of the velocity of the particles, or, what is the same, their energies, that is not δ-function-like (as we know, the distribution of the particles as the function of energy is the Maxwell-Boltzmann distribution, which is exponential). For details regarding thermal equilibrium and the equilibrium distribution functions, please consult the book Statistical Physics, by Landau and Lifshitz (Vol. 5, part 1).
I suppose that you speak of an ideal gas obeying the Maxwell-Boltzmann distribution. Indeed, taking the derivative of the distribution of velocities one gets that the most probable velocity corresponds to an energy of kT. On the other side, to find the distribution of the energies one doesn't just substitute mv2/2 = E, but also takes into account that 2dv/v = dE/E. One finds that the most probable energy is equal to kT/2. Energy and velocity are not equivalent degrees of freedom. On one and the same set of energies assigned to the molecules, one maps many possible sets of velocities, as for each energy are possible two velocities, one positive and one negative. Thus, the distributions of the two degrees of freedom are different.
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