InterviewSolution
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InterviewSolution
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A gas consists of molecules of mass `m` and is at a temperature `T`. Making use of the Maxwell velocity distribution function, find the corresponding distribution of the molecules over the kinetic energies `epsilon`. Determine the most probable value of a gas the kinetic energy `epsilon_p`. Does `epsilon_p` correspond to the most probable velocity ? |
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Answer» `dN (v) = N((m)/(2 pi kT))^(3//2) e^(-mv^2//2kT) 4 pi v^2 dv = dN(epsilon) = (dN(epsilon))/(d epsilon) d epsilon` or, `(dN(epsilon))/(d epsilon) = N((m)/(2 pi kT))^(3//2) e^(-mv^2//2RT) 4 pi v^2 (dv)/(d epsilon)` Now, `epsilon = (1)/(2) mv^2` so `(dv)/(d epsilon) = (1)/(mv)` or, `(dN(epsilon))/(d epsilon) = N((m)/(2 pi kT))^(3//2) e^(- epsilon//kT) 4 pi sqrt((2 epsilon)/(m))(1)/(m)` =`N (2)/(sqrt(pi))(kT)^(-3//2) e^(-epsilon//kT) epsilon^(1//2)` i.e., `dN (epsilon) = N(2)/(sqrt(pi)) (kT)^(-3//2) e^(- epsilon//kT) d epsilon` The most probable kinetic energy is given from `(d)/(d epsilon) (dN(epsilon))/(d epsilon) = 0` or, `(1)/(2) epsilon^(-1//2) e^(-epsilon//kT) - (epsilon^(1//2))/(kT) e^(epsilon//kT) = 0` or `epsilon = (1)/(2) kT = epsilon_(p r)` The corresponding velocity is `v = sqrt((kT)/(m)) = v_(pr)`. |
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