Question

The speed of a molecule in a uniform gas at equilibrium is a random variable V whose density
function is given by

f(v) = arre-buv > 0,

where and k, T and m denote Boltzmann’s constant, the absolute temperature, and
the mass of the molecule, respectively.

(a) Derive the distribution of , the kinetic energy of the molecule. (10)

(b) Find E (X). (4)

Answer 1

Let Vbe a random variable with density function is 0 otherwise Where, band k, Tand m denote Boltzmann's constant. Here we have to find the value of ‘a, such that, the given density is a proper density. Jo ave-by-dv = 1 Using substitution method, Let v-t Differentiate with respect to vwe get dt dv dt dt ate 27

Using gamma integration we get 2 b2 2b32 r(3/2) 3/2 2b Therefore, the proper density function of Vis given by v(3/2) 0; otherwise a) We have the derive the distribution of W-mV2 /2 nm1 1/2 1/2 Differentiate with respect to w, we get

1/2 I1 Now By the transformation method dh 1/2 r(3/2)" m 2b321 1-1/2 ,2x. ㄧㄨㄧㄨ 10. , w>0 1/2 0; otherwise

nm 2KT 1/2 2w 2AT m Therefore 3/2 0Otherwise bl We have to find E(W) By the definition of W is EW 3/2

using gamma density function :x"-'e-ardr=100 r(5/2) KT (3/2)T(3/2)x KT 312 (3/2)(kT) KT Therefore, E[w]=-kT