SHOW THAT THE AVERAGE TRANSLATIONAL KINETIC ENERGY OF THE MOLECULES OF A GAS IS DIRECTLY PROPORTIONAL TO ABSOLUTE TEMPERATURE



 QUESTION
Starting from the expression P = r 1/3  show that the Average translational
kinetic energy of the molecules of a gas is directly proportional to absolute temperature.
PROOF
According to the kinetic equation of pressure of a gas:
P = r 1/3 
But r = density of gas
                        r = density of gas = mass of gas / volume of gas
                        r = density of gas = mN / V

Putting the value of density (r)
P = 1/3 (mN / V ) 
V= 1/3 (mN) 
 But PV = nRT
putting the value of PV, we get,
nRT= 1/3 (mN) 
Since
number of mole (n) = molecules/Avogadro’s number
number of mole (n) = N/NA
Therefore,
[N/NA] R T= 1/3 (mN) 
NA R T= 1/3 (m) 
3 [NA R] T= (m) 
But NA R = Boltzman’s constant (K), thus,
3 K T= (m) 
Multiplying both sides by 1/2
(3/2) K T= (1/2) m
(1/2) m = (3/2) K T
But (1/2) m = average translational kinetic energy of gas molecules = (K.E)av
therefore,
(K.E)av = (3/2) K T
As (3/2) K is a constant term, therefore,
(K.E)av = (constant) T