SHOW THAT THE AVERAGE TRANSLATIONAL KINETIC ENERGY OF THE MOLECULES OF A GAS IS DIRECTLY PROPORTIONAL TO ABSOLUTE TEMPERATURE
SHOW THAT THE AVERAGE TRANSLATIONAL KINETIC ENERGY OF THE MOLECULES OF A GAS IS DIRECTLY PROPORTIONAL TO ABSOLUTE TEMPERATURE
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.
According to the kinetic equation of pressure of a gas:
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 V= 1/3 (mN)
But PV = nRT
putting the value of PV, we get,
Since
number of mole (n) = molecules/Avogadro’s number
number of mole (n) = N/NA Therefore,
NA R T= 1/3 (m)
3 [NA R] T= (m)
But NA R = Boltzman’s constant (K), thus,
Multiplying both sides by 1/2
But (1/2) m = average translational kinetic energy of gas molecules = (K.E)av
therefore,
As (3/2) K is a constant term, therefore,