Question

What is the average kinetic energy of a molecule at (a) 0°C and (b) 500°c? [5.66 x 10.21 12.64 x 1020 1)

Answer 1

ANSWER:

Boltzman constant, K_B = 1.38 x 10^-23 J/K

(a) 0 ^oC

T = 0 ^oC = 273.15 K

At this temperature only translation degree of freedom is taken into consideration. So,

Av. kinetic energy of a molecule = (3/2)K_BT

= (3/2) x 1.38 x 10^-23 J/K x 273.15 K

= 5.66 x 10^-21 J

Hence, the average kinetic energy of a molecule at 0 ^oC is 5.66 x 10^-21 J.

(b) 500 ^oC

T = 500 ^oC = 773.15 K

At this temperature rotational and translation degree of freedom both taken into the consideration. So,

Av. kinetic energy of a molecule = (5/2)K_BT

= (5/2) x 1.38 x 10^-23 J/K x 773.15 K

= 2.66 x 10^-20 J

Hence, the average kinetic energy of a molecule at 500 ^oC is 2.66 x 10^-20 J.