Question

Some ideal monoatomic gas A in an enclosure has a pressure P and the temparature T. Another ideal monoatomic gas B enclosed in a container of the same volume has a pressure 2P and temparature T/2. the ratio of average kinetic energy of molecule of gas A to gas B is

Answer 1

Given :

Temperature of gas A, Ta = T

Temperature of gas B, Tb = T/2

To find :

The ratio of the average kinetic energy of molecule of gas A to gas B.

Solution :

Average kinetic energy = (3/2) * k * (temperature)

average kinetic energy of gas A = (3/2) * k * T

average kinetic energy of gas B = (3/2) * k * (T/2)

ratio of the average kinetic energy of gas A to gas B

                        = [ (3/2) * k * T ] / [ (3/2) * k * (T/2) ]

                        = 2 : 1

The ratio of the average kinetic energy of molecule of gas A to gas B is 2:1 .