Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then

Options

A. No molecule can have a speed greater than
B. No molecule can have speed less than
C. `v_p lt vecv lt V_("m s")`
D. The average kinetic energy of a molecule is

Answer

C, D

Explanation

`V_("m s") = sqrt((3RT)/M),barv = sqrt((8)/pi (RT)/M) = sqrt((2.5 RT)/M )"and "V_rho = sqrt((2RT)/M)`
From these expression we can see that `v_p lt bar V lt v_("m s")`
Second, `V_("m s" ) = sqrt(3/2) v_P` and average kinetic energy of a gas molecule
`=1/2 "m v"_("m s")^2 = 1/2 m (sqrt(3/2)v_rho)^2 = 3/4 "m v"_rho^2`