Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is

Options

A. `[1/3]^(rd)`kinetic energy per unit volume of a gas
B. `[2/3]^(rd)`kinetic energy per unit volume of a gas
C. `[3/4]^(th)`kinetic energy per unit volume of a gas
D. `3/2xx` kinetic energy per unit volume of a gas

Answer

B

Explanation

According to kinetic theory of gases, the pressure exerted by the gas on the walls of container is
`p=p_o + p_1+p_2`
i.e. `p=p_o +1/(3)8V^2+3gh`
For a container,
`p_1=1/(3)rhov^2=1/3m/v.v^2xx2/2=2/(3).(1)/2mv^2.(1)/v`
`2/(3v)KE` `[because KE=1/(2)mv^2]`