Pressure exerted by an ideal gas on the walls of the container is equal to a) mean kinetic energy per unit volume. b) half of mean kinetic energy per unit volume. c) one-third of mean kinetic energy per unit volume. d) two-third of mean kinetic energy per unit volume.

Step-by-step solution

The pressure exerted by an ideal gas is given by P = 1 3 v_ rms ^2 , where is the density of the gas and v_ rms is the root mean square velocity. The mean kinetic energy per unit volume is given by E = 1 2 v_ rms ^2 . From these two equations, we can write P = 2 3 ( 1 2 v_ rms ^2 ) = 2 3 E . Therefore, the pressure exerted by an ideal gas is equal to two-thirds of the mean kinetic energy per unit volume. Answer: d