Question

The pressure exerted by a perfect gas is equal to
A. mean kinetic energy per unit volume.
B. half of mean kinetic energy per unit volume.
C. two-thirds of mean kinetic energy.
D. two-third of mean kinetic energy per unit volume.

Answer 1

Hint: The pressure exerted by an ideal gas is numerically equal to two-third of the mean kinetic energy of translation per unit volume of the gas.

Complete step by step solution:
The mean KE is given by for one $U = {c_v}T = {H_2}RT$
Mole of gas $ = \dfrac{3}{2}RT$
Since, $P = \dfrac{{hRT}}{V},$thus
$P = \dfrac{2}{3}\left( {\dfrac{U}{V}} \right)$

Hence, pressure is two-third of mean KE per unit volume.
Hence, option (D) is correct.

Additional Information: The total pressure of mixture of gases an be defined the sum of pressures of each individual gas:
\[{P_T} = {P_1} + {P_2} + ..... + {P_n}.\] The partial pressure of an individual gas is equal to the total press multiplied by the mole fraction of the gas.

Note: The pressure exerted by the gas is due to the continuous collision of the molecules against the walls of the container due to this continuous collision the walls experience a continuous force which is equal to the total momentum imparted to the walls per second.