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(B) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as:$P = \frac{1}{3} \rho v_{rms}^2$where $\rho = \frac{M}{V}$ is

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$(2/3)$ of kinetic energy per unit volume of a gas.

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(B) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as:$P = \frac{1}{3} \rho v_{rms}^2$where $\rho = \frac{M}{V}$ is the density of the gas and $v_{rms}$ is the root-mean-square speed.Substituting $\rho$ in the equation:$P = \frac{1}{3} \left( \frac{M}{V} \right) v_{rms}^2$Multiply and divide the right side by $2$:$P = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} v_{rms}^2 \right)$Since the total kinetic energy $K.E. = \frac{1}{2} M v_{rms}^2$,the kinetic energy per unit volume is $u = \frac{K.E.}{V} = \frac{1}{2} \rho v_{rms}^2$.Therefore,$P = \frac{2}{3} u$.

Assuming the expression for the pressure exerted by the gas,it can be shown that pressure is

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