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(B) The pressure $p$ exerted by an ideal gas is given by the kinetic theory of gases as:$p = \frac{1}{3} \rho \bar{v}^2$where $\rho$ is the density an

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$\frac{2 E}{3}$

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(B) The pressure $p$ exerted by an ideal gas is given by the kinetic theory of gases as:$p = \frac{1}{3} \rho \bar{v}^2$where $\rho$ is the density and $\bar{v}^2$ is the mean square speed.Since density $\rho = \frac{M}{V}$,we can write:$p = \frac{1}{3} \frac{M}{V} \bar{v}^2$Multiplying and dividing by $2$,we get:$p = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} \bar{v}^2 \right)$Here,$\frac{1}{2} M \bar{v}^2$ is the total kinetic energy of the gas molecules.Thus,$\frac{1}{2} \frac{M}{V} \bar{v}^2$ represents the kinetic energy per unit volume,which is given as $E$.Therefore,$p = \frac{2}{3} E$.

Kinetic energy per unit volume is $E$. The pressure exerted by the gas is given by

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