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(A) According to the kinetic theory of gases, the pressure $P$ exerted by an ideal gas is given by the expression:$P = \frac{1}{3} \rho v_{rms}^2$wher

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

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(A) According to the kinetic theory of gases, the pressure $P$ exerted by an ideal gas is given by the expression:$P = \frac{1}{3} \rho v_{rms}^2$where $\rho$ is the density of the gas and $v_{rms}$ is the root mean square velocity.We know that the kinetic energy per unit volume $E$ is given by:$E = \frac{1}{2} \rho v_{rms}^2$From this, we can write $\rho v_{rms}^2 = 2E$.Substituting this into the pressure expression:$P = \frac{1}{3} (2E) = \frac{2}{3} E$Thus, the pressure is $(2/3)$ rd of the kinetic energy per unit volume of the gas.

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

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