The relationship between the kinetic energy p | vedclass

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Question

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

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$P = \frac{2E}{3}$

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(A) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the relation: $P = \frac{1}{3} \rho \langle v^2 \rangle$,where $\rho$ is the density and $\langle v^2 \rangle$ is the mean square speed.We can rewrite this as $P = \frac{2}{3} \left( \frac{1}{2} \rho \langle v^2 \rangle \right)$.The kinetic energy per unit volume $(E)$ is defined as the total kinetic energy divided by the volume $V$,which is $E = \frac{1}{2} \frac{M}{V} \langle v^2 \rangle = \frac{1}{2} \rho \langle v^2 \rangle$.Substituting this into the pressure equation,we get $P = \frac{2}{3} E$.

The relationship between the kinetic energy per unit volume $(E)$ and the pressure $(P)$ for a given gas is .....

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