Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is
Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is
$\left(\frac{3}{4}\right)^{\text {th }}$ of kinetic energy per unit volume of a gas.
$\left(\frac{2}{3}\right)^{\text {rd }}$ of kinetic energy per unit volume of a gas.
$\left(\frac{1}{3}\right)^{\text {rd }}$ of kinetic energy per unit volume of a gas.
$\left(\frac{3}{2}\right)^{\text {rd }}$ of kinetic energy per unit volume of a gas.
Solution
Pressure exerted by the gas on wall of container is given by,
$\begin{aligned}
\quad P & =\frac{1}{3} \rho v^2 \\
\therefore \quad P & =\frac{1}{3}\left(\frac{M}{V}\right) v^2
\end{aligned}$
$(\mathrm{v}=$ r.m.s. speed $)$
Dividing and multiplying equation by 2 ,
$\begin{aligned}
& P=\frac{2}{3} \quad \frac{1}{2}\left(\frac{M}{V}\right) v^2 \\
& \left(V=\frac{2}{3}\left(\frac{\text { Volume of the gas })}{V}\right)\right. \\
& \therefore \quad \ldots\left(\because \text { K.E. }=\frac{1}{2} M v^2\right)
\end{aligned}$
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