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Kinetic Theory of Gases and Radiation — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsKinetic Theory of Gases and Radiation3 Marks
Question
Assuming the expression for the pressure $\mathrm{P}$ exerted by an ideal gas, prove that the kinetic energy per unit volume of the gas is $\frac{3}{2} P$.
✓

Answer


According to the kinetic theory of gases, the pressure $\mathrm{P}$ exerted by the gas is
$
\mathrm{P}=\frac{1}{3} \rho v_{\mathrm{rms}}^2=\frac{1}{3} \frac{M}{V} v_{\mathrm{rms}}^2
$
where $v_{\mathrm{rms}}$ is the rms speed (root-mean-square speed) of the gas molecules; $M, V$ and $\rho$ are the mass, volume and density of the gas, respectively.
$
\therefore P V=\frac{1}{3} M v_{r m s}^2=\frac{2}{3}\left(\frac{1}{2} M v_{r m s}^2\right)
$
Here, $\frac{1}{2} M v_{\text {rms }}^2$ is the total kinetic energy (KE) of all molecules of the gas.
$
\begin{aligned}
& \therefore P V=\frac{2}{3}(\mathrm{KE}) \\
& \therefore \mathrm{KE}=\frac{3}{2} P V \\
& \therefore \frac{\mathrm{KE}}{V}=\frac{3}{2} P
\end{aligned}
$
Thus, the kinetic energy per unit volume of an ideal gas is $\frac{3}{2} P$.

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