MHT CET · Physics · Kinetic Theory of Gases

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

A \(\left(\frac{3}{4}\right)^{\text {th }}\) of kinetic energy per unit volume of a gas.
B \(\left(\frac{2}{3}\right)^{\text {rd }}\) of kinetic energy per unit volume of a gas.
C \(\left(\frac{1}{3}\right)^{\text {rd }}\) of kinetic energy per unit volume of a gas.
D \(\left(\frac{3}{2}\right)^{\text {rd }}\) of kinetic energy per unit volume of a gas.

Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(\frac{2}{3}\right)^{\text {rd }}\) of kinetic energy per unit volume of a gas.

Step-by-step Solution

Detailed explanation

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 ,
\(\begin{aligned}
& P=\frac{2}{3} \quad \frac{1}{2}\left(\frac{M}{V}\right) v^2 \\
& \left(V=\frac{2}{3}\left(\frac{\text { Volume of the gas })}{V}\right)\right. \\
& \therefore \quad \ldots\left(\because \text { K.E. }=\frac{1}{2} M v^2\right)
\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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