Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is (a) (3/4) th of kinetic energy per unit volume of a gas. (b) (2/3) rd of kinetic energy per unit volume of a gas. (c) (1/3)rd of kinetic energy per unit volume of a gas. (d) (3/2) rd of kinetic energy per unit volume of a gas.

Answer: (b). P = 1/3ρ v² and KE per unit volume E = 1/2ρ v². P = 1/3(2E) = 2/3E So the answer is (b) 2/3 of KE per unit volume.

Assuming the expression for the pressure exerted by the gas, it can… | MHT CET 2024 (16 May, Shift 1)

Q22MHT CET 2024 · 16 MAY · SHIFT 1Easy

Molecular Speeds & Maxwell Distribution

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

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Answer: (b)

$P = \frac{1}{3}\rho \overline{v^2}$ and KE per unit volume $E = \frac{1}{2}\rho \overline{v^2}$ .
$P = \frac{1}{3}(2E) = \frac{2}{3}E$
So the answer is (b) $\frac{2}{3}$ of KE per unit volume.

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