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

Answer: (a). Kinetic theory gives P = 1/3ρ v². Kinetic energy per unit volume E = 1/2ρ v², so ρ v² = 2E. P = 1/3(2E) = 2/3E So the answer is (a) 2/3 of KE per unit volume.

Assuming the expression for the pressure exerted by the gas, it can… | MHT CET 2026 (19 Apr, Shift 2)

Q2MHT CET 2026 · 19 APR · SHIFT 2Easy

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: (a)

Kinetic theory gives $P = \frac{1}{3}\rho \overline{v^2}$ .
Kinetic energy per unit volume $E = \frac{1}{2}\rho \overline{v^2}$ , so $\rho \overline{v^2} = 2E$ .
$P = \frac{1}{3}(2E) = \frac{2}{3}E$
So the answer is (a) $\frac{2}{3}$ of KE per unit volume.

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Molecular Speeds & Maxwell Distribution question from MHT CET 2026 (19 Apr, Shift 2)

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