At 27 temperature, the mean kinetic energy of the atoms of an ideal gas is E₁. If the temperature is increased by 300 then the mean kinetic energy of the atoms will be (a) (E₁)/(√(2)) (b) √(2)E₁ (c) 2E₁ (d) E₁/2

Answer: (c). Mean KE ∝ T (kelvin). T₁ = 300 K; increasing by 300° gives T₂ = 600 K. E₂ = E₁ × (600)/(300) = 2E₁ So the answer is (c) 2E₁.

At 27 temperature, the mean kinetic energy of the atoms of an ideal… | MHT CET 2026 (12 May, Shift 2)

Q2MHT CET 2026 · 12 MAY · SHIFT 2Easy

Pressure & Kinetic Energy of a Gas

At $27^\circ\text{C}$ temperature, the mean kinetic energy of the atoms of an ideal gas is $E_1$ . If the temperature is increased by $300^\circ\text{C}$ then the mean kinetic energy of the atoms will be

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Answer: (c)

Mean KE $\propto T$ (kelvin).
$T_1 = 300$ K; increasing by $300^\circ$ gives $T_2 = 600$ K.
$E_2 = E_1 \times \frac{600}{300} = 2E_1$
So the answer is (c) $2E_1$ .

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