The average speed of the molecules of an ideal gas at − 2 3 ∘ C is 0.5 m/s. The average speed at 72 7 ∘ C will be -

u a vg ∝ T . T 1 = 250 K, T 2 = 1000 K. u 2 / u 1 = 1000/250 = 2 . u 2 = 2 × 0.5 = 1.0 m / s . Answer: (C).

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ChemistryIdeal GasKinetic Theory of GasesEasy2 minQB

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The average speed of the molecules of an ideal gas at $- 2 3^{\circ} C$ is 0.5 m/s. The average speed at $72 7^{\circ} C$ will be -

Answer:

Solution:

$u_{a vg} \propto T . T_{1} = 250$ K, $T_{2} = 1000$ K.

$u_{2} / u_{1} = 1000/250 = 2$ .

$u_{2} = 2 \times 0.5 = 1.0 m / s$ .

Answer: (C).

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Kinetic Theory of Gases

⏱ 2mℹ️ Source: QB

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