Each molecule of a gas registers 5 × 1 0 9 collisions per second with other molecules. If the gas contains 2 × 1 0 25 molecules per m 3 , the number of bimolecular collisions per m 3 per second is -

Collisions counted per molecule × molecules 5 × 1 0 9 × 2 × 1 0 25 = 1 × 1 0 35 . Each collision is shared by two molecules, so divide by 2. Collision frequency 5 × 1 0 34 m − 3 s − 1 . Answer: (C).

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Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasCollision ParametersEasy2 minQB

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Each molecule of a gas registers $5 \times 1 0^{9}$ collisions per second with other molecules. If the gas contains $2 \times 1 0^{25}$ molecules per $m^{3}$ , the number of bimolecular collisions per $m^{3}$ per second is -

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Answer:

Solution:

Collisions counted per molecule × molecules $5 \times 1 0^{9} \times 2 \times 1 0^{25} = 1 \times 1 0^{35}$ .

Each collision is shared by two molecules, so divide by 2.

Collision frequency $5 \times 1 0^{34} m^{- 3} s^{- 1}$ .

Answer: (C).

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Collision Parameters

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