In a rigid vessel the absolute temperature of a gas is raised to 4 times and 50% of its molecules are removed. Which option correctly gives the change in (mean free path, number of collisions made by

Question

In a rigid vessel the absolute temperature of a gas is raised to 4 times and 50% of its molecules are removed. Which option correctly gives the change in (mean free path, number of collisions made by one molecule per second, bimolecular collisions per unit volume per second)?

TestHub · Accepted Answer

Let N be the number of molecules per unit volume. V is fixed, so N becomes N/2 and T becomes 4T.

Mean free path is proportional to 1/N (it does not depend on T at constant V): it doubles.

Collisions per molecule per second is proportional to $N T = (1/2) (2) = 1$ : unchanged.

Bimolecular collisions per unit volume = (1/2) N × (collisions per molecule), proportional to $N^{2} T = (1/4) (2) = 1/2$ : it halves.

Answer: (C).

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