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Question

If molecular diameter of gas A is double of gas B, then mean free path of gas A for same molecular density is

A
14 times of gas B
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B
12 times of gas B
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C
4 times of gas B
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D
2 times of gas B
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Solution

The correct option is A 14 times of gas B
Given,
Diameter of gas A=2× diameter of gas B
dA=2dB(1)
We know, mean free path (λ)
λ=12nπd2
n molecular density
For constant value of n
λ1d2
For gas B, λB1d2B(2)
For gas A, λA14d2B [From (1)]
λA=λB4 [From (2)]
Mean free path of gas A is 14 times of gas B.

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