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Question

For a given gas the mean free path at a particular pressure is


A
independent of temperature
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B
decrease with rise in temperature.
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C
increases with rise in temperature.
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D
directly proportional to T2
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Solution

The correct option is A increases with rise in temperature.
Mean Free Path $$\mathrm{(\lambda_{mean}) = \cfrac{RT}{\sqrt2 \pi d^2 N_A P}}$$

where, 
R is ideal gas constant
T is the temperature in Kelvin
d is the collision diameter
$$\mathrm{N_A}$$ is the Avogadro's no.
P is pressure

So, the mean free path of a gas increases with
1) increase in temperature
2) decrease in pressure

Hence, Option "C" is the correct answer.

Chemistry

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