Kinetic Theory

JEE Main level, single correct, medium

30 of 168

Question 30 of 168

The mean free path of molecules of a certain gas at STP is 1500d1500d, where ddis the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373K373K is approximately:

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Answer: (C) 2049d2049d

The mean free path is given by the relation,
λ=RT2πd2NAP\lambda =\dfrac{RT}{\sqrt{2}\pi {d}^{2}{N}_{A}P}
So, λ∝T\lambda \propto T
Let λ′\lambda ' be the new mean free path.
Hence,
1500dλ′=273373\dfrac{1500d}{\lambda '}=\dfrac{273}{373}
⇒λ′=2049.45d≈2049d\Rightarrow {\lambda }^{'}=2049.45d\approx 2049d

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