Kinetic Theory

JEE Main level, single correct, medium

7 of 168

Question 7 of 168

Consider two boxes containing ideal gases AA and BB such that their temperatures, pressures and number densities are same. The molecular size of AA is half of that of BB and mass of molecule AA is four times that of BB. If the collision frequency in gas BB is 32×1018/s32 \times 10^{18} / \mathrm{s} then collision frequency in gas AA is ____\_\_\_\_ /s/ \mathrm{s}.

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Answer: (C) 4×1084 \times 10^{8}

Collision frequency (z)=2πd2N8RTπM(z) = \sqrt{2} \pi d^2 N \sqrt{\dfrac{8RT}{\pi M}}
Temp, N are same
Z∝d2MZ \propto \dfrac{d^2}{\sqrt{M}}
dA=dB2d_A = \dfrac{d_B}{2}
MA=4MBM_A = 4M_B
ZAZB=dA2dB2×MBMA=(MBMA)(dAdB)2\dfrac{Z_A}{Z_B} = \dfrac{d_A^2}{d_B^2} \times \dfrac{\sqrt{M_B}}{\sqrt{M_A}} = \left(\sqrt{\dfrac{M_B}{M_A}}\right) \left(\dfrac{d_A}{d_B}\right)^2
=(14)(12)2= \left(\sqrt{\dfrac{1}{4}}\right) \left(\dfrac{1}{2}\right)^2
ZAZB=12×14=18\dfrac{Z_A}{Z_B} = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}
⇒ZA=32×1088=4×108/s\Rightarrow Z_A = \dfrac{32 \times 10^8}{8} = 4 \times 10^8 /s

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