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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2026 · 22 January · Evening Shift · Physics · Kinetic Theory of Gases

Consider two boxes containing ideal gases A and B such that their...

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}.

Options

  1. A2×1082 \times 10^{8}
  2. B8×1088 \times 10^{8}
  3. C4×1084 \times 10^{8}
  4. D32×10832 \times 10^{8}

Correct answer

C. 4×1084 \times 10^{8}

Step-by-step solution

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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