Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half that of B and mass of molecule A is four times that of B. If the collision frequency

Question

Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $32 \times 10^{8} s^{- 1}$ then collision frequency in gas A is____________/s [2026]

Smart Achievers · Accepted Answer

(4)

$Collision frequency (Z) = \sqrt{2} π d^{2} N \sqrt{\frac{8 R T}{π M}}$

Temp, N are same

$Z \propto \frac{d^{2}}{\sqrt{M}}$

$d_{A} = \frac{d_{B}}{2}$

$M_{A} = 4 M_{B}$

$\frac{Z_{A}}{Z_{B}} = \frac{d_{A}^{2}}{\sqrt{M_{A}}} \times \frac{\sqrt{M_{B}}}{d_{B}^{2}} = (\sqrt{\frac{M_{B}}{M_{A}}}) {(\frac{d_{A}}{d_{B}})}^{2}$

$= (\sqrt{\frac{1}{4}}) {(\frac{1}{2})}^{2}$

$\frac{Z_{A}}{Z_{B}} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

$\Rightarrow Z_{A} = \frac{32 \times 10^{8}}{8} = 4 \times 10^{8} s^{- 1}$

Solution

1 $32 \times 10^{8}$

2 $8 \times 10^{8}$

3 $2 \times 10^{8}$

4 $4 \times 10^{8}$

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Solution

1 32×108

2 8×108

3 2×108

4 4×108

Ans. (4) Collision frequency (Z)=2πd2N8RTπM Temp, N are same Z∝d2M dA=dB2 MA=4MB ZAZB=dA2MA×MBdB2=(MBMA)(dAdB)2 =(14)(12)2 ZAZB=12×14=18 ⇒ZA=32×1088=4×108s-1

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1 $32 \times 10^{8}$

2 $8 \times 10^{8}$

3 $2 \times 10^{8}$

4 $4 \times 10^{8}$