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PhysicsHeat & ThermoKTGMedium2 minPYQ_2019

PhysicsMediumsingle choice

A $25 \times 10^{- 3} m^{3}$ volume cylinder is filled with $1$ mol of $O_{2}$ gas at room temperature $(300 K)$ . The molecular diameter of $O_{2}$ , and its root mean square speed, are found to be $0.3 n m$ and $200 m / s$ , respectively. What is the average collision rate (per second) for an $O_{2}$ molecule?

Options:

Answer:

Solution:

Average collision rate $= n \bar{v} d^{2} π$
$= \frac{6.023 \times 10^{23}}{25 \times 10^{- 3}} \times \sqrt{\frac{8}{3 π}} \times 200 \times {(0.3 \times 10^{- 9})}^{2} π \approx 4 \times 10^{- 12}$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2019

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