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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 nm and 200 m/s, respectively.

What is the average collision rate ( per second) for an $O_{2}$ molecule?

Option 1)
$\approx 10^{12}$

Option 2)
$\approx 10^{11}$

Option 3)
$\approx 10^{10}$

Option 4)
$\approx 10^{13}$

Answers (1)

best_answer

Formula for mean free path -

$Y= \frac{KT}{\sqrt{2}\pi \sigma ^{2}p}$

- wherein

$\sigma =$ Diameter of the molecule

$p =$ pressure of the gas

$T =$ temperature

$K =$ Boltzmann's Constant

$V_{rms}=200m/s$

$D=0.3nm$

$n=1mole$

$volume=25\times 10^{-3}m^{3}$

$V_{av}=V_{rms}\sqrt{\frac{8}{3\pi}}$

f = collision rate (per second)

$=\frac{V_{rms}}{RT}\sqrt2\pi d^{2}N_A\cdot P\times \sqrt{\frac{8}{3\pi}}$

$=\frac{200\times \sqrt2\times \pi\times (0.3)^{2}\times 10^{-9}\times 10^{-9}\times 6.023\times 10^{-23}\times \sqrt8}{25\times 10^{-3}\times \sqrt{3\pi}}$

$\approx 10^{10}$

Option 1)

$\approx 10^{12}$

Option 2)

$\approx 10^{11}$

Option 3)

$\approx 10^{10}$

Option 4)

$\approx 10^{13}$

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