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)

 

 

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}

Exams
Articles
Questions