Question

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1.5 atm pressure and -0.77°C? Take the diameter of an oxygen molecule to be 8.4 x 10^-8 cm and the speed of sound to be 330 m/s.

Answer 1

Answer:

$\vartheta = 41.31 GHz$

Given:

Pressure, $P_{O} = 1.5 atm = 1.5\times 10^{5} Pa$

Temperature, T = $- 0.77^{\circ}$ = 273 + (- 0.77) = 272.23 K

Diameter of oxygen molecule, $d_{O} = 8.4\times 10^{- 8} cm = 8.4\times 10^{-10} m$

Speed of sound, v = 330 m/s

$k_{B} = 1.38\times 10^{- 23} J/s$

Solution:

Mean free path of oxygen is given by:

$L_{O} = \frac{k_{B}T}{\sqrt{2}P_{O}\pi d_{O}^{2}}$

Now, substituting the given values in the above formula:

$L_{O} = \frac{1.38\times 10^{- 23}\times 272.23}{\sqrt{2}\times 1.5\times 10^{5}\pi\times (8.4\times 10^{- 10})^{2}}$

$L_{O} = 7.99\times 10^{-8} m$

Now, the frequency, $\vartheta$ is given by:

$\vartheta = \frac{c}{\lambda}$  

Since, mean free path = wavelength = $7.99\times 10^{-9} m$      

Therefore,

$\vartheta = \frac{v}{L_{O}}$  

$\vartheta = \frac{330}{7.99\times 10^{-8}}$  

$\vartheta = 4.131\times 10^{10} Hz = 41.31 GHz$