Respuesta :
Answer:
[tex]\vartheta = 41.31 GHz[/tex]
Given:
Pressure, [tex]P_{O} = 1.5 atm = 1.5\times 10^{5} Pa[/tex]
Temperature, T = [tex]- 0.77^{\circ}[/tex] = 273 + (- 0.77) = 272.23 K
Diameter of oxygen molecule, [tex]d_{O} = 8.4\times 10^{- 8} cm = 8.4\times 10^{-10} m[/tex]
Speed of sound, v = 330 m/s
[tex]k_{B} = 1.38\times 10^{- 23} J/s[/tex]
Solution:
Mean free path of oxygen is given by:
[tex]L_{O} = \frac{k_{B}T}{\sqrt{2}P_{O}\pi d_{O}^{2}}[/tex]
Now, substituting the given values in the above formula:
[tex]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}}[/tex]
[tex]L_{O} = 7.99\times 10^{-8} m[/tex]
Now, the frequency, [tex]\vartheta [/tex] is given by:
[tex]\vartheta = \frac{c}{\lambda}[/tex]
Since, mean free path = wavelength = [tex]7.99\times 10^{-9} m[/tex]
Therefore,
[tex]\vartheta = \frac{v}{L_{O}}[/tex]
[tex]\vartheta = \frac{330}{7.99\times 10^{-8}}[/tex]
[tex]\vartheta = 4.131\times 10^{10} Hz = 41.31 GHz[/tex]
