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Setler [38]
3 years ago
14

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 an

d -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.
Physics
1 answer:
quester [9]3 years ago
5 0

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

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