At a temperature of 314 K and a pressure of 100 kPa , the speed of sound in a gas is 1380 `ms^(-1)`. The radius of each gas molecule is `0.5 Å`. The frequency of sound at which the wavelength of sound wave in the gas becomes equal to the mean free path of the gas molecules is
( Boltzmann constant `= 1.38 xx 10^(-23) JK^(-1)`. )

Options

A. 1000 MHz
B. `1000sqrt(2)` MHz
C. `1000/sqrt(2)` MHz
D. 500 MHz

Answer

B

Explanation

Given, temperature, `T = 314 K,`
pressure, `p = 100 k "Pa" = 1.0 xx 10^(5) ` Pa, speed of sound, `v = 1380 ms^(-1)` and diameter of gas
molecule, `d = 10^(-10)` m or radius `r = 1/2 xx 10^(-10) m = 1/2 Å`
As, mean free path, `overline(lambda) = (kT)/(sqrt(2) pi d^(2) p)" " ...(i)`
and frequency `,v = v/lambda" "...(ii)`
So, from Eqs. (i) and (ii) we get
`v = (sqrt(2) pi d^(2) p xxv)/(kT) `
Now, putting the given values, we get
`= (sqrt(2) xx 3.14 xx 10^(-20) xx 10^(5) xx 1380)/(1.38xx 10^(-23)xx314)`
`v = sqrt(2) xx 10^(9) Hz rArr v = 1000 sqrt( 2)` MHz
hence, lthe correct option is (b)