At a temperature of 314 ,K and a pressure of | vedclass

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Question

(B) Given: Temperature $T = 314 \,K$, pressure $p = 100 \,kPa = 1.0 	imes 10^5 \,Pa$, speed of sound $v = 1380 \,ms^{-1}$, and radius of gas molecule

Vedclass · Accepted Answer

$1000 \sqrt{2} \,MHz$

Vedclass · Answer

(B) Given: Temperature $T = 314 \,K$, pressure $p = 100 \,kPa = 1.0 	imes 10^5 \,Pa$, speed of sound $v = 1380 \,ms^{-1}$, and radius of gas molecule $r = 0.5 \,Å = 0.5 	imes 10^{-10} \,m$. The diameter $d = 2r = 10^{-10} \,m$.The mean free path $\lambda$ is given by $\lambda = \frac{kT}{\sqrt{2} \pi d^2 p}$.The frequency $
u$ of the sound wave is given by $
u = \frac{v}{\lambda}$.Substituting $\lambda$ into the frequency formula, we get $
u = \frac{v \sqrt{2} \pi d^2 p}{kT}$.Substituting the given values:$
u = \frac{1380 	imes \sqrt{2} 	imes 3.14 	imes (10^{-10})^2 	imes 1.0 	imes 10^5}{1.38 	imes 10^{-23} 	imes 314}$$
u = \frac{1380 	imes \sqrt{2} 	imes 3.14 	imes 10^{-20} 	imes 10^5}{1.38 	imes 10^{-23} 	imes 314}$$
u = \frac{1380 	imes \sqrt{2} 	imes 3.14 	imes 10^{-15}}{1.38 	imes 314 	imes 10^{-23}}$$
u = \frac{1380 	imes \sqrt{2} 	imes 3.14 	imes 10^8}{433.32} \approx 10 	imes \sqrt{2} 	imes 10^8 \,Hz = \sqrt{2} 	imes 10^9 \,Hz = 1000 \sqrt{2} \,MHz$.Thus, the correct option is $B$.

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