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AP EAMCET · PHYSICS · Kinetic Theory of Gases

At a temperature of \(314 \mathrm{~K}\) and a pressure of \(100 \mathrm{kPa}\), the speed of sound in a gas is \(1380 \mathrm{~ms}^{-1}\). 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 \(\left(\right.\) Boltzmann constant \(=1.38 \times 10^{-23} \mathrm{JK}^{-1}\).)

  1. A \(1000 \mathrm{MHz}\)
  2. B \(1000 \sqrt{2} \mathrm{MHz}\)
  3. C \(\frac{1000}{\sqrt{2}} \mathrm{MHz}\)
  4. D \(500 \mathrm{MHz}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1000 \sqrt{2} \mathrm{MHz}\)

Step-by-step Solution

Detailed explanation

Given, temperature, \(T=314 \mathrm{~K}\), pressure, \(p=100 \mathrm{kPa},=1.0 \times 10^5 \mathrm{~Pa}\), speed of sound, \(v=1380 \mathrm{~ms}^{-1}\) and diameter of gas molecule, \(d=10^{-10} \mathrm{~m}\) or radius \(r=\frac{1}{2} \times 10^{-10} \mathrm{~m}=\frac{1}{2} Å\)…
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