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 \(=1.38 \times 10^{-23} \mathrm{JK}^{-1}\).)
- A \(1000 \mathrm{MHz}\)
- B \(1000 \sqrt{2} \mathrm{MHz}\)
- C \(\frac{1000}{\sqrt{2}} \mathrm{MHz}\)
- D \(500 \mathrm{MHz}\)
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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