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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

If the speed of sound in a mixture of 2 moles of Helium and 2 moles...

If the speed of sound in a mixture of 2 moles of Helium and 2 moles of Hydrogen at temperature 9725 K\frac{972}{5} \mathrm{~K} is n×100 ms−1n \times 100 \mathrm{~ms}^{-1}, then the value of nn is (Take, \left.R=\frac{25}{3} \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)

Options

  1. A9
  2. B10
  3. C100
  4. D90

Correct answer

A. 9

Step-by-step solution

Given, T=9725 K,R=253 J mol−1 K−1T=\frac{972}{5} \mathrm{~K}, R=\frac{25}{3} \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} Molecular mass of the mixture =nHeMHe+nHMHnHe+nH∵nHe=2 moles, nH=2 moles  and MHe=4,MH=2\begin{aligned}& \quad=\frac{n_{\mathrm{He}} M_{\mathrm{He}}+n_H M_H}{n_{\mathrm{He}}+n_H}\\ & \because n_{\mathrm{He}}=2 \text { moles, }n_H=2 \text { moles }\\ \text{ and } M_{\mathrm{He}}=4, M_H=2 \end{aligned}\thereforeMolecularmassofthemixtures,Molecular mass of the mixtures,M=2×4+2×22+2=8+44=3gm/mole\begin{aligned}M & =\frac{2 \times 4+2 \times 2}{2+2}\\ & =\frac{8+4}{4}=3 \mathrm{gm}/ \mathrm{mole} \end{aligned}∵\because Degree of freedom of mixture f=2×3+2×52+2=4∵fmix =1+2f=1.5∵ Speed of sound =fmix RTM=1.5×253×97253×10−3=900=n×100 m/s\begin{aligned}f=\frac{2 \times 3+2 \times 5}{2+2}& =4 \\ \because \quad f_{\text {mix }}=1+\frac{2}{f}& =1.5 \\ \because \text { Speed of sound }& =\sqrt{\frac{f_{\text {mix }} R T}{M}}\\ & =\sqrt{\frac{1.5 \times \frac{25}{3} \times \frac{972}{5}}{3 \times 10^{-3}}}\\ & =900=n \times100 \mathrm{~m}/ \mathrm{s} \end{aligned}Hence, $n=9
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