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  • Figure shows a parabolic graph between and <img alt="\mathrm{\frac{1}{V}}" src="https://entrancecorner

Figure shows a parabolic graph between \mathrm{T}and \mathrm{\frac{1}{V}} for a mixture of a gas undergoing an adiabatic process. What is the ratio of \mathrm{v_{r m s}} and speed of sound in the mixture?

Option: 1

\sqrt{\frac{3}{2}}


Option: 2

\sqrt{2}


Option: 3

\sqrt{\frac{2}{3}}


Option: 4

\sqrt{3}


Answers (1)

best_answer

Equation of the parabola given in the graph is

\mathrm{y}^{2} =4 \mathrm{ax}
\mathrm{\left(T_{o}\right)^{2} =4 a\left(\frac{1}{V_{o}}\right)}

\mathrm{\Rightarrow \quad T_{o}^{2} V_{o}= constant}
\mathrm{T_{o} V_{o}^{1 / 2}= constant \quad [\text{for an adiabatic process }\mathrm{TV}^{\gamma-1}= constant]}

\mathrm{\therefore \quad \gamma-1=1 / 2 \Rightarrow \gamma=3 / 2}
\mathrm{\frac{v_{r m s}}{v_{\text {sound }}}=\frac{\sqrt{\left(\frac{3 R T}{M}\right)}}{\sqrt{\frac{\gamma R T}{M}}}=\sqrt{\frac{3}{\gamma}}=\sqrt{2}}

\therefore  (b) is correct.
 

Posted by

Rishabh

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