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A mixture of two diatomic gases is obtained by mixing ${m_1}$ mass of first gas and ${m_2}$ mass of the second gas. The speeds of sound in two gases are ${v_1}$ and ${v_2}$ respectively. The speed of sound in the given mixture is
(A) $\sqrt {\dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{2\left( {{m_1} + {m_2}} \right)}}} $
(B) $\sqrt {\dfrac{{2\left( {{m_1}{v_1}^2 + {m_2}{v_2}^2} \right)}}{{\left( {{m_1} + {m_2}} \right)}}} $
(C) $2\sqrt {\dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{{m_1} + {m_2}}}} $
(D) $\sqrt {\dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{{m_1} + {m_2}}}} $

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Last updated date: 20th Jun 2024
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Answer
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Hint To answer this question, we have to use the Laplace correction formula for the speed of the sound. Then, calculating the parameters of the mixture in terms of those of the individual gases, we will get the velocity of sound in the given mixture.

Formula Used The formula used to solve this question is
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $, here $v$ is the velocity of sound in a gaseous medium having pressure $P$, density $\rho $ and ratio of specific heats at constant pressure and at constant volume as $\gamma $.

Complete step by step answer
Let the speed of sound in the given mixture be ${v_m}$.
We know that the speed of sound in a gaseous medium is given by
$\Rightarrow v = \sqrt {\dfrac{{\gamma P}}{\rho }} $
Substituting $\rho = \dfrac{m}{V}$
$\Rightarrow v = \sqrt {\dfrac{{\gamma PV}}{m}} $
From the ideal gas equation, we have
$\Rightarrow PV = nRT$
$\Rightarrow v = \sqrt {\dfrac{{\gamma nRT}}{m}} $
Since both the gases are diatomic, so $\gamma $ will be the same for both the gases. Since the gases are mixed, they both will attain a common equilibrium temperature $T$. Also $R$ is a universal constant for all gases. So, combining all these three constants into $k$, we get
$\Rightarrow v = k\sqrt {\dfrac{n}{m}} $ …...(i)
Let the number of moles of the first gas be ${n_1}$ and that of the second gas be ${n_2}$.
For the first gas, according to the question, the speed of sound is
$\Rightarrow {v_1} = k\sqrt {\dfrac{{{n_1}}}{{{m_1}}}} $
Squaring both the sides, we get
$\Rightarrow {v_1}^2 = {k^2}\dfrac{{{n_1}}}{{{m_1}}}$
So that we get the number of moles of the first gas as
$\Rightarrow {n_1} = \dfrac{{{v_1}^2{m_1}}}{{{k^2}}}$ ……..(ii)
Similarly, we get the number of moles of the second gas
$\Rightarrow {n_2} = \dfrac{{{v_2}^2{m_2}}}{{{k^2}}}$ ……...(iii)
As the gases are being mixed so the total number of moles of the mixture will be equal to the sum of the number of moles of the individual gases. So, the number of moles of the mixture
$\Rightarrow {n_m} = {n_1} + {n_2}$
From (ii) and (iii)
$\Rightarrow {n_m} = \dfrac{{{v_1}^2{m_1}}}{{{k^2}}} + \dfrac{{{v_2}^2{m_2}}}{{{k^2}}}$
$\Rightarrow {n_m} = \dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{{k^2}}}$ …….(iv)
Also, the total mass of the mixture will be equal to the sum of the masses of the individual gases, that is
$\Rightarrow M = {m_1} + {m_2}$ ……..(v)
From (i)
$\Rightarrow v = k\sqrt {\dfrac{n}{m}} $
So the velocity of sound of the mixture is given by
$\Rightarrow {v_m} = k\sqrt {\dfrac{{{n_m}}}{M}} $
Substituting (iv) and (v)
$\Rightarrow {v_m} = k\sqrt {\dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{{k^2}\left( {{m_1} + {m_2}} \right)}}} $
$\Rightarrow {v_m} = \sqrt {\dfrac{{{k^2}\left( {{m_1}{v_1}^2 + {m_2}{v_2}^2} \right)}}{{{k^2}\left( {{m_1} + {m_2}} \right)}}} $
So, finally we get
$\Rightarrow {v_m} = \sqrt {\dfrac{{{m_1}{v_1}^2 + {m_2}{v_2}^2}}{{{m_1} + {m_2}}}} $
Hence, the correct answer is option (D).

Note
If we do not remember the formula for the speed of sound, then also we can attempt this question very easily. We can solve this question using the energy conservation principle. We know that the sound is propagated as the energy transfer of the gaseous molecules. As both the gases are mixed, the sum of the kinetic energies of the molecules of the two gases will be equal to that of the gaseous mixture.