
At a given temperature if ${v_{rms}}$ is the root mean square velocity of the molecules of a gas and ${v_s}$ is the velocity of sound in it. Then these are related as $\left( {\gamma = \dfrac{{{C_P}}}{{{C_V}}}} \right)$
A. ${v_{rms}} = {v_s}$
B. ${v_{rms}} = \sqrt {\dfrac{3}{\gamma }} \times {v_s}$
C. ${v_{rms}} = \sqrt {\dfrac{\gamma }{3}} \times {v_s}$
D. ${v_{rms}} = \dfrac{3}{\gamma } \times {v_s}$
Answer
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Hint:Here , the velocity of sound can be found out using the equation,
${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $
Where $\gamma $ is the ratio of specific heat at a constant pressure to specific heat at constant volume.
P is the pressure and $\rho $ is the density
With the help of an ideal gas equation we can write this equation in terms of temperature and mass and universal gas constant.
Then on comparing this equation with the equation of root mean square velocity we can reach the final answer.
Complete step-by-step solution:
We need to find the relationship between root mean square velocity and the speed of sound at a given temperature.
We know that the root mean square velocity is given by the equation
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ (1)
Where, R is the universal gas constant, T is the temperature in Kelvin and M is the molecular weight.
We know that the velocity of sound can be found out using the equation,
${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $ (2)
Where, $\gamma $ is the ratio of specific heat at a constant pressure to specific heat at constant volume.
P is the pressure and $\rho $ is the density.
We know that density is equal to mass divided by volume. That is,
$\rho = \dfrac{M}{V}$
Let us substitute this in equation 2 .
${v_s} = \sqrt {\dfrac{{\gamma PV}}{M}} $ (3)
From the ideal gas equation, we have
$PV = nRT$
Let us take the number of moles n to be one. Then we can write PV is equal to RT.
Substituting this in the equation 3 we get
${v_s} = \sqrt {\dfrac{{\gamma RT}}{M}} $ (4)
Now let us divide equation 4 with equation 1.
Then we get,
$\dfrac{{{v_s}}}{{{v_{rms}}}} = \dfrac{{\sqrt {\dfrac{{\gamma RT}}{M}} }}{{\sqrt {\dfrac{{3RT}}{M}} }}$
Therefore,
$\dfrac{{{v_s}}}{{{v_{rms}}}} = \dfrac{{\sqrt \gamma }}{{\sqrt 3 }}$
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{3}{\gamma }} {v_s}$
This is the relation between RMS velocity and velocity of sound at a given temperature.
So the correct answer is option B.
Note:-
Alternative method –
Velocity of sound is given as ${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $ .
Where $\rho $ is the density and P is the pressure.
The relation between pressure P and root mean square velocity ${v_{rms}}$ is given as
$P = \dfrac{1}{3}\rho {\left( {{v_{rms}}} \right)^2}$
On substituting this in the first equation we get
${v_s} = \sqrt {\dfrac{{\gamma \dfrac{1}{3}\rho {{\left( {{v_{rms}}} \right)}^2}}}{\rho }} $
$\therefore {v_s} = \sqrt {\dfrac{\gamma }{3}} {v_{rms}}$ .
${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $
Where $\gamma $ is the ratio of specific heat at a constant pressure to specific heat at constant volume.
P is the pressure and $\rho $ is the density
With the help of an ideal gas equation we can write this equation in terms of temperature and mass and universal gas constant.
Then on comparing this equation with the equation of root mean square velocity we can reach the final answer.
Complete step-by-step solution:
We need to find the relationship between root mean square velocity and the speed of sound at a given temperature.
We know that the root mean square velocity is given by the equation
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ (1)
Where, R is the universal gas constant, T is the temperature in Kelvin and M is the molecular weight.
We know that the velocity of sound can be found out using the equation,
${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $ (2)
Where, $\gamma $ is the ratio of specific heat at a constant pressure to specific heat at constant volume.
P is the pressure and $\rho $ is the density.
We know that density is equal to mass divided by volume. That is,
$\rho = \dfrac{M}{V}$
Let us substitute this in equation 2 .
${v_s} = \sqrt {\dfrac{{\gamma PV}}{M}} $ (3)
From the ideal gas equation, we have
$PV = nRT$
Let us take the number of moles n to be one. Then we can write PV is equal to RT.
Substituting this in the equation 3 we get
${v_s} = \sqrt {\dfrac{{\gamma RT}}{M}} $ (4)
Now let us divide equation 4 with equation 1.
Then we get,
$\dfrac{{{v_s}}}{{{v_{rms}}}} = \dfrac{{\sqrt {\dfrac{{\gamma RT}}{M}} }}{{\sqrt {\dfrac{{3RT}}{M}} }}$
Therefore,
$\dfrac{{{v_s}}}{{{v_{rms}}}} = \dfrac{{\sqrt \gamma }}{{\sqrt 3 }}$
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{3}{\gamma }} {v_s}$
This is the relation between RMS velocity and velocity of sound at a given temperature.
So the correct answer is option B.
Note:-
Alternative method –
Velocity of sound is given as ${v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $ .
Where $\rho $ is the density and P is the pressure.
The relation between pressure P and root mean square velocity ${v_{rms}}$ is given as
$P = \dfrac{1}{3}\rho {\left( {{v_{rms}}} \right)^2}$
On substituting this in the first equation we get
${v_s} = \sqrt {\dfrac{{\gamma \dfrac{1}{3}\rho {{\left( {{v_{rms}}} \right)}^2}}}{\rho }} $
$\therefore {v_s} = \sqrt {\dfrac{\gamma }{3}} {v_{rms}}$ .
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