
If ${v_s}$ is the velocity of sound in air and v is the RMS value, then the relation between C and ${C_s}$ will be?
(A) ${v_s} > v$
(B) ${v_s} = v$
(C) ${v_s} = v{\left( {\dfrac{\gamma }{3}} \right)^{\dfrac{1}{2}}}$
(D) ${v_s} = v{\left( {\dfrac{3}{\gamma }} \right)^{\dfrac{1}{2}}}$
Answer
563.1k+ views
Hint
We know the ideal gas equation for 1 mol of gas. We also know that density means mass of a unit volume of a material substance. So we will put the value of density in the real gas equation and we will get the equation for velocity of sound. Then we will write the rms value of the velocity of sound. Then will take the ratio between the velocities and get the required solution.
Complete step by step answer
For an ideal gas we know that $PV = nRT$ [ $\because P = $ pressure on the gas, V= volume of the gas, T= temperature, n= number of moles, R= molar gas constant]
For 1 mol the formula becomes $PV = RT....(i)$
We know that the velocity of sound in air is ${v_s}$
M= mass of the gas
Again, let us assume that the density is $(\rho ) = \dfrac{M}{V}....(ii)$
Velocity of sound in gas or air is directly proportional to the square root of the temperature in absolute scale of the gas.
$\Rightarrow {v_s} \propto \sqrt T $
Putting the value of $\rho $ in equation (i) we get $\dfrac{P}{\rho } = \dfrac{{RT}}{M}$
$\Rightarrow{v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $
$\Rightarrow{v_s} = \sqrt {\dfrac{{\gamma RT}}{M}} $
We know that the RMS value of sound be $(v) = \sqrt {\dfrac{{3RT}}{M}} $
Now we take the ratio between ${v_s}$ and v, we can get that
$\Rightarrow\dfrac{{{v_s}}}{v} = \sqrt {\dfrac{\gamma }{3}} $
$\Rightarrow{v_s} = v{\left( {\dfrac{\gamma }{3}} \right)^{\dfrac{1}{2}}}$
Hence the required solution is ${v_s} = v{\left( {\dfrac{\gamma }{3}} \right)^{\dfrac{1}{2}}}$ (Option (C)).
Note
Sound propagation requires the disturbances to be transferred from one molecule to the other. In a gas, since the intermolecular forces are negligible compared to solids, this transfer can happen only if the molecules are sufficiently close or collide. Thus, the disturbance cannot move faster than their rms speed. Thus, the speed of sound is nearly the same as the rms speed but they can’t be equal so option “B” is not correct. After taking the ratio we can see that option “C” is the correct option. For air the ratio of specific heat $\gamma = 1.4$ because the air is mainly diatomic gas. Therefore at NTP velocity of sound is 332m/s.
We know the ideal gas equation for 1 mol of gas. We also know that density means mass of a unit volume of a material substance. So we will put the value of density in the real gas equation and we will get the equation for velocity of sound. Then we will write the rms value of the velocity of sound. Then will take the ratio between the velocities and get the required solution.
Complete step by step answer
For an ideal gas we know that $PV = nRT$ [ $\because P = $ pressure on the gas, V= volume of the gas, T= temperature, n= number of moles, R= molar gas constant]
For 1 mol the formula becomes $PV = RT....(i)$
We know that the velocity of sound in air is ${v_s}$
M= mass of the gas
Again, let us assume that the density is $(\rho ) = \dfrac{M}{V}....(ii)$
Velocity of sound in gas or air is directly proportional to the square root of the temperature in absolute scale of the gas.
$\Rightarrow {v_s} \propto \sqrt T $
Putting the value of $\rho $ in equation (i) we get $\dfrac{P}{\rho } = \dfrac{{RT}}{M}$
$\Rightarrow{v_s} = \sqrt {\dfrac{{\gamma P}}{\rho }} $
$\Rightarrow{v_s} = \sqrt {\dfrac{{\gamma RT}}{M}} $
We know that the RMS value of sound be $(v) = \sqrt {\dfrac{{3RT}}{M}} $
Now we take the ratio between ${v_s}$ and v, we can get that
$\Rightarrow\dfrac{{{v_s}}}{v} = \sqrt {\dfrac{\gamma }{3}} $
$\Rightarrow{v_s} = v{\left( {\dfrac{\gamma }{3}} \right)^{\dfrac{1}{2}}}$
Hence the required solution is ${v_s} = v{\left( {\dfrac{\gamma }{3}} \right)^{\dfrac{1}{2}}}$ (Option (C)).
Note
Sound propagation requires the disturbances to be transferred from one molecule to the other. In a gas, since the intermolecular forces are negligible compared to solids, this transfer can happen only if the molecules are sufficiently close or collide. Thus, the disturbance cannot move faster than their rms speed. Thus, the speed of sound is nearly the same as the rms speed but they can’t be equal so option “B” is not correct. After taking the ratio we can see that option “C” is the correct option. For air the ratio of specific heat $\gamma = 1.4$ because the air is mainly diatomic gas. Therefore at NTP velocity of sound is 332m/s.
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