Tag: speed of sound in gas

Questions Related to speed of sound in gas

A Sound wave with an amplitude of $ 3 \mathrm { cm }$ starts towards right from origin and gets reflected at a rigid wall after a second. If the velocity of  the wave is $ 340 \mathrm { ms } ^ { - 1 }$  and it has a wavelength of $ 2 \mathrm { m } $, the equations of incident and reflected waves respectively could be

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\begin{array} { l } { y = 3 \times 10 ^ { - 2 } \sin \pi ( 340 t - x ) } \ { y = - 3 \times 10 ^ { - 2 } \sin \pi ( 340 t + x ) \text { towards left } } \end{array}$

  2. $\begin{array} { l } { y = 3 \times 10 ^ { - 2 } \sin \pi ( 340 t + x ) } \ { y = - 3 \times 10 ^ { - 2 } \sin \pi ( 340 t + x ) \text { towards left } } \end{array}$

  3. $\begin{array} { l } { y = 3 \times 10 ^ { - 2 } \sin \pi ( 340 t - x ) } \ { y = - 3 \times 10 ^ { - 2 } \sin \pi ( 340 t - x ) \text { towards left } } \end{array}$

  4. $\begin{array} { l } { y = 3 \times 10 ^ { 2 } \sin \pi ( 340 t - x ) } \ { I = 3 \times 10 ^ { - 2 } \sin \pi ( 340 t + x ) \text { towards left } } \end{array}$

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Correct Option: A
Explanation:

$\begin{array}{l} y=\left( { 3\times { { 10 }^{ -2 } } } \right) \sin  \left( { \omega t-kx } \right)  \ \lambda =2m=\frac { { 2\pi  } }{ k }  \ \Rightarrow k=\pi  \ v=340\, m/s \ w=vk \ =240\pi  \ Now, \ y=\left( { 3\times { { 10 }^{ -2 } } } \right) \sin  \left( { 340\pi t-\pi x } \right) \, \, towards\, right \ and, \ y=\left( { 3\times { { 10 }^{ -2 } } } \right) \sin  \left( { 340\pi t+\pi x+\pi  } \right)  \ =-\left( { 3\times { { 10 }^{ -2 } } } \right) \sin  \left( { 340\pi t+\pi x } \right) \, \, \, \, \, \, \, towards\, \, left \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

The isothermal elasticity of a medium is $E _i$ and the adiabatic elasticity is $E _a$. The velocity of the sound in the medium is proportional to :

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\sqrt{E _i}$

  2. $E _a$

  3. $\sqrt{E _a}$

  4. $E _i$

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Correct Option: C
Explanation:

For longitudinal sound waves in gasses velocity of sound v, 

$v=\sqrt{\dfrac{\gamma E _i}{d}}$, 

$\gamma=C _P/C _V$, 

$E _i$  isothermal elasticity of medium,

$d$ density of the medium.

$E _a=\gamma E _i$

Option "C" is correct.

Sound waves are propagating in a medium. The moduli of isothermal and adiabatic elasticity of the medium are $E _T$ and $E _S$ respectively. The velocity of sound wave is proportional to

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\sqrt{E _T}$

  2. $\sqrt{E _S}$

  3. $E _T$

  4. $\displaystyle\frac{E _S}{E _T}$

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Correct Option: B
Explanation:

Velocity of sound $V _s$ is given by


${V _s}^2={[\dfrac{\delta{p}}{\delta{\rho}}]} _S=E _S$

$V _s \propto \sqrt{E _S}$

Option 'B' is correct.

The density of air at NTP is $1.293\space kgm^{-3}$ and density of mercury at $0^{\small\circ}\space C$ is $13.6\times10^3 \space kgm^{-3}$. If $C _p = 0.2417\space calkg^{-10}C^{-1}$ and $C _v = 0.1715$, the speed of sound in air at $100^{\small\circ}\space C$ will be $(g = 9.8\space Nkg^{-1})$

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $260\space ms^{-1}$

  2. $332\space ms^{-1}$

  3. $350.2\space ms^{-1}$

  4. $369.4\space ms^{-1}$

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Correct Option: D
Explanation:
NTP conditions:        $T = 25  ^o C=  298.15  K                P =  1  bar  =  10^5    Pa$

Given:   Density of air at NTP  $\rho = 1.293     kg /m^3$

$\gamma =  \dfrac{C _p}{C _v} = \dfrac{0.2417}{0.1715} = 1.4$

Speed of sound in air at NTP,      $v _{25^o C} =  \sqrt{\dfrac{\gamma  P}{\rho} }  = \sqrt{\dfrac{1.4  \times 10^5}{1.293}}  = 330.15   m/s$

Let speed of sound in air at $100^o  C$ be  $v _{100^o  C}$

As     $v   \propto  \sqrt{T}$


Thus   $\dfrac{v _{100^o  C}}{v _{25^o  C} } = \sqrt{\dfrac{373.15}{298.15}} = 1.118$

$\implies  v _{100^o  C} = 1.118 \times  330.15 = 369.35   m/s$

Velocity of sound in a gas proportional to

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. square root of isothermal elasticity

  2. isothermal elasticity

  3. square root of adiabatic elasticity

  4. adiabatic elasticity

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Correct Option: C
Explanation:

According to the Laplace's formula for velocity of sound in gases,


$v = \sqrt {\dfrac{E}{\rho}}$

where, $E = \gamma p$ is the adiabatic elasticity and $\rho$ is the  density of gas.


This is because the compression and rarefaction occurs rapidly one after  another without exchanging the thermal energy with surrounding hence, this  the process becomes adiabatic and not the isothermal. Hence, velocity of sound in a gas proportional to square root of adiabatic elasticity

Two gases with different densities and same ratio of specific heats $(\gamma)$ are mixed in proportions $V _1$ and $V _2$ by volume. The velocity $C$ of sound in mixture will be given by $(C _1, \space C _2$ are velocities in individual gases$)$

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\displaystyle\frac{C _1+C _2}{2}$

  2. $\sqrt{C _1C _2}$

  3. $\displaystyle\frac{C _1C _2\sqrt{(V _1+V _2)}}{\sqrt{(V _1C _2^2+V _2C _1^2)}}$

  4. $\displaystyle\frac{C _1C _2\sqrt{(V _1+V-2)}}{\sqrt{(V _1C _1^2+V _2C _2^2)}}$

Show answer
Correct Option: C
Explanation:
$C _1=\sqrt{\dfrac{\gamma P}{\rho _1}}$ and $C _2=\sqrt{\dfrac{\gamma P}{\rho _2}}$

$\rho _1=\dfrac{\gamma P}{{C _1}^2}$

$\rho _2=\dfrac{\gamma P}{{C _2}^2}$
Mixture density, 
$\rho=\dfrac{\rho _1 \times V _1 + \rho _2 \times V _2}{V _1 + V _2}$

$\rho=\dfrac{\dfrac{\gamma P}{{C _1}^2} \times V _1 + \dfrac{\gamma P}{{C _2}^2} \times V _2}{V _1 + V _2}$

$\rho={\gamma P} \dfrac{\dfrac{1}{{C _1}^2} \times V _1 + \dfrac{1}{{C _2}^2} \times V _2}{V _1 + V _2}$

$C=\sqrt{\dfrac{\gamma P}{\rho}} = \sqrt{\dfrac{1}{\dfrac{\dfrac{1}{{C _1}^2} \times V _1 + \dfrac{1}{{C _2}^2} \times V _2}{V _1 + V _2}}}$

$C = C _1 C _2 \sqrt{\dfrac{V _1+V _2}{V _1 C _2^2+V _2 C _1^2}}$

Standing waves of frequency 5.0 KHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities ${ C } _{ p }$   and ${ C } _{ v }$ of the gas.

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $20.7J/molK,29.0J/molK$

  2. $29.0J/molK,20.7J/molK$

  3. $2.90J/molK,2.07J/molK$

  4. none of these

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Correct Option: A

The echo of a gunshot is heard 8 s after the gun is fired. How far from a person is the surface that reflects the sound (velocity of sound in air = $350 m/s)$?

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. 1400 m

  2. 2800 m

  3. 700 m

  4. 350 m

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Correct Option: B

The speed of sound in hydrogen at $  N T P,  $ is 1270 $ \mathrm{m} / \mathrm{s} .$ Then the speed in a mixture of hydrogen and oxigen in the ratio $  4 : 1  $ by volume, (in $  m / s )  $ will be

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. 635

  2. 318

  3. 158

  4. 1270

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Correct Option: A

The speed of sound in an ideal gas at ${ T } _{ 1 }$ K and   ${ T } _{ 2 }$K  are $ { V } _{ 1 }$ and $ { V } _{ 2 }$ respectively. if the root mean square velocity of molecules of same gas at these temperatures are  $  { v } _{ rms1 }  $ and${ v } _{ rms1 }$ respectively, then 

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. ${ v } _{ rms2 }={ v } _{ rms1 }\left( \dfrac { { v } _{ 2 } }{ { v } _{ 1 } } \right) $

  2. ${ v } _{ rms2 }={ v } _{ rms1 }\left( \dfrac { { v } _{ 1 } }{ { v } _{ 2 } } \right) $

  3. $ { v } _{ rms2 }={ v } _{ rms1 }\left( \sqrt { \dfrac { { v } _{ 2 } }{ { v } _{ 1 } } } \right) $

  4. $ { v } _{ rms2 }={ v } _{ rms1 }\left( \sqrt { \dfrac { { v } _{ 1 } }{ { v } _{ 2 } } } \right) $

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Correct Option: D