Tag: speed of sound in gas

Questions Related to speed of sound in gas

As per Newton's formula velocity of sound , at NTP is 

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

  1. 340 m/s

  2. 332.3 m/s

  3. 279.9m/s

  4. 290 m/s

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Correct Option: B

The relation between velocity of sound in gas $(v)$ and r.m.s velocity of molecules of gas $v _{r.m.s}$ is

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

  1. $v=v _{r.m.s}(\gamma/ 3)^{1/2} $

  2. $v _{r.m.s}=v(2/3)^{1/2} $

  3. $v=v _{r.m.s} $

  4. $ v=v _{r.m.s}(3/\gamma)^{1/2}$

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Correct Option: A
Explanation:

Velocity of sound in a gas is

$v=\sqrt{\dfrac{\gamma P}{\rho}}$

and from $P=\dfrac{1}{3}\rho v _{rms}^2$

$v _{rms}=\sqrt{\dfrac{3P}{\rho}}$
Thus

$\dfrac{v}{v _{rms}}=\sqrt{\dfrac{\gamma}{3}}$

Ans: A

The velocity of sound in air is $330$ m/s. The r.m.s velocity of air molecules $(\gamma=1.4) $ is approximately equal to

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

  1. 400 m/s

  2. 471.4 m/s

  3. 231 m/s

  4. 462 m/s

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Correct Option: B
Explanation:

$v _{air}=\sqrt{\dfrac{\gamma RT}{M}}=330m/s$

$v _{rms}=\sqrt{\dfrac{3RT}{M}}$
$=\sqrt{\dfrac{3}{\gamma}}\times 330m/s$
$\gamma=1.4$
$\implies v _{rms}=471.4m/s$

The velocity of sound in a gas at pressure $P$ and density $d$ is

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

  1. $\displaystyle v= \sqrt {\frac {\gamma P}{d}}$

  2. $\displaystyle v= \sqrt {\frac {P}{\gamma d}}$

  3. $\displaystyle v= \gamma \sqrt {\frac {P}{d}}$

  4. $\displaystyle v= \sqrt {\frac {2 P}{d}}$

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Correct Option: A
Explanation:

$\displaystyle v= \sqrt {\frac {\gamma RT}{M}}$

$PV=RT$

$\displaystyle P\frac {M}{d}=RT$

$\displaystyle \frac {P}{d} = \frac {RT}{M}$

$\displaystyle v= \sqrt {\frac {\gamma P}{d}}$

Does the sound of an explosion travel faster than the sound produced by a humming bee?

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

  1. True

  2. False

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Correct Option: B
Explanation:

$No$


The speed of sound depends only on the physical conditions of the medium in which the sound is travelling and the speed and direction of the wind present if any.
The speed of the sound doesn't depend on its loudness.

Hence although the sound of explosion is much louder than the humming of a bee, both sounds travel with equal speed.

Ultrasonic, infrasonic and audio waves travel through a medium with speeds $V _{u}, V _{i}$ and $V _a$ respectively then,

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

  1. $V _{u}, V _{i}$ and $V _{a}$ are equal

  2. $V _{u} > V _{a}> V _{i}$

  3. $V _{u} < V _{a} < V _{i}$

  4. $ V _{a}< V _{u} $ and $V _{u} \approx V _{i} $

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Correct Option: A
Explanation:

Velocity of sound wave in a medium is given by
$v= \sqrt{\frac{K}{\rho}}$ where K is the bulk modulus and $\rho$ is the density.
The classification of sound waves based on wavelength($\lambda$) is independent of speed of sound in the medium( speed depends on properties of a medium).
Hence, $V _u$, $V _i$ and $V _a$ are all equal.

The extension in a string obeying Hooke's law $v$ is $x$. The speed of sound in the stretched string is $v$. If the extension in the string is increased to $1.5\ x$, the speed of sound will be

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

  1. $1.22\ v$

  2. $0.61\ v$

  3. $1.50\ v$

  4. $0.75\ v$

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Correct Option: A
Explanation:

Speed of sound in stretched string
$v = \dfrac {\overline {T}}{\mu} ..... (i)$
where $T$ is the tension in the string and $\mu$ is mass per unit length.
According to Hooke's law, $F\propto X$
$\therefore T\propto X$ .... (ii)
From Eqs. (i) and (ii)
$v\propto$
$\therefore v' = \overline {1.5V} = 1.22\ V$.

According to Newton's formula, the speed of sound in air at STP is:
(Take the mass of $1$ mole of are is $29 \times 10^{-3} \,\,kg)$

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

  1. $250 \,\, m \,\,s^{-1}$

  2. $260 \,\, m \,\,s^{-1}$

  3. $270 \,\, m \,\,s^{-1}$

  4. $280 \,\, m \,\,s^{-1}$

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Correct Option: D
Explanation:

$1$ mole of any gas occupies $22.4$ liters at STP.
Therefore, the density of air at STP is
$\rho = \dfrac{\text{Mass of one mole of air}}{\text{Volume of one mole of air at STP}}$

$= \dfrac{29 \times 10^{-3} \,\,kg}{22.4 \times 10^{-3} \,\,m^3} = 1.29 \,\,kg \,\,m^{-3}$

At STP, $P = 1\,\,atm = 1.01 \times 10^5 \,\,N \,\,m^{-2}$

$V =\sqrt{\left( \dfrac { P }{ \rho  }\right)}=\sqrt { \dfrac {1.01 \times 10^5 \,\,N \,\,m^{-2}  }{ 1.29 \times kg \,\,m^{-3} }  } = 280 \,\,m \,\,s^{-1} $

The velocities of sound at the same temperature in two monoatomic gases of densities $p _1$ and $p _2$ are $v _1$ and $v _2$ respectively. If $p _1/p _2 = 4$, then the value of $v _1/v _2$ is

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

  1. $\dfrac{1}{4}$

  2. $2$

  3. $\displaystyle \dfrac {1}{2}$

  4. $4$

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Correct Option: A,C
Explanation:

For velocity of sound in gas
$\displaystyle v=\sqrt {\frac {p\gamma }{p}}$

$[P$ is pressure and $p$ is density of gas, $\gamma$ is $C _p/C _v]$

Here, $\displaystyle v _1 = \sqrt {\frac {\gamma P}{p _1}}$ and $v _2 = \sqrt {\frac {\gamma P}{p _2}}$

$\displaystyle \frac {v _1}{v _2} = \sqrt {\frac {p _2}{p _1}} = \sqrt {\frac {1}{4}}=\frac {1}{2}$

The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. What is this constant ratio ?

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

  1. $\sqrt{\dfrac{\gamma}{3}}$

  2. $\gamma -1$

  3. $\sqrt{\dfrac{2\gamma}{3}}$

  4. $\gamma$

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Correct Option: A
Explanation:

$V _{sound}=\sqrt{\dfrac{\gamma RT}{M}}$ and $V _{rms}=\sqrt{\dfrac{3RT}{M}}$
$\Rightarrow\dfrac{V _{sound}}{V _{rms}}=\sqrt{\dfrac{\gamma}{3}}$
Hence (A) is correct.