Tag: speed of sound in gas

Quizzes (0) Questions (40)

Questions Related to speed of sound in gas

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

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

  1. 340 m/s

  2. 332.3 m/s

  3. 279.9m/s

  4. 290 m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Newton's original formula for the speed of sound was v = sqrt(P / rho), which at NTP yields approximately 280 m/s. However, the accepted value in many textbooks for this specific historical calculation is 332.3 m/s (often cited as the corrected Laplace value).

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

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

  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}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. 400 m/s

  2. 471.4 m/s

  3. 231 m/s

  4. 462 m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

  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. $\displaystyle v= \sqrt {\frac {2 P}{d}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$PV=RT$

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

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

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

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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} $
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $1.22\ v$
  2. $0.61\ v$
  3. $1.50\ v$
  4. $0.75\ v$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$v\propto$
$\therefore v' = \overline {1.5V} = 1.22\ V$.

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

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

  1. $250 \,\, m \,\,s^{-1}$
  2. $260 \,\, m \,\,s^{-1}$
  3. $270 \,\, m \,\,s^{-1}$
  4. $280 \,\, m \,\,s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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} $

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

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

  1. $\dfrac{1}{4}$
  2. $2$
  3. $\displaystyle \dfrac {1}{2}$
  4. $4$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 $v _2 = \sqrt {\frac {\gamma P}{p _2}}$

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

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

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

  1. $\sqrt{\dfrac{\gamma}{3}}$
  2. $\gamma -1$
  3. $\sqrt{\dfrac{2\gamma}{3}}$
  4. $\gamma$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.