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

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$)$

  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. $\displaystyle\frac{C _1C _2\sqrt{(V _1+V-2)}}{\sqrt{(V _1C _1^2+V _2C _2^2)}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, 
$\rho=\dfrac{\rho _1 \times V _1 + \rho _2 \times V _2}{V _1 + V _2}$

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

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

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

$C = C _1 C _2 \sqrt{\dfrac{V _1+V _2}{V _1 C _2^2+V _2 C _1^2}}$
Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance between nodes is lambda/2 = 3.3 cm, so lambda = 6.6 cm = 0.066 m. Frequency f = 5000 Hz. Speed v = f * lambda = 5000 * 0.066 = 330 m/s. For oxygen (diatomic), v = sqrt(gamma * R * T / M). Gamma = v^2 * M / (R * T) = 330^2 * 0.032 / (8.314 * 300) = 1.4. Cp = (gamma * R) / (gamma - 1) = 1.4 * 8.314 / 0.4 = 29.1 J/molK. Cv = Cp - R = 29.1 - 8.314 = 20.8 J/molK.

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

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

  1. 635

  2. 318

  3. 158

  4. 1270
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Speed of sound v = sqrt(gamma * R * T / M). For a mixture, M_mix = (n1M1 + n2M2) / (n1 + n2). With 4:1 ratio, M_mix = (4*2 + 1*32) / 5 = 40/5 = 8. Since v is inversely proportional to sqrt(M), v_mix = v_H2 * sqrt(M_H2 / M_mix) = 1270 * sqrt(2 / 8) = 1270 * 0.5 = 635 m/s.

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

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 

  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) $
Reveal answer Fill a bubble to check yourself
D Correct answer
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.