Speed of sound in gas - class-XII

For an open pipe, frequency f = v / (2L). Beats = f1 - f2 = (v/2) * (1/L1 - 1/L2). Substituting L1 = 0.2m, L2 = 0.201m, and beats = 10, we get 10 = (v/2) * (1/0.2 - 1/0.201) = (v/2) * (0.001 / 0.0402). Solving for v gives 402 m/s.

Velocity of wave is equal to product of its wavelength and frequency

Newton's formula for velocity of sound in gas is:

$\displaystyle v= \sqrt {\frac {P}{\rho}}$, where $P$ is pressure & $\rho$ is density of gas

Newton assumed that sound propagation in a gas takes under isothermal condition.

The sound propogation according to Newton takes place in isothermal condition

The correct option is (b)

The sound propogation according to Newton takes place in isothermal condition. Hence no temperature change happens or temperature change is zero

The correct option is (a)

According to Newton, when sound waves propagate in air, compression and rarefaction are formed. He assumed that the process is very slow and the heat produced during compression is given to surrounding and heat loss during compression is gained from surrounding. So the temperature remains constant and sound waves propagate through an isothermal process. 

When ${M} _{1}=0.004kg/mol$) is mixed with ${n} _{2}$ moles of (${M} _{2}=0.020kg/mol$), the equivalent molar mass of mix would be:

Power P is proportional to (pressure amplitude)^2 * frequency^2. Given sqrt(P1/P2) = a and (deltaP1/deltaP2) = b, then (a)^2 = (b)^2 * (omega1/omega2)^2. Rearranging gives omega1/omega2 = a/b. The ratio of frequencies is proportional to the ratio of angular frequencies.

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

For longitudinal sound waves in gasses velocity of sound v, 

Velocity of sound $V _s$ is given by

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

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.

Speed of sound v = sqrt(gamma * R * T / M). For a mixture, M_mix = (n1M1 + n2M2) / (n1 + n2). With 4:1 ratio, M_mix = (42 + 132) / 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.

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).

Velocity of sound in a gas is

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

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

$PV=RT$

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

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

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

$No$

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.

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)
$v\propto$
$\therefore v' = \overline {1.5V} = 1.22\ V$.

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

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 $v _2 = \sqrt {\frac {\gamma P}{p _2}}$

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

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

RMS velocity is independent of pressure. Hence velocity does not change with variations in pressure

The correct option is (c)

Speed of sound in air, ${{C} _{s}}=\sqrt{\dfrac{\gamma P}{\rho }}\,\ldots \ldots \,(1)$

 $ Where, $

$ \gamma =specific\,heat\,ratio $

$ P=\,pressure $

$ \rho =\,density $

RMS velocity of air molecule, $C=\sqrt{\dfrac{3\overline{R}T}{{{M} _{o}}}}=\sqrt{\dfrac{3P}{\rho }}\,\ldots \ldots \,(2)$

$ where,\, $

$ \overline{R}=\text{universal}\,\text{gas}\,\text{constant} $

$ {{M} _{o}}=Molecular\,mass $

$ T=temperature $

From (1) and (2)

${{C} _{s}}=C{{\left( \dfrac{\gamma }{3} \right)}^{1/2}}$ 

Both RMS speed and speed of sound in gas are directly proportional to temperature. Thus, both the speeds increases with temperature

The correct option is (a)

Molecular mass of Nitrogen molecule=$14$ g/mol

The RMS speed of sound is directly proportional to square root of temperature. Hence, it is independent of the properties of the medium

$\displaystyle v _{mix} = v _{sound} \sqrt \frac{3}{\gamma} = 400 m/s$

root mean square speed of molecules > average speed of molecules > most probable speed of molecules 

Velocity v = sqrt(gamma * P / rho). Since pressure P and gamma are constant, v is inversely proportional to sqrt(rho). Thus, v1 / v2 = sqrt(rho2 / rho1). Given rho1 / rho2 = 4, then rho2 / rho1 = 1/4. So v1 / v2 = sqrt(1/4) = 1/2.

v_rms = sqrt(3RT/M_mix). v_sound = sqrt(gamma_mix * RT / M_mix). Given v_rms = sqrt(2) * v_sound, then 3RT/M_mix = 2 * gamma_mix * RT / M_mix, so gamma_mix = 1.5. For a mixture, gamma = (n1Cp1 + n2Cp2) / (n1Cv1 + n2Cv2). With 2 moles H2 (gamma=1.4, Cv=2.5R) and n moles He (gamma=1.67, Cv=1.5R), solving for n yields 2.

$\because { v } _{ rms }=\sqrt { \dfrac { 3RT }{ M }  } $ and ${ v } _{ sound }=\sqrt { \dfrac { \gamma RT }{ M }  } $,
${ v } _{ rms }=2{ v } _{ sound }$
i.e. $\gamma =\dfrac { 3 }{ 2 } =$ ratio of $\dfrac { { C } _{ p } }{ { C } _{ V } } $ for the mixture
${ C } _{ V }=\dfrac { { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
and ${ C } _{ p }=\dfrac { { n } _{ 1 }{ c } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
$\therefore \gamma =\dfrac { { C } _{ p } }{ { C } _{ V } } =\dfrac { { n } _{ 1 }{ C } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } } $
$\therefore \dfrac { 3 }{ 2 } =\dfrac { 2\left( \dfrac { 5 }{ 2 } R \right) +n\left( \dfrac { 7 }{ 2 } R \right)  }{ 2\left( \dfrac { 3 }{ 2 } R \right) +n\left( \dfrac { 5 }{ 2 } R \right)  } $
$\Rightarrow \dfrac { 3 }{ 2 } =\dfrac { 10+7n }{ 6+5n } $
$\Rightarrow n=2$