91Ó°ÊÓ

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\mathrm{ÒÏ}}}$ ……. (1)

Here, is the bulk modulus, is the density of the medium.

Expression for bulk modulus is

$B=\frac{\mathrm{Δ}P}{−\left(\frac{\mathrm{Δ}V}{V}\right)}$ …… (2)

Here,$\mathrm{Δ}P$ is the change in pressure,$\mathrm{Δ}V$ is the change in volume, and minus sign in the denominator is there because an increase in pressure causes a decrease in volume.

For an isothermal process, the expression for pressure is

$P=\frac{NkT}{V}$ …… (3)

Here,$k$ is Boltzmann constant,$N$ is the number of molecules,$V\text{ a²Ô»å }T$ are volume and temperature of the gas respectively.

For adiabatic process,

$P=\text{C}{V}^{−\mathrm{γ}}$

Where C is constant.

As$T$ is constant for an isothermal process, change in pressure can be written as

$\mathrm{Δ}P=NkT\mathrm{Δ}\left(\frac{1}{V}\right)$

$\mathrm{Δ}P=−NkT\left(\frac{\mathrm{Δ}V}{V^2}\right)$ …… (4)

Substitute$P$ for$\frac{NkT}{V}$ in equation (4)

$\begin{array}{l}\mathrm{Δ}P=−NkT\left(\frac{\mathrm{Δ}V}{V^2}\right)\\ \mathrm{Δ}P=−\left(\frac{NkT}{V}\right)\left(\frac{\mathrm{Δ}V}{V}\right)\\ \mathrm{Δ}P=−P\left(\frac{\mathrm{Δ}V}{V}\right)\end{array}$

$\left(\frac{\mathrm{Δ}V}{V}\right)=−\frac{\mathrm{Δ}P}{P}$ …… (5)

Substitute$\frac{\mathrm{Δ}P}{P}$ for$\left(−\frac{\mathrm{Δ}V}{V}\right)$ in equation (2)

$B=\frac{\cancel{\mathrm{Δ}P}}{\frac{\cancel{\mathrm{Δ}P}}{P}}$

$\boxed{B=P}$

For adiabatic compression,

$P=\text{C}{V}^{−\mathrm{γ}}$ …… (6)

Change in pressure is calculated as

$\begin{array}{l}\mathrm{Δ}P=C\mathrm{Δ}\left(V^{−\mathrm{γ}}\right)\\ \mathrm{Δ}P=C\left(−\mathrm{γ}{V}^{−\mathrm{γ}−1}\mathrm{Δ}V\right)\end{array}$

$\mathrm{Δ}P=−\mathrm{γ}\text{C}\left(\frac{V^{−\mathrm{γ}}}{V}\mathrm{Δ}V\right)$ …… (7)

Equation (7) can be simplified as

$\mathrm{Δ}P=−\mathrm{γ}\text{C}{V}^{−\mathrm{γ}}.\left(\frac{1}{V}\mathrm{Δ}V\right)$ …… (8)

Substitute$P$ for$\text{C}{V}^{−\mathrm{γ}}$ in equation (8)

$\mathrm{Δ}P=−\mathrm{γ}P\left(\frac{1}{V}\mathrm{Δ}V\right)$

$\left(\frac{1}{V}\mathrm{Δ}V\right)=−\frac{\mathrm{Δ}P}{\mathrm{γ}P}$ …… (9)

Substitute $−\frac{\mathrm{Δ}P}{\mathrm{γ}P}$for$\left(\frac{1}{V}\mathrm{Δ}V\right)$ in equation (2)

$\boxed{B_{adiabatic}=\mathrm{γ}P}$

Hence, the required bulk modulus for the isothermal process is $\boxed{B=P}$and for adiabatic compression is $\boxed{B_{adiabatic}=\mathrm{γ}P}$.

Isothermal compression is so slow that the temperature of the gas doesn’t rise at all and in the adiabatic compression, the process is so fast that no heat escapes from the gas during the process. Most real compression processes will be somewhere between these extremes usually closer to the adiabatic approximation.

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\mathrm{ÒÏ}}}$ ……. (1)

Here, is the bulk modulus, is the density of the medium.

Since the sound wave is a kind of pressure wave that propagates through the medium at a high speed at, pressure changes occur here and heat exchange proceeds very slowly, so the adiabatic bulk modulus is the perhaps proper one to use in calculating. The speed of sound can be calculated from the bulk modulus and the material's mass density as $c_s=\sqrt{\frac{B_{adiabatic}}{\mathrm{ÒÏ}}}$.

Hence, the speed of sound in terms of bulk modulus and density of the medium is $c_s=\sqrt{\frac{B_{adiabatic}}{\mathrm{ÒÏ}}}$.

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\mathrm{ÒÏ}}}$ ……. (1)

Here,$B$ is the bulk modulus,$\mathrm{ÒÏ}$ is the density of the medium.

Speed of sound in terms of the bulk modulus is

$c_s=\sqrt{\frac{B_{adiabatic}}{\mathrm{ÒÏ}}}$ …… (2)

For an ideal gas, the expression for pressure is

$P=\frac{NkT}{V}$ …… (3)

Here,$k$ is Boltzmann constant,$N$ is the number of molecules,$V\text{ a²Ô»å }T$ are volume and temperature of the gas respectively.

Average molecular mass can be expressed as

$m=\frac{\text{³¾²¹²õ²õ o´Ú o²Ô±ð″¾´Ç±ô±ð o´Ú a¾±°ù}}{{\text{²Ô³Ü³¾²ú±ð°ù o´Ú″¾´Ç±ô±ð³¦³Ü±ô±ð²õ i²Ô o²Ô±ð″¾´Ç±ô±ð N}}_{\text{A}}}$ …… (4)

Density can be expressed in terms of mass and volume is

$\mathrm{ÒÏ}=\frac{Nm}{V}$ …… (5)

Here,$N$ is a number of molecules.

Substitute$\frac{Nm}{V}$ for$\mathrm{ÒÏ}$ in equation (2)

$c_s=\sqrt{\frac{B_{adiabatic}}{\mathrm{ÒÏ}}}$

$c_s=\sqrt{\frac{V{B}_{aiabatic}}{Nm}}$ …… (6)

Substitute$\mathrm{γ}P$ for$B_{adiabatic}$ in equation (5)

$c_s=\sqrt{\frac{VPV}{Nm}}$ …… (7)

But from equation (3),$PV=NkT$

Substitute $NkT$for$PV$ in equation (7)

$c_s=\sqrt{\frac{VNkT}{Nm}}=\sqrt{\frac{\mathrm{γ}kT}{m}}$

$\boxed{c_s=\sqrt{\frac{\mathrm{γ}kT}{m}}}$ …… (8)

The root mean square speed of the molecules in an ideal gas is given by

$v_{rms}=\sqrt{\frac{3kT}{m}}$ …… (9)

Divide equation (8) by equation (9)

$\frac{c_s}{v_{rms}}=\sqrt{\frac{\mathrm{γ}}{3}}$ …… (10)

But for air,$\mathrm{γ}=1.4$

Substitute$1.4$ for$\mathrm{γ}$ in equation (11)

$\begin{array}{l}\frac{c_s}{v_{rms}}=\sqrt{\frac{1.4}{3}}\\ \boxed{\frac{c_s}{v_{rms}}=0.68}\end{array}$

Substitute 0.0289 g for the mass of one mole of air and$6.022×{10}^{23}$ for$N_A$ in equation (4)

$\begin{array}{c}m=\frac{0.0289}{6.022×{10}^{23}}\\ m=4.81×{10}^{−26}\text{ k²µ}\end{array}$

Substitute$4.81×{10}^{−26}\text{ k²µ}$ for $m$,$1.38×{10}^{−23}{\text{″¾}}^{\text{2}}{\text{.kg.s}}^{\text{-2}}{\text{.K}}^{\text{-1}}$ for $k$, 1.4 for $\mathrm{γ}$,$293\text{‿é}$ for$T$ in equation (8)

$\begin{array}{l}{c}_s=\sqrt{\frac{\mathrm{γ}kT}{m}}\\ =\sqrt{\frac{1.4×1.38×{10}^{−23}×293}{4.81×{10}^{−26}}}\\ \boxed{c_s=343{\text{″¾²õ}}^{\text{-1}}}\end{array}$

Hence, the required expression for the speed of sound is, ratio between root mean square speed and speed of sound in air is and speed of sound in air in room temperature is.

The speed of sound is dependent on the density of the medium and the temperature of the medium. It also depends on the adiabatic exponent. It directly doesn’t depend on altitude but indirectly it depends.

Speed of sound wave is given by

$c_s=\sqrt{\frac{\mathrm{γ}kT}{m}}$ ……. (1)

Here,$\mathrm{γ}$ is the adiabatic exponent,$k$ is Boltzmann constant,$T$ is the temperature in Kelvin, and$m$ its mass of gas.

Since the temperature decreases by about$9.8\text{ }°\text{C}$ for every 1000 m if it is sunny or dry day and it decreases by$6°\text{C}$ if it’s snowing or raining. Since the speed of sound is proportional to the square root of temperature, so the speed of sound seems to decrease as the temperature decreases, from where it can be concluded that the speed of sound decreases as altitude increases.

Hence, the speed of sound decreases with increasing altitude.