Answer
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Hint:
-Newton formulated the velocity of sound in gas by assuming that the sound wave going through the gas medium in the isothermal process. That means in the time of propagation of the sound the pressure and the volume of many parts of the gas medium can be changed due to the compression and rarefaction but the temperature of the medium remains constant.
-Laplace corrected the assumption of Newton by telling that in the time of propagation of the sound the temperature can not be constant. So, the propagation of sound in the gas medium is an adiabatic process.
Complete step by step answer:
Newton’s Calculation: The velocity of sound that propagates in a gas medium was first formulated by Newton.
He stated the formula of the velocity of the propagation of sound in a gas medium by assuming that the sound wave going through the gas medium in the isothermal process.
In the time of propagation of the sound, the pressure and the volume of many parts of the gas medium can be changed due to the compression and rarefaction but the temperature of the medium remains constant.
Let us consider, for a part of the gas a certain amount of pressure is $P$ and volume is $V$. During the propagation of the sound, the pressure becomes $(P + {P_1})$ by increasing and hence the volume becomes $(V - v)$ by decreasing.
Since the temperature is constant, from Boyle's law,
$PV = (P + {P_1})(V - v)$
On multiplying the term and we get,
$ \Rightarrow PV = PV + {P_1}V - Pv - {P_1}v$
$ \Rightarrow 0 = {P_1}V - Pv$ [The term ${P_1}v$ is too small and hence neglected]
$ \Rightarrow Pv = {P_1}V$
$ \Rightarrow \dfrac{{{P_1}V}}{v} = P$
On rewriting the terms and we get,
$ \Rightarrow P = \dfrac{{{P_1}}}{{\dfrac{v}{V}}} = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
${\text{bulk modulus = }}k = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
$\therefore P = k$
We know, if the sound propagating in a gas medium of density $\rho $ and bulk modulus $k$, the velocity of sound is,
$c = \sqrt {\dfrac{k}{\rho }} $
So, from Newton’s calculation, we get, $c = \sqrt {\dfrac{P}{\rho }} $.
Laplace’s Correction: Laplace first told that during the propagation of sound in the gas medium the temperature does not constant. The assumption of Newton of the isothermal process is not right.
Laplace assumed that during the compression and rarefaction the expansion and compression of the layers are so fast to keep the temperature constant. Hence the process should be the adiabatic process instead of the isothermal process.
In the adiabatic process, if a part of the gas has a certain amount of pressure is $P$ and volume is $V$,
$P{V^\gamma } = {\text{constant}}$.
Hence, During the propagation of the sound, if the pressure becomes $(P + {P_1})$ by increasing and hence the volume becomes $(V - v)$ by decreasing.
$P{V^\gamma } = (P + {P_1}){(V - v)^\gamma }$
$ \Rightarrow P{V^\gamma } = (P + {P_1}){V^\gamma }{\left( {1 - \dfrac{v}{V}} \right)^\gamma }$
$ \Rightarrow P = (P + {P_1})\left( {1 - \dfrac{{\gamma v}}{V}} \right)$ [ for Binomial expansion the other terms are neglected]
$ \Rightarrow P = P + {P_1} - \dfrac{{P\gamma v}}{V} - \dfrac{{{P_1}\gamma v}}{V}$
$ \Rightarrow {P_1} = \dfrac{{\gamma Pv}}{V}$ [The term ${P_1}v$ is too small and hence neglected]
\[ \Rightarrow \gamma P = \dfrac{{{P_1}}}{{\dfrac{v}{V}}} = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}\]
${\text{bulk modulus = }}k = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
$\therefore \gamma P = k$
So, from Laplace's Correction we get, $c = \sqrt {\dfrac{{\gamma P}}{\rho }} $.
Note:
-From Newton’s calculation, we get, $c = \sqrt {\dfrac{P}{\rho }} $.
-From Laplace's Correction we get, $c = \sqrt {\dfrac{{\gamma P}}{\rho }} $.
-Newton formulated the velocity of sound in gas by assuming that the sound wave going through the gas medium in the isothermal process. That means in the time of propagation of the sound the pressure and the volume of many parts of the gas medium can be changed due to the compression and rarefaction but the temperature of the medium remains constant.
-Laplace corrected the assumption of Newton by telling that in the time of propagation of the sound the temperature can not be constant. So, the propagation of sound in the gas medium is an adiabatic process.
Complete step by step answer:
Newton’s Calculation: The velocity of sound that propagates in a gas medium was first formulated by Newton.
He stated the formula of the velocity of the propagation of sound in a gas medium by assuming that the sound wave going through the gas medium in the isothermal process.
In the time of propagation of the sound, the pressure and the volume of many parts of the gas medium can be changed due to the compression and rarefaction but the temperature of the medium remains constant.
Let us consider, for a part of the gas a certain amount of pressure is $P$ and volume is $V$. During the propagation of the sound, the pressure becomes $(P + {P_1})$ by increasing and hence the volume becomes $(V - v)$ by decreasing.
Since the temperature is constant, from Boyle's law,
$PV = (P + {P_1})(V - v)$
On multiplying the term and we get,
$ \Rightarrow PV = PV + {P_1}V - Pv - {P_1}v$
$ \Rightarrow 0 = {P_1}V - Pv$ [The term ${P_1}v$ is too small and hence neglected]
$ \Rightarrow Pv = {P_1}V$
$ \Rightarrow \dfrac{{{P_1}V}}{v} = P$
On rewriting the terms and we get,
$ \Rightarrow P = \dfrac{{{P_1}}}{{\dfrac{v}{V}}} = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
${\text{bulk modulus = }}k = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
$\therefore P = k$
We know, if the sound propagating in a gas medium of density $\rho $ and bulk modulus $k$, the velocity of sound is,
$c = \sqrt {\dfrac{k}{\rho }} $
So, from Newton’s calculation, we get, $c = \sqrt {\dfrac{P}{\rho }} $.
Laplace’s Correction: Laplace first told that during the propagation of sound in the gas medium the temperature does not constant. The assumption of Newton of the isothermal process is not right.
Laplace assumed that during the compression and rarefaction the expansion and compression of the layers are so fast to keep the temperature constant. Hence the process should be the adiabatic process instead of the isothermal process.
In the adiabatic process, if a part of the gas has a certain amount of pressure is $P$ and volume is $V$,
$P{V^\gamma } = {\text{constant}}$.
Hence, During the propagation of the sound, if the pressure becomes $(P + {P_1})$ by increasing and hence the volume becomes $(V - v)$ by decreasing.
$P{V^\gamma } = (P + {P_1}){(V - v)^\gamma }$
$ \Rightarrow P{V^\gamma } = (P + {P_1}){V^\gamma }{\left( {1 - \dfrac{v}{V}} \right)^\gamma }$
$ \Rightarrow P = (P + {P_1})\left( {1 - \dfrac{{\gamma v}}{V}} \right)$ [ for Binomial expansion the other terms are neglected]
$ \Rightarrow P = P + {P_1} - \dfrac{{P\gamma v}}{V} - \dfrac{{{P_1}\gamma v}}{V}$
$ \Rightarrow {P_1} = \dfrac{{\gamma Pv}}{V}$ [The term ${P_1}v$ is too small and hence neglected]
\[ \Rightarrow \gamma P = \dfrac{{{P_1}}}{{\dfrac{v}{V}}} = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}\]
${\text{bulk modulus = }}k = \dfrac{{{\text{volume stress}}}}{{{\text{volume strain}}}}$
$\therefore \gamma P = k$
So, from Laplace's Correction we get, $c = \sqrt {\dfrac{{\gamma P}}{\rho }} $.
Note:
-From Newton’s calculation, we get, $c = \sqrt {\dfrac{P}{\rho }} $.
Now for STP, the value of the velocity will be $c = \sqrt {\dfrac{{76 \times 13.6 \times 980}}{{0.001293}}} \simeq 28000cm/s \simeq 280m/s$. But the value obtained from many experiments is $332m/s$.
-From Laplace's Correction we get, $c = \sqrt {\dfrac{{\gamma P}}{\rho }} $.
Now for STP, the value of the velocity will be $c = \sqrt {\dfrac{{1.4 \times 76 \times 13.6 \times 980}}{{0.001293}}} = 33117cm/s = 331.17m/s$. This value is very much nearer to the experimental value(i.e. $332m/s$).
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