How is the speed of sound dependent on pressure change in this formula? [closed]








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In the following equation,
$$\text{speed of sound}=c=\sqrt{\frac{\gamma RT}{M}}$$


The pressure term is missing or I guess is integrated in the adiabatic constant $\gamma$. But I don't know how to even derive the adiabatic constant for a certain gas.




The best way to understand an equation in my view is to derive it from scratch ... along the way you pick up the details as to why you have certain conditions or restrictions that limit the usage of the equation. With this in mind I will outline an easy way to derive the equation in your book. Along the way you may think "how on earth would you first think about doing this derivation"! The answer is that it is a longer yet much more basic way of arriving at the same result as opposed to the "proper way. 


If you wish me to describe it the proper way I will add it in at the end, its much shorter but possibly a little harder maths/concepts.


Okay here goes:


Longitudinal sound waves cause small oscillations in the distribution of the gas molecules (hard spheres) which propagate in the direction of the oscillations. These compressions (regions of higher pressure than usual) and rarefactions (regions of lower pressure than usual) effects the local density and pressure of the gas molecules in a given region of space with respect to their equilibrium values. 


If we imagine that a cylinder of gas, which has a cross sectional area of $A$ and a length of $vdt$ where $v$ is the velocity of the gas molecules and $dt$ is the time they spend in the cylinder, has a sound wave propagating through it length ways. Where we have a compression of the molecules: 


1) their pressure increases from equilibrium $P\_0$ to $P$ by $dP$, where $P=P\_0+dP$. 


2) the local density increases from equilibrium $\rho \_0$ to $\rho$ by $d\rho$, where $\rho=\rho \_0 +d\rho$. 


3) due to all the extra collisions of lots of molecules in a small space the average velocity of each molecule $v$ is less than it is at equilibrium $v\_o$ by $dv$ where $dv$ represents a deceleration (so is negative). 


This is the opposite of the rarefied regions which experience an acceleration, and occupy a localised region of lower $P$ and $\rho$ than equilibrium. 


The extra force exerted on the molecules in the compressed zone is given by $F=dPA$. The volume of the cylinder $V$ is its length multiplied by its cross sectional area giving us $V=Avdt$. 


The mass is just volume times density (equilibrium density), $m=\rho \_0 Avdt$, notice that this part is independent of local changes since rarefactions/compressions manifest themselves by making $v$ larger/smaller and correspondingly $dt$ smaller/larger.


We can now substitute that into Newton's second law $F=ma$ where $a$ is the acceleration of the gas molecules. We can go one of two ways here either compression or rarefaction ... lets pick a compression. This means that we will have a deceleration in the velocity of the gas molecules. Therefore the acceleration is now given by:
\begin{equation}
a=-\frac{dv}{dt}
\end{equation}
Further we can substitute the mass into Newton's second law: 
\begin{equation}
F=-\rho \_0Avdt\frac{dv}{dt}
\end{equation}
But remember from the start we said that the force is also given by $F=AdP$. 
\begin{equation}
AdP=-\rho \_0Avdt\frac{dv}{dt}
\end{equation}
Cancel the $A$s and $dt$s and re-arrange:
\begin{equation}
-\frac{dP}{dv}=\rho \_0v
\end{equation}
Multiply both sides by $v$,
\begin{equation}
-v\frac{dP}{dv}=\rho \_0v^2
\end{equation}
Right lets put this result to one side and think about the change in volume. In fact we need to consider the fractional change in volume since volume is an extensive property. Therefore we will be interested not in $dV$ but in the quantity $dV/V\_0$. We already know that the equilibrium volume is given by $V\_0=Avdt$, the change in this volume under a compression is given by $dV=Advdt$. Therefore,
\begin{equation}
\frac{dV}{V\_0}=\frac{Advdt}{Avdt}
=\frac{dv}{v}
\end{equation}
Substituting this result into the above equation:
\begin{equation}
-\frac{V\_0}{dV}dP=-\frac{dP}{\bigg(\frac{dV}{V\_0}\bigg)}=\rho\_0v^2
\end{equation}
Now define the quantity in the middle as $the$ *[bulk modulus](http://en.wikipedia.org/wiki/Bulk_modulus)*, $B$. It is the ratio of an infinitesimal pressure change to the relative change in volume, it gives an indication of the compressibility of a material. 


Now rearrange this for the velocity:
\begin{equation}
v= \sqrt{\frac{B}{\rho \_0}}
\end{equation}


At this point you have your equation in the most general form, the [Newton-Laplace equation](http://www.thermaxxjackets.com/newton-laplace-equation-sound-velocity/). From here you can pick what equation of state you wish to use and tailor this equation as needed. You can clearly see how the pressure and density information is tied up in this expression. But that being said its not so great for the experimentalists. Think about how you would take these readings and so on... 


Therefore we are not quite done!


If we use [Boyle's law](https://en.wikipedia.org/wiki/Boyle%27s_law) $P\_1V\_1=P\_2V\_2$ as Newton did then you will find that although close to experimental values your always slightly off. Therefore we will use a different equation of state... one that inherently assumes constant entropy (some people say that adiabatic means that "no heat is transferred", this is the easy way to think about things .... what they really mean is that the process occurs at constant entropy .... an adiabatic process means constant entropy... (maybe if you study statistical mechanics and phase space this will be more apparent). 


The equation we should use is the adiabatic version of Boyle's law. The reason that a sound wave travelling through a medium is considered adiabatic is because "no heat is exchanged between rarefactions and compressions" i.e the equilibrium temperature is maintained because rarefaction velocity change cancels out compressions velocity change on average meaning that the kinetic energy is pretty much constant and hence so is $T$ from $E\_k=\frac{3}{2} NKT$. So instead of Boyle's law we use $PV^{\gamma}=P\_0V\_0^{\gamma}$. But for a given temperature the entire right hand side is just a constant so lets instead use $PV^{\gamma}=K$ where $K$ is the constant. 


\begin{equation}
P=KV^{-\gamma}
\end{equation}


Now lets differentiate $P$ with respect to $V$.... lets find the rate of change oof pressure with respect to the change in volume...


\begin{equation}
\frac{dP}{dV}=-\gamma KV^{-\gamma-1}
\end{equation}


Multiply both sides by $-V$ and you will notice we have an expression for $B$,


\begin{equation}
-V\frac{dP}{dV}=KV\gamma V^{-\gamma-1}=VK\gamma V^{-\gamma}V^{-1}=\frac{K\gamma}{V^{\gamma}}
\end{equation}


Let $V=V\_0$ the equilibrium volume which we are free to do. Now show the equilibrium volume in terms of the equilibrium density $\rho\_0$.


\begin{equation}
\rho\_0=\frac{nM}{V\_0}
\end{equation}


If we now go back and put these in $v=\sqrt{\frac{B}{\rho\_0}}$ and simplify then we have the equation stated in your book. 


\begin{equation}
v=\sqrt{\frac{B}{\rho\_0}}=\sqrt{\frac{K\gamma V\_0}{V\_0^{\gamma}nM}}
\end{equation}


Using $K/V\_0^{\gamma}=P\_0$ and then the [ideal gas law](https://en.wikipedia.org/wiki/Ideal_gas_law) $P\_0V\_0=nRT$,


\begin{equation}
v=\sqrt{\frac{\gamma P\_0V\_0}{nM}}=\sqrt{\frac{\gamma RT}{M}}
\end{equation}


So you see now how pressure is tied up in the entire expression, along with the density. The assumption we had to assume was that the sound wave travelling through is adiabatic and there we have the speed of the wave in a gas of non-interacting (except for collisions) hard spheres. 


If you'd like I can derive the original formula in a much more professional way from elementary fluid mechanics ... but although quicker and more sophisticated it is a little harder. (if I've made any errors feel free to edit away) :) 


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