Speed of Sound in a Gas (2.1.03)

As this fighter jet reaches the speed of sound, it leaves behind a
trail of water vapor. This is because of the drop in air pressure!

Question: The speed of sound in a gas $c_s$ is related to the pressure $P$ and density $\rho$ of the gas. Use dimensional analysis to figure out the form of this relationship.

Ahhhh... dimensional analysis! It is one of the most useful tools we have to make sure that any problem we are solving actually makes sense. Because if the units don't make sense, then you messed up somewhere.

Dimensional analysis can also be used to figure out the relationships between various physical properties. In this problem, we are given that the speed of sound is related to the pressure $P$ and density $\rho$ of a gas. But how do pressure and density relate to each other, and how can we make sure that they are accurately representing the relationship to the speed of sound in a gas?

Lets start by defining pressure $P$, density $\rho$, and the speed of sound, $c_s$ in terms of their physical units. To keep things consistent, let's use $d$ for distance (in centimeters), $t$ for time (in seconds), and $m$ for mass (in grams):

• $P = \frac{Force}{Area} = \frac{Newtons}{d^2} = \frac{\frac{m \cdot d}{t^2}}{d^2}$
• $\rho = \frac{Mass}{Volume} = \frac{m}{d^3}$
• $c_s = velocity = \frac{d}{t}$

Based on this, we can see that the distance unity, $d$ needs to end up in the numerator, and the time unit, $t$ needs to be in the denominator in order for the units to make sense relative to the speed of sound. Therefore, since only $P$ has a time unit, $t$ in should be in the numerator. Since $\rho$ has a larger magnitude of distance, $d^3$ in it's denominator than either of the distance units in $P$, it will contribute more by being the denominator of the ratio between $P$ and $\rho$ so that the distance unit is actually in the numerator when simplified. Let's check to see if this reasoning makes sense mathematically using dimensional analysis:  

$\begin{align*}
c_s & \propto \frac{P}{\rho}\\
\frac{d}{t} & \propto  \frac{\frac{\frac{m \cdot d}{t^2}}{d^2}}{\frac{m}{d^3}}\\
\frac{d}{t} & \propto \frac{\frac{m \cdot d}{d^2 \cdot t^2}}{\frac{m}{d^3}}\\
\frac{d}{t} & \propto \frac{m \cdot d^4}{m \cdot d^2 \cdot t^2}\\
\frac{d}{t} & \propto \frac{d^2}{t^2}\\

\therefore c_s & \propto \sqrt{\frac{P}{\rho}}
 \end{align*}$

As you can see, the the speed of sound in a gas is proportional to the square root of the ratio of $P$ to $\rho$.