# Adiabatic Process and the speed of sound

Hi all,

Sorry if this post is a bit wordy but I've been going round in circles and I thought I'd see if anyone on here can help, it's also my first post so be nice...

I've been trying to understand how Newton miscalculated the speed of sound. I know that he thought that the propagation of a sound wave was an isothermal process and he worked out the speed based on air's elasticity and density. I'm also aware that it is actually an adiabatic process. I understand this to the point that there is a local change in temperature due to the compression and rarefaction half - cycles of the waveform and that the these changes occur so quickly that no heat can enter or leave the cycle, hence an adiabatic process. What I don't understand is why this heating and cooling cause the speed of sound to increase? I have a few ideas but I haven't been able to find an explanation anywhere and am beginning to think this because the answer is so obvious it doesn't need saying!!! Any help with this will be greatly appreciated.

Thanks

Justin

Andrew Mason
Homework Helper
I understand this to the point that there is a local change in temperature due to the compression and rarefaction half - cycles of the waveform and that the these changes occur so quickly that no heat can enter or leave the cycle, hence an adiabatic process. What I don't understand is why this heating and cooling cause the speed of sound to increase? I have a few ideas but I haven't been able to find an explanation anywhere and am beginning to think this because the answer is so obvious it doesn't need saying!!! Any help with this will be greatly appreciated.
Newton found that the speed was related to the rate of change of pressure with density:

$$v^2 = \frac{dP}{d\rho}$$

This was based on Newton's correct analysis of the mechanical forces that are needed to accelerate the medium (which has mass).

If one applies the relationship between pressure and volume at constant temperature: PV=constant = nRT. So:

$P = \frac{n}{V}RT$

Since $\rho = nM/V$ where M = molecular weight of the gas, $P = MRT\rho$

$$\frac{dP}{d\rho} = MRT$$

so $$v^2 = MRT$$

What Newton did not take into account was the fact that when a gas is compressed (adiabatically), its temperature also increases. This increase in temperature makes the pressure higher. So the rate of pressure with density is higher when one takes into account this increase in temperature. Applying the adiabatic condition:

$$PV^\gamma = \text{constant}$$

If you work out the rate of change of P with density (mass/V = nM/V) you get:

$$\frac{dP}{d\rho} = \gamma MRT$$

so $$v = \sqrt{\gamma MRT}$$

For air, $\gamma$ = 1.4 so the difference is a factor of $\sqrt{1.4} = 1.18$

AM

Last edited:
Andrew Mason
Homework Helper
Since $\rho = nM/V$ where M = molecular weight of the gas, $P = MRT\rho$

$$\frac{dP}{d\rho} = MRT$$

so $$v^2 = MRT$$
I made a sloppy error in this post and then discovered it was too late to correct it so I will post this correction. The RT term must be divided by M (molecular wt in Kg/mol):

Since $\rho = nM/V$, the ideal gas law can be written:

$$P = \frac{n}{V}RT = \frac{\rho}{M}RT$$

Differentiating with respect to $\rho$ we get:

$$\frac{dP}{d\rho} = RT/M$$

so $$v^2 = \frac{RT}{M}$$

Similarly, when correcting for $\gamma$, it should be

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

AM

Hi Andrew,

Thanks for responding. Although I can follow the maths I don't really know what a lot of the terms mean, but that's okay as I was just trying to understand why there is a change. But I do now get that if the speed of sound is the rate of change of pressure with density and that as a gas is compressed it increases in temperature, I can see how the rate of change of pressure would also increase and therefore so would the speed of sound. Am I right in saying that during the rarefaction half cycle the cooling of air would decrease the pressure and therefore increase the rate of change of pressure with density?

Thanks

Justin

Andrew Mason
Yes. If the change in density is negative (expansion), the greater drop in pressure resulting from the temperature decrease makes $dP/d\rho$ greater.