0
$\begingroup$

Recently, I have learned that the velocity of sound in air is increased or decreased by temperatures, assumed from the formula V = 331 + 0.6 x (TC), TC being the temperature in Celsius. Then, I realized that if the Celsius is low a number enough (negatively), the speed of sound would be negative!

If there is a place in the universe cold enough with an abundance of oxygen to meet such conditions, or if we say theoretically it is existent, how would the negative velocity act like? Would sound travel behind where it is intended to go?

This may be a stupid question, so please feel free to downvote and explain/comment if my question is illogical or doesn't make sense. Thanks!

$\endgroup$
1
  • $\begingroup$ The minimum possible degree Celsius value is $-273.15\;^\circ\mathrm{C}$, so that formula will never reach a negative result. $\endgroup$
    – Steeven
    Commented Mar 20, 2018 at 6:19

3 Answers 3

2
$\begingroup$

Temperatures can't get arbitrarily cold; the coldest possible temperature is called absolute zero, which occurs at 0 degrees Kelvin, which is -273.15 degrees Celsius. At that temperature, the equation you gave says the velocity of sound in air will be about 167 m/s, which is still a positive velocity.

However, the equation you gave will fail to give correct answers at some low temperature, even before absolute zero is reached. If nothing else, all of the gases in air will liquefy at a temperature higher than absolute zero, and the equation you gave certainly won't continue to apply for liquid air.

$\endgroup$
1
$\begingroup$

Recently, I have learned that the velocity of sound in air is increased or decreased by temperatures, assumed from the formula V = 331 + 0.6 x (TC), TC being the temperature in Celsius.

This is a first order approximation of the ideal gas approximation of the speed of sound in dry air. The ideal gas approximation of the speed of sound in dry air is $$c = \sqrt{\frac{\gamma R T}M}$$ where

  • $\gamma$ is the unitless adiabatic index of the gas, about 1.399 for dry air,
  • $R$ is the universal gas constant, 8.3144598 joule/mole/kelvin,
  • $T$ is the temperature in kelvins, and
  • $M$ is the mean molar of the gas, about 28.965 grams/mole for dry air.

With temperature in degrees Celsius rather than kelvins, this becomes $$c = \sqrt{\frac{\gamma R (273.15 + T)}M}$$ A first order linearization of this is $(331.2 + 0.606 T)\,\text{m/s}$. This linearization is only valid for temperatures within a few dozens of degrees about 0° Celsius, and the ideal gas approximation itself is only valid for temperatures from about -50° Celsius to a few hundred degrees Celsius.

$\endgroup$
0
$\begingroup$

There's a few reasons why you wont see this behavior. The first is that equations like that are typically simplified to exclude factors that typically don't matter. If you got cold enough, you'd find the air starts to liquefy, but few people worry about when it is that cold. Similarly, that model probably does not function properly at temperatures where the air is so hot that it ionizes into plasma. Charged particles tend to transmit sound differently.

The bigger issue, however, is that the temperature where this model would yield negative velocities is too low. You'd have to be at -551.7 C before that model predicts a negative velocity. As it turns out, there is a minimum temperature. 0 Kelvin, or "absolute zero," is -273.15 C. You can't go colder than that.

Well, almost. There's some funny tricks with metastability you can use, but we typically ignore that. For the situations you are interested in, 0K is the limit.

$\endgroup$
1
  • $\begingroup$ Indeed. that formula is an approximation, good for a convenient range of temperatures. brings to mind a prediction made by someone in the fender electric guitars marketing department, looking at trends in buying age versus disposable income for stratocasters. he had a straight line which indicated in ten years, the biggest market opportunity would be in selling electric guitars to millionaire fetuses. $\endgroup$ Commented Mar 20, 2018 at 5:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.