-3
$\begingroup$

Formula for calculating speed of sound in dry air is

$$V(t)=V(0)+0.61t$$

The temperature here is always taken in Celsius .Why don't we use kelvin?

$\endgroup$

closed as unclear what you're asking by ACuriousMind, user36790, Gert, CuriousOne, Qmechanic Feb 27 '16 at 0:53

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It is a useful approximation for those who work in degree Celsius. en.wikipedia.org/wiki/… $\endgroup$ – Farcher Feb 25 '16 at 15:56
  • 2
    $\begingroup$ ...because Kelvin would have a different formula (obtained by $t\mapsto t + 273.15$)? I don't know what your question is. $\endgroup$ – ACuriousMind Feb 25 '16 at 16:00
  • $\begingroup$ It's an approximate scaling equation for small changes in temperature near zero Celsius. The temperature actually scales like the square root of the absolute temperature. $\endgroup$ – Bill N Feb 26 '16 at 21:57
1
$\begingroup$

The speed of sound in an ideal gas is given by:

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

Suppose we want to calculate the speed near some reference temperature $T_0$, i.e. the temperature is $T = T_0 + \delta T$ where $\delta T$ is small. We rewrite the equation for the velocity as:

$$ v = \sqrt{\gamma R(T_0 + \delta T} $$

and then rearrange this to:

$$ v = \sqrt{\gamma R T_0} \left(1 + \frac{\delta T}{T_0}\right)^{1/2}$$

Then because $\delta T \ll T_0$ we can expand the square root using a binomial expansion to get:

$$ v \approx \sqrt{\gamma R T_0} \left(1 + \frac{1}{2}\frac{\delta T}{T_0}\right) $$

But the term $\sqrt{\gamma R T_0}$ is just the velocity at the temperature $T_0$ so our equation becomes:

$$ v \approx v(T_0) + \frac{1}{2}\sqrt{\frac{\gamma R}{T_0}}\delta T $$

and that's how we get the approximate equation you cite. The particular case when $T$ is given in Celcius just comes from taking $T_0 = 273.15$K so there is nothing special about it - it's just a convenient choice of $T_0$.

$\endgroup$
  • $\begingroup$ I think you're missing the molar mass in the denominator. $\endgroup$ – Bill N Feb 26 '16 at 21:59

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