Question

the temperature at which the speed of sound in air becomes double of its value at 27 c is

Answer 1

Use the formula 
velocity of sound waves v = sqrt (P/rho) 

P is the pressure and rho is the density.Write this as 

v1 = root(p1/rho) 
v2= root (p2/rho) 

square both sides 

v1^2 = p1 /rho 

v2^2 = p2/rho 

Take the ration 

v1^2/v2^2 = p1/p2 

Put the values v1=1 and v2 = 2 

YOu get 

1/4 = p1/p2 

p1 = n kb T1 and p2= n kb T2. Here n is the number density. kb Boltzmann constant and T1 and T2 are temperatures. 

If you use it in the above equation you get 

T2= 4 T1 

T1 is 27 degrees= 273 + 27 = 300 

T2= 1200 degrees = 1200-273 = 927 degrees.
 

Answer 2

Answer:

Concept:

Speed of sound is the rate at which sound waves move through various substances. Modern estimates of the speed of sound, specifically for dry air at a temperature of 0 °C (32 °F), are 331.29 meters (1,086.9 feet) per second.

Explanation:

White noise is a complicated signal or sound that spans the full audible frequency spectrum with equal intensity at each frequency. White noise is comparable to white light, which has approximately equal intensities of all visible light wavelengths. The static that may be heard between FM radio stations is a decent representation of white noise.

Speed of sound in air is given as $\sqrt{\frac{3 R T}{M_{0}}}$

Thus the speed   $V \propto \sqrt{T}$

Thus velocity becomes double, the temperature must go to four times its initial value.

Initial temperature  $= 27+273 = 300 \mathrm{~K}$

Hence final temperature  $=1200 \mathrm{~K} = 927^{\circ} \mathrm{C}$.

The temperature at which the speed of sound in the air becomes double its value at 27 c is  27° C.

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