The temperature at which the speed of sound in air becomes double of its value at 27 degee celsius?   (a)= -123 deg cel 
(b)= 927 deg cel 
(c)= 327 deg cel 
(d)= 54 deg cel

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.
 
  • 7
Find the answer in the picture below.

  • 2
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.
  • 1
What are you looking for?