. The temperature at which the speed of sound
in air becomes double of its value at 27°cis:
Answers
Answered by
1
Answer:
927 degrees
Explanation:
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)
squaring both sides
v1^2 = p1 /rho
v2^2 = p2/rho
Taking the ratio
v1^2/v2^2 = p1/p2
Put the values v1=1 and v2 = 2
WE 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.
use it in the above equation we get
T2 = 4 T1
T1 = 27 degrees = 273 + 27 = 300
T2= 1200 degrees = 1200-273 = 927 degrees.
Answered by
0
Answer:
108°c
Explanation:
v= _/`nRT/M
M = Molar mass
R = gas constant
T = Temperature
n = number of moles
v1/2v1 = _/`T1/_/`T2
1/2 = _/`T1/T2
1/4 = T1/T2
1/4 = 27°c/T2
T2 = 27°c×4
T2 = 108°c ans
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