velocity of sound in air in 27C is 340m/s what will be the velocity if the temperature is increased by 50C and pressure of air is doubled?
Here,
T1 = 300 K
T2 = 350 K
v1 = 340 ms-1
P1 = P
P2 = 2P
Let initial density = ρ1
Final density = ρ2
Assuming a perfect gas law remains valid in the atmosphere,
The equation of state of the gas is given by
M = equivalent molar mass of air
P1V1 = nRT1
=> P1 V1 = (m/M) RT1
=> P1 = (m/V1)RT1/M
=> P1 = ρ1RT1/M
=> P1/ρ1 = RT1/M
For the first state
P/ρ1 = RT1/M
Multiplying both sides by γ
γP/ρ1 = γRT1/M
v1 = √ (γP/ρ1)
=> v1 = √ (γRT1/M) ---1.
For the second state
2P/ρ2 = RT2/M
=> ρ2 = 2PM/(RT2)
Now,
v2 = √(γP2/ρ2)
=> v2 = √(γ2P/ρ2)
=> v2 = √(γ2P/{2PM/(RT2)})
=> v2 = √(γRT2/M) ---2.
v2/v1 = √(T2/T1)
v2 = v1√(T2/T1)
= 340×√(350/300)
≈ 367 m/s
T1 = 300 K
T2 = 350 K
v1 = 340 ms-1
P1 = P
P2 = 2P
Let initial density = ρ1
Final density = ρ2
Assuming a perfect gas law remains valid in the atmosphere,
The equation of state of the gas is given by
M = equivalent molar mass of air
P1V1 = nRT1
=> P1 V1 = (m/M) RT1
=> P1 = (m/V1)RT1/M
=> P1 = ρ1RT1/M
=> P1/ρ1 = RT1/M
For the first state
P/ρ1 = RT1/M
Multiplying both sides by γ
γP/ρ1 = γRT1/M
v1 = √ (γP/ρ1)
=> v1 = √ (γRT1/M) ---1.
For the second state
2P/ρ2 = RT2/M
=> ρ2 = 2PM/(RT2)
Now,
v2 = √(γP2/ρ2)
=> v2 = √(γ2P/ρ2)
=> v2 = √(γ2P/{2PM/(RT2)})
=> v2 = √(γRT2/M) ---2.
v2/v1 = √(T2/T1)
v2 = v1√(T2/T1)
= 340×√(350/300)
≈ 367 m/s