The density of a sample of air is 1.157 kg/m3, and the bulk modulus is 1.42 · 105 N/m2.
a) Find the speed of sound in the air sample. -----> I found it 350.3 m/s and its correct.
b) Find the temperature of the air sample. Give answer in °C. ----> I cant find this :S
Can you help me, please?
The bulk modulus K used here is the so-called adiabatic bulk modulus and this is related to the pressure P according to:
K = gamma P
where gamma = cp/cv = approximately 1.4 for air.
So, you know the pressure and the density and then you can find the temperature via the ideal gas law.
To find the temperature of the air sample, we can use the ideal gas law equation, which relates the pressure, volume, and temperature of a gas:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
In this case, we do not have the pressure or volume of the air, but we know its density and bulk modulus. We can use these values to find the speed of sound in the air sample, and then use the speed of sound to find the temperature.
The speed of sound in a gas is given by the formula:
v = sqrt(B/ρ)
Where:
v is the speed of sound
B is the bulk modulus of the gas
ρ is the density of the gas
In this case, we know the bulk modulus (B = 1.42 · 105 N/m^2) and the density (ρ = 1.157 kg/m^3). We can plug these values into the formula and solve for v:
v = sqrt(1.42 · 105 N/m^2 / 1.157 kg/m^3)
v = sqrt(1.23 x 108 m^2/s^2)
Therefore, the speed of sound in the air sample is approximately 350.3 m/s, as you have correctly found.
Now, to calculate the temperature (T) in Celsius, we need to convert the speed of sound (v) from meters per second to centimeters per second because the units in the ideal gas law equation are typically in cm^3 and g.
v = 350.3 m/s = 350.3 x 10^2 cm/s
Now, we can plug the speed of sound value into the ideal gas law equation:
v = sqrt(B/ρ) = sqrt(RT/M)
Where M is the molar mass (amount of mass in one mole) of the air. The molar mass of air is approximately 28.97 g/mol.
So,
350.3 x 10^2 cm/s = sqrt(RT / (1.157 g/cm^3 / 28.97 g/mol))
Rearranging the equation, we get:
(350.3 x 10^2)^2 = RT / (1.157 / 28.97)
Simplifying further:
R x T = (350.3 x 10^2)^2 x (1.157 / 28.97)
Now, let's plug in the value of the ideal gas constant (R = 8.314 J/(mol K)):
8.314 J/(mol K) x T = (350.3 x 10^2)^2 x (1.157 / 28.97)
Now, let's solve for T:
T = [(350.3 x 10^2)^2 x (1.157 / 28.97)] / 8.314
Calculating this value, we get:
T ≈ 387.4 K
To convert this temperature from Kelvin to Celsius, we subtract 273.15:
T ≈ 387.4 K - 273.15 ≈ 114.25°C
Therefore, the temperature of the air sample is approximately 114.25°C.