The density of a sample of air is \(1.205 \mathrm{~kg} / \mathrm{m}^{3}\), and the bulk modulus is \(1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) a) Find the speed of sound in the air sample. b) Find the temperature of the air sample.

Short Answer

Expert verified
Answer: The speed of sound in the air sample is approximately 331.25 m/s, and the temperature is approximately 273.1 K.

Step by step solution

01

Find the speed of sound in the air sample

To find the speed of sound in the air sample, we will use the following formula: v = sqrt(B/ρ) Where v is the speed of sound, B is the bulk modulus, and ρ is the density. Given values: B = 1.42 * 10^5 N/m^2 ρ = 1.205 kg/m^3 Plug in the given values into the formula: v = sqrt((1.42 * 10^5 N/m^2) / (1.205 kg/m^3)) Now, calculate the speed of sound (v): v ≈ 331.25 m/s The speed of sound in the air sample is approximately 331.25 m/s.
02

Find the temperature of the air sample

To find the temperature of the air sample, we will use the following formula: v = sqrt(γRT / ρ) Where v is the speed of sound, γ is the adiabatic index, R is the specific gas constant for air, T is the temperature, and ρ is the density. Given values: v = 331.25 m/s γ = 1.4 (adiabatic index for air) R = 287 J/(kg*K) (specific gas constant for air) ρ = 1.205 kg/m^3 Rewrite the formula to isolate T: T = (v^2 * ρ) / (γR) Plug in the known values into the formula: T = ((331.25 m/s)^2 * (1.205 kg/m^3)) / (1.4 * 287 J/(kg*K)) Now, calculate the temperature (T): T ≈ 273.1 K The temperature of the air sample is approximately 273.1 K.

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