Question

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

Answer: The speed of sound in the air sample is approximately 341.2 m/s, and the temperature of the air sample is about 292.8 K.

Step-by-step solution

Calculate the speed of sound in the air sample

Use the speed of sound equation, \(v = \sqrt{\frac{B}{\rho}}\), where \(B = 1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) and \(\rho = 1.205 \mathrm{~kg} / \mathrm{m}^{3}\). \(v = \sqrt{\frac{1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m} ^{2}}{1.205 \mathrm{~kg} / \mathrm{m}^{3}}}\) Calculate the value of v using the given values of B and rho: \(v \approx 341.2 \mathrm{~m/s}\) The speed of sound in the air sample is about 341.2 m/s.

Calculate the temperature of the air sample

Use the equation relating the bulk modulus, temperature, and density of an ideal gas, \(B = \rho R T\), where \(B = 1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}\), \(\rho = 1.205 \mathrm{~kg} / \mathrm{m}^{3}\), and \(R = 287 \mathrm{~J} / (\mathrm{kg·K})\). \(1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2} = 1.205 \mathrm{~kg} / \mathrm{m}^{3} \times 287 \mathrm{~J} / (\mathrm{kg·K}) \cdot T\) Rearrange the equation and solve for T: \(T = \frac{1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{1.205 \mathrm{~kg} / \mathrm{m}^{3} \times 287 \mathrm{~J} / (\mathrm{kg·K})}\) Calculate the value of T using the given values of B, rho, and R: \(T \approx 292.8 \mathrm{~K}\) The temperature of the air sample is about 292.8 K.