Question

An adiabatic cylindrical tube of cross-sectional area 1 cm² is closed at one end and fitted with a piston at the other end. The tube contains 0.03 g of an ideal gas. At 1 atm pressure and at the temperature of the surrounding, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure of the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.

Answer 1

Given:
Area of the tube, A = 1 cm² = 1 × 10⁻⁴ m²
Mass of the gas, M = 0.03 g = 0.03 × 10⁻³ kg
Initial pressure, P = 1 atm = 10⁵ Pascal
Initial length of the mercury column, L = 40 cm = 0.4 m
Final length of the mercury column, L₁ = 80 cm = 0.8 m
Final pressure, P' = 0.355 atm

The process is adiabatic. So,
P(V)^γ = P'(V')^γ
⇒ 1 × (A × 0.4)^γ = 0.355 × (A × 0.8)^γ
⇒ 1 × 1 = 0.355 × 2^γ
$$\begin{gathered} \Rightarrow \frac{1}{0.355}=2^{\gamma } \\ \gamma \log 2=\log \left(\frac{1}{0.355}\right) \end{gathered}$$
⇒ γ = 1.4941

Speed of sound in the gas,
$$\begin{gathered} v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma P}{m\mathit{/}v}} \\ v=\sqrt{\frac{1.4941\times {10}^5}{m/v}}=\sqrt{\frac{1.4941\times {10}^5}{\left({\displaystyle \frac{0.03\times {10}^{-3}}{{10}^{-4}\times 1\times 0.4}}\right)}} \\ \Rightarrow v=446.33\approx 447\mathrm{m}/\mathrm{s} \end{gathered}$$