# 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.

Grade:upto college level

## 1 Answers

Jitender Pal
askIITians Faculty 365 Points
10 years ago
Sol. A = 1 cm^2 = 1 × 10^–4 m^2, M = 0.03 g = 0.03 × 10^–3 kg, P = 1 atm = 10^5 pascal, L= 40 cm = 0.4 m. L base 1 = 80 cm = 0.8 m, P = 0.355 atm The process is adiabati P(V)^γ = P(V’)^γ = ⇒ 1 * (AL)^γ = 0.355 * (A2L)^γ ⇒ 1 1 = 0.355 2 ^γ ⇒ 1/0.355 = 2^γ = γ log 2 = log (1/0.355) = 1.4941 V = √γP/f = √1.4941 * 10^5/m / v = √1.4941 * 10^5/(0.03 * 10^-3/10^-4 * 1 * 0.4) = √1.441 * 10^5 * 4 * 10^-5/3 * 10^-5 = 446.33 = 447 m/s

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