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`        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.`
5 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
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

```
5 years ago
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