An adibatic cylindrical tube of cross-sectional area `1 cm^(2)` is closed at one end 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 temperture of the surrounding, the length of the gas column. The pressure of the gas falls to `0.355 atm`. Find th speed of sound in the gas at atmospheric temperture.

Explanation

`A=1cm^2= 1 xx 10^(-4) m^2`,
`M=0.03 g = 0.03 xx 10 ^(-3) kg.`
`P=1 atm = 10^5 Pascal`
`L = 40 cm = 0.4 m`, ` L_1=80cm=0.8 m`
`P=0.355 atm`
The process is adiabatic
`P(V)^gamma = P^'(V^')^gamma`
implies `1 xx (A xx 0.4)^gamma = 0.355 xx (A xx 0.8)^gamma`
implies `1xx1 = 0.355 xx 2 ^gamma`
`implies (1)/(0.355) = 2gamma`
`gamma log gamma = log (1/0.355)`
implies `gamma= 1.4041`
Hence `V= (gamma p)/rho`
`=(1.4941 xx 105)/ (m//v)
`=446.33`
=447 m//s`.