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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.
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Solution
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 × 2γ
`=> 1/0.355 = 2^gamma`
`gamma "log" 2 = "log"(1/0.355)`
⇒ γ =1.4941
Speed of sound in the gas,
`"v" = sqrt((gamma"P")/ρ) = sqrt ((gamma"P")/("m"/"v"))`
`"v" = sqrt ((1.4941 xx 10^5)/("m"/"v")) = sqrt ((1.4941 xx 10^5)/(( 0.03 xx 10 ^-3)/(10 ^-4 xx 1 xx 0.4)`
⇒ v = 446.33 ≈ 447 m/s
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