Closed at one end and fitted with a piston at the other, an adiabatic cylindrical tube has a cross-sectional area of 1 cm^2 . The tube holds 0.03 g of an ideal gas. At a pressure of 1 atm and at the surrounding

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Correct answer: A. 447 m/s

For an adiabatic process, the relation between pressure and volume is given by

P_1 V_1^\gamma = P_2 V_2^\gamma

.

Given

P_1 = 1 \text{ atm}

,

V_1 = A L_1 = 1 \times 40 = 40 \text{ cm}^3

.

After the piston is pulled out,

P_2 = 0.355 \text{ atm}

and

V_2 = A L_2 = 1 \times 80 = 80 \text{ cm}^3

.

Substituting the values:

1 \times (40)^\gamma = 0.355 \times (80)^\gamma

2^\gamma = \dfrac{1}{0.355} \approx 2.817

Since

2^{1.5} = 2\sqrt{2} \approx 2.828

, we get

\gamma \approx 1.5

.

The density of the gas at atmospheric temperature is:

\rho = \dfrac{m}{V_1} = \dfrac{0.03 \times 10^{-3}}{40 \times 10^{-6}} = 0.75 \text{ kg/m}^3

The speed of sound in the gas is given by:

v = \sqrt{\dfrac{\gamma P_1}{\rho}}

Taking

1 \text{ atm} = 10^5 \text{ Pa}

:

v = \sqrt{\dfrac{1.5 \times 10^5}{0.75}} = \sqrt{2 \times 10^5} \approx 447 \text{ m/s}

Closed at one end and fitted with a piston at the other, an...

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