In a tube filled with oxygen at 300 K , standing waves of frequency 5.0 kHz are generated. The separation between successive nodes is 3.3 cm . Determine the specific heat capacities C_p and C_V of the gas.

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Correct answer: A. C_p = 29.0 J K^ -1 mol^ -1 , C_V = 20.7 J K^ -1 mol^ -1

The separation between successive nodes in a standing wave is

\dfrac{\lambda}{2}

.

\dfrac{\lambda}{2} = 3.3 \text{ cm} \Rightarrow \lambda = 6.6 \text{ cm} = 0.066 \text{ m}

The speed of sound in the gas is:

v = f \lambda = 5000 \times 0.066 = 330 \text{ m/s}

Using the formula for the speed of sound in an ideal gas,

v = \sqrt{\dfrac{\gamma R T}{M}}

:

\gamma = \dfrac{v^2 M}{R T} = \dfrac{(330)^2 \times 0.032}{8.3 \times 300} \approx 1.4

We know that

C_p - C_V = R

and

\dfrac{C_p}{C_V} = \gamma

.

C_V = \dfrac{R}{\gamma - 1} = \dfrac{8.3}{1.4 - 1} = \dfrac{8.3}{0.4} = 20.75 \approx 20.7 \text{ J K}^{-1} \text{ mol}^{-1}

C_p = \gamma C_V = 1.4 \times 20.75 = 29.05 \approx 29.0 \text{ J K}^{-1} \text{ mol}^{-1}

In a tube filled with oxygen at 300 K , standing waves of frequency...

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