A Kundt tube is filled with an ideal gas having a density of 1.7 × 10^ -3 g cm^ -3 at a pressure of 1.5 × 10^5 Pa . When the gas is resonated at a frequency of 3.0 kHz , nodes form at a separation of 6.0 cm . Determine

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Quantrex Academy · Accepted Answer

Correct answer: A. 26 J K^ -1 mol^ -1 and 17.7 J K^ -1 mol^ -1

The distance between two consecutive nodes in a Kundt tube is

\dfrac{\lambda}{2}

.

\dfrac{\lambda}{2} = 6.0 \text{ cm} \Rightarrow \lambda = 12.0 \text{ cm} = 0.12 \text{ m}

The speed of sound in the gas is given by:

v = f \lambda = 3000 \text{ Hz} \times 0.12 \text{ m} = 360 \text{ m s}^{-1}

The density of the gas is

\rho = 1.7 \times 10^{-3} \text{ g cm}^{-3} = 1.7 \text{ kg m}^{-3}

.

The speed of sound is also given by

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

.

Squaring both sides and solving for

\gamma

:

\gamma = \dfrac{v^2 \rho}{P} = \dfrac{(360)^2 \times 1.7}{1.5 \times 10^5} = \dfrac{129600 \times 1.7}{150000} = 1.4688

Using the relation

C_p - C_V = R

and

\dfrac{C_p}{C_V} = \gamma

:

C_V = \dfrac{R}{\gamma - 1} = \dfrac{8.314}{1.4688 - 1} = \dfrac{8.314}{0.4688} \approx 17.7 \text{ J K}^{-1} \text{ mol}^{-1}

C_p = C_V + R = 17.7 + 8.314 \approx 26.0 \text{ J K}^{-1} \text{ mol}^{-1}

Thus,

C_p = 26 \text{ J K}^{-1} \text{ mol}^{-1}

and

C_V = 17.7 \text{ J K}^{-1} \text{ mol}^{-1}

.

A Kundt tube is filled with an ideal gas having a density of 1.7 ×...

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