Question: The molar mass of iodine is 127 g /mol. When sound at frequencyis introduced to a tube of iodine gas at 400 K, an internal acoustic standing wave is set up with nodes separated by 9.57 cm.What is $\gamma$ for the gas? (Hint: See Problem 91)

1. The molar mass of iodine is,
   $M=127\text{g}/\text{mol}=127\times {10}^{-3}\text{kg}/\text{mol}$
2. The frequency of sound is$f=1000\text{Hz}$.
3. The temperature is $T=400\text{K}$.
4. The nodes of the standing wave are separated by 9.57 cm.

By using the equation for the speed of sound in gas from the problems 19-91, andthe velocity of sound wave from Eq. (16-13), find$\gamma$for the gas.

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\sqrt{\frac{\mathbf{\gamma RT}}{\mathbf{M}}}$

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{\lambda f}$

where, Ris the gas constant,Tis the temperature, vis velocity, is wavelength, f is frequency andMis the molar mass.

From problems 19-91, the speed of the sound in the gas is,

$v_s=\sqrt{\frac{\gamma RT}{M}}$

Where$\gamma ={\text{C}}_p/{\text{C}}_V$ ,T is the temperature ,R is gas constant, and Mis the molar mass.

Squaring both sides,

$v_s^2=\frac{\gamma RT}{M}$

Therefore,

$\gamma =\frac{M{v}_s^2}{RT}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

From Eq. (16-13) the velocity of sound wave is,

$v_s=\lambda f\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(16-13\right)$

Since the nodes of the standing wave are separated by half a wavelength, the wavelength is given by,

$$\begin{gathered} \lambda =2\times \left(9.57\text{cm}\right) \\ =19.14\text{cm} \\ =0.1914\text{m} \end{gathered}$$

Putting in Eq. (16-13),

$$\begin{gathered} v_s=0.1914\text{m}\times 1000\text{Hz} \\ =191.4\text{m}/\text{s} \end{gathered}$$

Putting this in Eq. (1),

$$\begin{gathered} \gamma =\frac{127\times {10}^{-3}\text{g}/\text{mol}\times {\left(191.4\text{m}/\text{s}\right)}^2}{8.31\text{J}/\text{mol}.\text{k}\times 400\text{K}} \\ =1.40 \end{gathered}$$

Hence, $\gamma$for the gas is $\gamma =1.40$.

By using the equation for the speed of sound in gas from the problems 19-91, andthe equation for the velocity of sound wave, this problem can be solved.