Question

The atomic mass of iodine is $127\text{g /mol}$. A standing wave in iodine vapour at $400\text{K}$ has nodes that are $6.77\text{cm}$ apart when the frequency is $1000\text{Hz}$. At this temperature iodine vapour is

• A. Monoatomic
• B. Diatomic
• C. Triatomic
• D. None of the above

Answer 1

The correct option is B Diatomic

We know that,
Distance between two successive nodes $=\frac{\lambda }{2}$
where, $\lambda$ is wavelength.

Given, $\frac{\lambda }{2}=6.77\text{cm}$

$\Rightarrow \lambda =2\times 6.77\text{cm}$

We also know, $v=\lambda f$

where, $v$ is velocity of wave and $f$ is frequency of wave.
As, $f=1000\text{Hz}$

So, $v=2\times 6.77\times {10}^{-2}\times 1000=135.4\text{m/s}$

Further, $v=\sqrt{\frac{\gamma RT}{M}}......\left(1\right)$

where,
Gas constant, $R=8.3\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$
Temperature, $T=400\text{K}$
Atomic mass, $M=127\text{g/mol}$
Specific heat ratio, $\gamma$

If iodine vapour is monoatomic,
Substituting values in $\left(1\right)$,

$135.4=\sqrt{\frac{\gamma \times 8.3\times 400}{127\times {10}^{-3}}}$

$\Rightarrow \gamma =0.7\nless 1$

Which is not possible as,
$c_p-c_v=R$

$\Rightarrow c_P>c_V$

$\Rightarrow \frac{c_P}{c_V}>1$

$\therefore \gamma >1$ (Always$)$

Now, considering iodine vapour as diatomic,

$135.4=\sqrt{\frac{\gamma \times 8.3\times 400}{2\times 127\times {10}^{-3}}}$

$\Rightarrow \gamma =1.4=$ value of $\gamma$ for diatomic gas.

Hence, option $\left(b\right)$ is correct answer.