A student is performing an experiment using a resonance column and a tuning fork of frequency \[244\,{{\text{s}}^{ - 1}}\] . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is \[\left( {0.350 \pm 0.005} \right)\,{\text{m}}\] , the gas in the tube is:
(Useful information: \[\sqrt {167RT} = 640\,{{\text{J}}^{\dfrac{1}{2}}}\] ; \[\sqrt {140RT} = 590\,{{\text{J}}^{\dfrac{1}{2}}}{\text{mol}}{{\text{e}}^{ - \dfrac{1}{2}}}\] . The molar masses \[M\] in grams are given in the option. Take the values of \[\sqrt {\dfrac{{10}}{M}} \] for each gas given)
A. Neon (\[M = 20\] ; \[\sqrt {\dfrac{{10}}{{20}}} = \dfrac{7}{{10}}\])
B. Nitrogen (\[M = 28\] ; \[\sqrt {\dfrac{{10}}{{28}}} = \dfrac{3}{5}\])
C. Oxygen (\[M = 32\] ; \[\sqrt {\dfrac{{10}}{{32}}} = \dfrac{9}{{16}}\])
D. Argon (\[M = 36\] ; \[\sqrt {\dfrac{{10}}{{36}}} = \dfrac{{17}}{{32}}\])

Hint: First of all, we will use the concept where the wavelength is four times the length of the air column. After that we will find the range of velocity by taking the error count. We will further find the velocity of sound using the ideal gas model and match with the range. The gas which falls in this range is the correct answer.

Complete step by step answer:
In the given question, we are supplied with the following information:
The observer is performing an experiment using a resonance column. The tuning fork that he is using has a frequency of \[244\,{{\text{s}}^{ - 1}}\] .The minimum height at which resonance occurs is \[\left( {0.350 \pm 0.005} \right)\,{\text{m}}\] .We are asked which one of the following gases are put inside the tube.

To begin with, let us try to find the range of velocity which is possible in the given situation taking the error into count.
Let the length of the resonance column be \[l\] and wavelength of the sound wave be \[\lambda \] .
Then we can write:
$l = \dfrac{\lambda }{4} \\
\Rightarrow \lambda = 4l \\$
We can now find the velocity of the sound wave, which is given by:
\[v = \lambda n\] …… (1)
Where,
\[v\] indicates the velocity of the sound wave.
\[n\] indicates the frequency of the wave.

Now, the equation (1) becomes:
\[v = 4l \times n\]
\[ \Rightarrow v = 244 \times 4 \times l\] …… (2)

When,
$l = \left( {0.350 + 0.005} \right)\,{\text{m}} \\
\Rightarrow l = 0.355\,{\text{m}} \\$
The velocity of the sound wave becomes:
$v = 244 \times 4 \times l \\
\Rightarrow v = 244 \times 4 \times 0.355 \\
\Rightarrow v = 346.5\,{\text{m}}\,{{\text{s}}^{ - 1}}$

When,
$l = \left( {0.350 - 0.005} \right)\,{\text{m}} \\
\Rightarrow l = 0.345\,{\text{m}} \\
The velocity of the sound wave becomes:
v = 244 \times 4 \times l \\
\Rightarrow v = 244 \times 4 \times 0.345 \\
\Rightarrow v = 336.5\,{\text{m}}\,{{\text{s}}^{ - 1}}$

Again, we know, for the velocity of a sound wave is given by:
\[v = \sqrt {\dfrac{{\gamma P}}{\rho }} \] or
\[v = \sqrt {\dfrac{{\gamma PV}}{M}} \] or
\[v = \sqrt {\dfrac{{\gamma RT}}{M}} \]
Where,
\[M\] indicates molar mass of the gas in grams.
If we convert it into kilograms, then the expression becomes:
$v = \sqrt {\dfrac{{\gamma RT}}{{M \times {{10}^{ - 3}}}}} \\
\Rightarrow v = \sqrt {\dfrac{{1000 \times \gamma RT}}{M}} \\$
\[ \Rightarrow v = \sqrt {100 \times \gamma RT} \times \sqrt {\dfrac{{10}}{M}} \]…… (3)
We know, neon and argon are monatomic gases while oxygen and nitrogen and oxygen are diatomic gases.
For monoatomic gases, \[\gamma = 1.67\] .
For diatomic gases, \[\gamma = 1.4\] .

Velocity of sound in neon:
${v_{{\text{Ne}}}} = \sqrt {100 \times 1.67 \times RT} \times \sqrt {\dfrac{{10}}{{20}}} \\
\Rightarrow {v_{{\text{Ne}}}} = \sqrt {167RT} \times \sqrt {\dfrac{{10}}{{20}}} \\
\Rightarrow {v_{{\text{Ne}}}} = 640 \times \dfrac{7}{{10}} \\
\Rightarrow {v_{{\text{Ne}}}} = 448\,{\text{m}}\,{{\text{s}}^{ - 1}}$

Velocity of sound in argon:
${v_{{\text{Ar}}}} = \sqrt {100 \times 1.67 \times RT} \times \sqrt {\dfrac{{10}}{{36}}} \\
\Rightarrow {v_{{\text{Ar}}}} = \sqrt {167RT} \times \dfrac{{17}}{{32}} \\
\Rightarrow {v_{{\text{Ar}}}} = 640 \times \dfrac{{17}}{{32}} \\
\Rightarrow {v_{{\text{Ar}}}} = 340\,{\text{m}}\,{{\text{s}}^{ - 1}}$

Velocity of sound in oxygen:
${v_{{{\text{O}}_{\text{2}}}}} = \sqrt {100 \times 1.4 \times RT} \times \sqrt {\dfrac{{10}}{{32}}} \\
\Rightarrow {v_{{{\text{O}}_{\text{2}}}}} = \sqrt {140RT} \times \dfrac{9}{{16}} \\
\Rightarrow {v_{{{\text{O}}_{\text{2}}}}} = 590 \times \dfrac{9}{{16}} \\
\Rightarrow {v_{{{\text{O}}_{\text{2}}}}} = 331.8\,{\text{m}}\,{{\text{s}}^{ - 1}}$

Velocity of sound in nitrogen:
${v_{{{\text{N}}_{\text{2}}}}} = \sqrt {100 \times 1.4 \times RT} \times \sqrt {\dfrac{{10}}{{28}}} \\
\Rightarrow {v_{{{\text{N}}_{\text{2}}}}} = \sqrt {140RT} \times \dfrac{3}{5} \\
\Rightarrow {v_{{{\text{N}}_{\text{2}}}}} = 590 \times \dfrac{3}{5} \\
\Rightarrow {v_{{{\text{N}}_{\text{2}}}}} = 354\,{\text{m}}\,{{\text{s}}^{ - 1}}$

From the above calculation it can be seen that:
\[336.5\,{\text{m}}\,{{\text{s}}^{ - 1}} < {v_{{\text{Ar}}}} = 340\,{\text{m}}\,{{\text{s}}^{ - 1}} < 346.5\,{\text{m}}\,{{\text{s}}^{ - 1}}\]
So, the possible choice is argon.

Hence, the tube is filled with argon.The correct option is D.

Note: It is important to note that resonance happens only where the air column length is proportional to one-fourth of the sound wavelength with a frequency equal to the tuning fork frequency. It is the most common part where most of the students tend to make mistakes. Mistakes in this section can cost the whole problem and ruin it. It is important to take the error count as without it we cannot get the range of the velocity of the sound wave.