A student is performing an experiment using a resonance column and tuning fork of frequency $244 s^{- 1}$ . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 $(0.350 \pm 0.005) m$ , the gas(es) in the tube is (Useful information: $\sqrt{167 R T} = 640 J^{1 / 2} m o l e^{- 1 / 2}$ ; $\sqrt{140 R T} = 590 J^{1 / 2} m o l e^{- 1 / 2}$ . The molar masses $M$ in grams are given in the options. Take the values of $\sqrt{\frac{10}{M}}$ for each gas as given there.)

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$A r g o n (M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32})$

$O x y g e n (M = 32, \sqrt{\frac{10}{32}} = \frac{9}{16})$

$N i t r o g e n (M = 28, \sqrt{\frac{10}{28}} = \frac{3}{5})$

$N e o n (M = 20, \sqrt{\frac{10}{20}} = \frac{7}{10})$

answer is D.

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Detailed Solution

Here $υ = \frac{v}{4 l} = \sqrt{\frac{γ R T}{M \times 10^{- 3}}} \times \frac{1}{4 l} \Rightarrow v = υ λ = υ \times 4 l$
$\Rightarrow v (244) \times 4 \times l = 336.7 m / s to 346.5 m / s [∴ l = 0.350 \pm 0.005]$
For monatomic gas $γ = 1.67$
$∴ v = \sqrt{\frac{γ R T}{M \times 10^{- 3}}} = \sqrt{100 γ R T} \times \sqrt{\frac{10}{M}} = \sqrt{167 R T} \times \sqrt{\frac{10}{M}} = 640 \sqrt{\frac{10}{M}}$
For Neon $M = 20 ∴ v = 640 \times \frac{7}{10} = 448 m s^{- 1}$
For argon $M = 36, ∴ v = 640 \times \frac{17}{32} = 340 m s^{- 1}$
For diatomic gas $γ = 1.4$
$v = \sqrt{140 R T} \sqrt{\frac{10}{M}} = 590 \times \sqrt{\frac{10}{M}}$
For Oxygen $M = 32 ∴ v = 590 \times \frac{9}{16} = 331.87 m s^{- 1}$
For Nitrogen $M = 28 ∴ v = 590 \times \frac{3}{5} = 354 m s^{- 1}$