JEE Challenger
More from Waves
PhyHardAdv 2014 • Paper 1

Identify Gas Using Resonance Column and Tuning Fork Frequency

A student carries out an experiment utilizing a resonance column and a tuning fork of frequency 244 s−1244\text{ s}^{-1}. It is known that the air inside the tube has been replaced by a different gas (assume the column remains completely filled with this gas). If the minimum tube height at which resonance takes place is (0.350±0.005) m(0.350 \pm 0.005)\text{ m}, determine the gas inside the tube.

(Useful information:\textbf{Useful information:} 167RT=640 J1/2mole−1/2\sqrt{167RT} = 640\text{ J}^{1/2}\text{mole}^{-1/2}; 140RT=590 J1/2mole−1/2\sqrt{140RT} = 590\text{ J}^{1/2}\text{mole}^{-1/2}. Molar masses MM in grams are given in the options. Take the values of 10M\sqrt{\frac{10}{M}} for each gas as given there.)

Options

A

Neon (M=20,1020=710)\left(M = 20, \sqrt{\frac{10}{20}} = \frac{7}{10}\right)

B

Nitrogen (M=28,1028=35)\left(M = 28, \sqrt{\frac{10}{28}} = \frac{3}{5}\right)

C

Oxygen (M=32,1032=916)\left(M = 32, \sqrt{\frac{10}{32}} = \frac{9}{16}\right)

D

Argon (M=36,1036=1732)\left(M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32}\right)

Correct

Topics & Concepts

WavesSound Waves

Step-by-Step Solution

To determine the gas inside the resonance column, we relate the resonant frequency of the closed pipe in its fundamental mode to the speed of sound in the gas.

The wavelength λ\lambda for the fundamental mode of resonance in a tube closed at one end is given by: ℓ=λ4  ⟹  λ=4ℓ\ell = \frac{\lambda}{4} \implies \lambda = 4\ell

Given that the frequency of the tuning fork is f=244 s−1f = 244\text{ s}^{-1}, the speed of sound vv in the gas is: v=fλ=4fℓv = f \lambda = 4 f \ell

Given ℓ=(0.350±0.005) m\ell = (0.350 \pm 0.005)\text{ m}: v=4×244×(0.350±0.005)=976×(0.350±0.005) m/sv = 4 \times 244 \times (0.350 \pm 0.005) = 976 \times (0.350 \pm 0.005)\text{ m/s} v=341.6±4.88 m/sv = 341.6 \pm 4.88\text{ m/s}

Thus, the experimental speed of sound falls within the range: 336.72 m/s≤v≤346.48 m/s336.72\text{ m/s} \le v \le 346.48\text{ m/s}

The theoretical speed of sound in an ideal gas is given by: v=γRTMkg=γRTM×10−3=1000γRTM=100γRT×10Mv = \sqrt{\frac{\gamma RT}{M_{\text{kg}}}} = \sqrt{\frac{\gamma RT}{M \times 10^{-3}}} = \sqrt{\frac{1000 \gamma RT}{M}} = \sqrt{100\gamma RT} \times \sqrt{\frac{10}{M}}

For a monoatomic gas, γ≈1.67\gamma \approx 1.67, so: 100γRT=167RT=640 J1/2mole−1/2\sqrt{100\gamma RT} = \sqrt{167 RT} = 640\text{ J}^{1/2}\text{mole}^{-1/2}

For a diatomic gas, γ≈1.40\gamma \approx 1.40, so: 100γRT=140RT=590 J1/2mole−1/2\sqrt{100\gamma RT} = \sqrt{140 RT} = 590\text{ J}^{1/2}\text{mole}^{-1/2}

Now, let us calculate the theoretical speed of sound for the gases given in each option:

  1. Option A: Neon (monoatomic, M=20M = 20) v=167RT×1020=640×710=448 m/sv = \sqrt{167RT} \times \sqrt{\frac{10}{20}} = 640 \times \frac{7}{10} = 448\text{ m/s}

  2. Option B: Nitrogen (diatomic, M=28M = 28) v=140RT×1028=590×35=354 m/sv = \sqrt{140RT} \times \sqrt{\frac{10}{28}} = 590 \times \frac{3}{5} = 354\text{ m/s}

  3. Option C: Oxygen (diatomic, M=32M = 32) v=140RT×1032=590×916=331.875 m/sv = \sqrt{140RT} \times \sqrt{\frac{10}{32}} = 590 \times \frac{9}{16} = 331.875\text{ m/s}

  4. Option D: Argon (monoatomic, M=36M = 36) v=167RT×1036=640×1732=340 m/sv = \sqrt{167RT} \times \sqrt{\frac{10}{36}} = 640 \times \frac{17}{32} = 340\text{ m/s}

Comparing the theoretical values with the experimental range [336.72 m/s,346.48 m/s][336.72\text{ m/s}, 346.48\text{ m/s}], only the speed of sound in Argon (340 m/s340\text{ m/s}) lies within the specified limits.

Therefore, the correct option is D.

PreviousNext