Question

Consider a system of gas of a diatomic molecule in which the speed of sound at $0^\circ\text{C}$ is 1260 $\text{ms}^{-1}$. Then, the molecular weight of the gas is (Given the gas constant $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$)

• A. \( 2\text{g} \)
• B. \( 2\text{mg} \)
• C. \( 4\text{g} \)
• D. \( 10\text{g} \)
• E. \( 20\text{g} \)

Hint:

Standard values for \(\gamma\): - Monatomic gas: \(1.67\) (\(5/3\)) - Diatomic gas: \(1.4\) (\(7/5\)) - Polyatomic gas: \(1.33\) (\(4/3\))

Answer 1

The Correct Option is A

Concept: The speed of sound \(v\) in an ideal gas is given by the formula: \[ v = \sqrt{\frac{\gamma RT}{M}} \] where:
• \(\gamma\) is the adiabatic index (ratio of specific heats). For diatomic gases, \(\gamma = 1.4\) or \(\frac{7}{5}\).
• \(R\) is the universal gas constant (8.314 J/mol.K).
• \(T\) is the absolute temperature (in Kelvin).
• \(M\) is the molar mass (molecular weight) of the gas in kg/mol.

Step 1: List the given parameters and convert units.
Speed \(v = 1260 \, \text{m/s}\) Temperature \(T = 0^\circ\text{C} = 273 \, \text{K}\) Adiabatic index \(\gamma = 1.4\) (for diatomic molecules)

Step 2: Rearrange the formula to solve for M.
Squaring both sides of the speed of sound equation: \[ v^2 = \frac{\gamma RT}{M} \quad \Rightarrow \quad M = \frac{\gamma RT}{v^2} \]

Step 3: Substitute and calculate.
\[ M = \frac{1.4 \times 8.314 \times 273}{(1260)^2} \] \[ M = \frac{3177.6}{1587600} \approx 0.002001 \, \text{kg/mol} \] Convert kg to grams: \[ M \approx 0.002 \times 1000 = 2 \, \text{g/mol} \]