JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 9)

The speed of sound in oxygen at S.T.P. will be approximately: (given, $$R=8.3 \mathrm{~JK}^{-1}, \gamma=1.4$$)
341 m/s
333 m/s
325 m/s
315 m/s

Explanation

The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:

$$ v = \sqrt{\gamma \frac{R T}{M}} $$

Where:

  • $$ v $$ = speed of sound in the gas
  • $$ \gamma $$ = adiabatic index (ratio of specific heats, $$ C_p/C_v $$)
  • $$ R $$ = universal gas constant
  • $$ T $$ = temperature in Kelvin
  • $$ M $$ = molar mass of the gas in kilograms per mole (kg/mol)

Given:

  • $$ R = 8.3 \, J \cdot mol^{-1} \cdot K^{-1} $$

  • $$ \gamma = 1.4 $$ (For diatomic gases such as oxygen)

  • $$ T = 273.15 \, K $$ (Standard temperature, 0°C in Kelvin)

  • Molar mass of Oxygen ($$ O_2 $$) $$ M = 32 \times 10^{-3} \, kg/mol $$ (32 g/mol converted to kg/mol)

Plugging these values into the formula:

$$ v = \sqrt{1.4 \times \frac{8.3 \times 273.15}{32 \times 10^{-3}}} $$

Calculating the values inside the square root:

$$ v = \sqrt{1.4 \times \frac{2268.745}{0.032}} $$

$$ v = \sqrt{1.4 \times 70896.40625} $$

$$ v = \sqrt{99304.96875} $$

$$ v \approx 315 \, m/s $$

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