Question:
The speed of sound in oxygen at STP will be approximately:(Given, R = 8.3J\((K)^−1\), \(γ = 1.4)\)
The speed of sound in oxygen at STP will be approximately:(Given, R = 8.3J\((K)^−1\), \(γ = 1.4)\)
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The speed of sound in a gas depends on the temperature, molar mass, and the adiabatic index of the gas.
Updated On: Jan 13, 2026
- 315 m/s
- 333 m/s
- 341 m/s
- 325 m/s
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The Correct Option is A
Approach Solution - 1
Step 1: {Given data}
Temperature, \( T = 273 \, {K} \) Molecular mass of oxygen, \( M = 32 \times 10^{-3} \, {kg} \)
Step 2: {Calculating the speed of sound}
\[ v = \sqrt{\frac{\gamma RT}{M}} = \sqrt{\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}} \approx 315 \, {m/s} \] Thus, the speed of sound in oxygen at STP is approximately 315 m/s.
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Approach Solution -2
Step 1: Use the formula for speed of sound in a gas
The speed of sound in a gas is given by:
v = √(γRT / M)
where:
- γ = 1.4 (for diatomic gases like oxygen)
- R = 8.3 J/mol·K
- T = 273 K (at STP)
- M = 32 g/mol = 0.032 kg/mol (for oxygen)
Step 2: Plug in the values
v = √[(1.4 × 8.3 × 273) / 0.032]
= √(3175.62 / 0.032)
= √99238.125
≈ 315 m/s
Final Answer: 315 m/s
The speed of sound in a gas is given by:
v = √(γRT / M)
where:
- γ = 1.4 (for diatomic gases like oxygen)
- R = 8.3 J/mol·K
- T = 273 K (at STP)
- M = 32 g/mol = 0.032 kg/mol (for oxygen)
Step 2: Plug in the values
v = √[(1.4 × 8.3 × 273) / 0.032]
= √(3175.62 / 0.032)
= √99238.125
≈ 315 m/s
Final Answer: 315 m/s
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