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Heat and Thermodynamics question

2024 · 31 Jan · Shift 2 · Q69
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  5. /2024 · 31 Jan · Shift 2 · Q69

Heat and Thermodynamics question

2024 · 31 Jan · Shift 2 · Q69

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The speed of sound in oxygen at S.T.P. will be approximately: (given, R=8.3 JK−1,γ=1.4R=8.3 \mathrm{~JK}^{-1}, \gamma=1.4R=8.3 JK−1,γ=1.4)
  1. A
    341 m/s
  2. B
    333 m/s
  3. C
    325 m/s
  4. D
    315 m/s
View written solutionFree

Correct answer: D

  1. Use the formula for speed of sound in a gas

For an ideal gas,

v=γRTMv = \sqrt{\frac{\gamma R T}{M}}v=MγRT​​

where:

  • γ=1.4\gamma = 1.4γ=1.4
  • R=8.3 J mol−1K−1R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1
  • T=273 KT = 273\,\text{K}T=273K at S.T.P.
  • MMM = molar mass of oxygen gas O2=32 g/mol=0.032 kg/molO_2 = 32\,\text{g/mol} = 0.032\,\text{kg/mol}O2​=32g/mol=0.032kg/mol
  1. Substitute the values

v=1.4×8.3×2730.032v = \sqrt{\frac{1.4 \times 8.3 \times 273}{0.032}}v=0.0321.4×8.3×273​​

First calculate the numerator:

1.4×8.3=11.621.4 \times 8.3 = 11.621.4×8.3=11.62

11.62×273=3172.2611.62 \times 273 = 3172.2611.62×273=3172.26

So,

v=3172.260.032v = \sqrt{\frac{3172.26}{0.032}}v=0.0323172.26​​

3172.260.032=99133.125\frac{3172.26}{0.032} = 99133.1250.0323172.26​=99133.125

Thus,

v=99133.125≈315 m/sv = \sqrt{99133.125} \approx 315\,\text{m/s}v=99133.125​≈315m/s

  1. Match with the options

The closest option is:

  • A: 341 m/s341\,\text{m/s}341m/s
  • B: 333 m/s333\,\text{m/s}333m/s
  • C: 325 m/s325\,\text{m/s}325m/s
  • D: 315 m/s315\,\text{m/s}315m/s

Hence, the correct option is D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They match.

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