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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Specific Heat Capacities of Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Specific Heat Capacities of Gases

At STP, 4.0 g of helium occupies a volume of 22400 cm^3 . Given...

At STP, 4.0 g4.0 \text{ g} of helium occupies a volume of 22400 cm322400 \text{ cm}^3. Given that the specific heat capacity of helium at constant pressure is 5.0 cal K−1 mol−15.0 \text{ cal K}^{-1} \text{ mol}^{-1}, determine the speed of sound in helium at STP.

Options

  1. A945 m s−1945 \text{ m s}^{-1}
  2. B988 m s−1988 \text{ m s}^{-1}
  3. C972 m s−1972 \text{ m s}^{-1}
  4. D960 m s−1960 \text{ m s}^{-1}

Correct answer

C. 972 m s−1972 \text{ m s}^{-1}

Step-by-step solution

Given mass of helium, m=4.0 g=4.0×10−3 kgm = 4.0 \text{ g} = 4.0 \times 10^{-3} \text{ kg}

Volume at STP, V=22400 cm3=22.4×10−3 m3V = 22400 \text{ cm}^3 = 22.4 \times 10^{-3} \text{ m}^3

Density of helium, ρ=mV=4.0×10−322.4×10−3=15.6 kg m−3\rho = \dfrac{m}{V} = \dfrac{4.0 \times 10^{-3}}{22.4 \times 10^{-3}} = \dfrac{1}{5.6} \text{ kg m}^{-3}

Molar heat capacity at constant pressure, Cp=5.0 cal K−1 mol−1C_p = 5.0 \text{ cal K}^{-1} \text{ mol}^{-1}

Universal gas constant, R≈2.0 cal K−1 mol−1R \approx 2.0 \text{ cal K}^{-1} \text{ mol}^{-1}

Molar heat capacity at constant volume, Cv=Cp−R=5.0−2.0=3.0 cal K−1 mol−1C_v = C_p - R = 5.0 - 2.0 = 3.0 \text{ cal K}^{-1} \text{ mol}^{-1}

Ratio of specific heats, γ=CpCv=53\gamma = \dfrac{C_p}{C_v} = \dfrac{5}{3}

Pressure at STP, P=1.013×105 N m−2P = 1.013 \times 10^5 \text{ N m}^{-2}

The speed of sound in a gas is given by Laplace's formula:

v=γPρv = \sqrt{\dfrac{\gamma P}{\rho}}

Substituting the values:

v=53×1.013×105×5.6v = \sqrt{\dfrac{5}{3} \times 1.013 \times 10^5 \times 5.6}

v=28×1.013×1053v = \sqrt{\dfrac{28 \times 1.013 \times 10^5}{3}}

v=9.454×105≈972 m s−1v = \sqrt{9.454 \times 10^5} \approx 972 \text{ m s}^{-1}
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