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Difficulty: HardKinetic Theory of Matter and Pressure of Gases

A sample of gas enclosed in a rigid container exerts a pressure of 3.0×105 N m−23.0 \times 10^5 \text{ N m}^{-2} with a density of 0.40 kg m−30.40 \text{ kg m}^{-3}. If the gas is heated at constant volume until its pressure increases to 1.2×106 N m−21.2 \times 10^6 \text{ N m}^{-2}, what is the final root-mean-square (r.m.s.) speed of the gas molecules?

  1. 3000 m s−13000 \text{ m s}^{-1}Answer
  2. B
    6000 m s−16000 \text{ m s}^{-1}
  3. C
    1500 m s−11500 \text{ m s}^{-1}
  4. D
    750 m s−1750 \text{ m s}^{-1}

Answer

The final root-mean-square speed of the gas molecules is 3000 m s−13000 \text{ m s}^{-1}.
Using the kinetic theory equation P=13ρvrms2P = \frac{1}{3}\rho v_{\text{rms}}^2, the initial speed is v1=3×3.0×1050.40=1500 m s−1v_1 = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.40}} = 1500 \text{ m s}^{-1}. When heated at constant volume, density remains unchanged. The new pressure 1.2×106 N m−21.2 \times 10^6 \text{ N m}^{-2} is 4 times the initial pressure, so the new speed is v2=4×v1=2×1500 m s−1=3000 m s−1v_2 = \sqrt{4} \times v_1 = 2 \times 1500 \text{ m s}^{-1} = 3000 \text{ m s}^{-1}.

Step-by-Step Solution

1
Express the relationship between pressure, density, and root-mean-square speed using kinetic theory.
P=13ρvrms2  ⟹  vrms=3PρP = \frac{1}{3}\rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}
According to the kinetic theory of gases, the pressure exerted by a gas is related to its density and molecular r.m.s. speed.
2
Calculate the initial root-mean-square speed v1v_1 using initial pressure P1=3.0×105 N m−2P_1 = 3.0 \times 10^5 \text{ N m}^{-2} and density ρ=0.40 kg m−3\rho = 0.40 \text{ kg m}^{-3}.
v1=3×(3.0×105)0.40=9.0×1050.40=2.25×106=1500 m s−1v_1 = \sqrt{\frac{3 \times (3.0 \times 10^5)}{0.40}} = \sqrt{\frac{9.0 \times 10^5}{0.40}} = \sqrt{2.25 \times 10^6} = 1500 \text{ m s}^{-1}
Establishes the baseline molecular speed prior to heating.
3
Determine the final root-mean-square speed v2v_2 after the pressure increases to P2=1.2×106 N m−2P_2 = 1.2 \times 10^6 \text{ N m}^{-2} at constant volume.
v2=3×(1.2×106)0.40=3.6×1060.40=9.0×106=3000 m s−1v_2 = \sqrt{\frac{3 \times (1.2 \times 10^6)}{0.40}} = \sqrt{\frac{3.6 \times 10^6}{0.40}} = \sqrt{9.0 \times 10^6} = 3000 \text{ m s}^{-1}
Since the volume is rigid, density ρ\rho remains constant while pressure increases due to heating.

Key Concept

Root-Mean-Square Speed and Pressure Relation in Kinetic Theory
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Topics

Kinetic Theory of Matter and Pressure of Gases

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