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Molecular kinetic theory questions
The gas laws were found by experiment; kinetic theory derives them from molecules bouncing elastically off the walls of a container. The derivation of pV as one third of Nm times the mean square speed is examinable in full, and comparing it with pV = NkT is what identifies temperature as a measure of mean molecular kinetic energy.
19 original questions · 53 marks · the molecular kinetic theory notes · Thermal physics
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State two assumptions of the kinetic theory model of an ideal gas.
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Any two of the following, one mark each (2): molecules are in constant random motion; collisions with the walls and each other are perfectly elastic; the volume of the molecules is negligible compared with the container; the time of collisions is negligible; and there are no forces between molecules except during collisions.Calculate the average kinetic energy of a gas molecule at 300 K, using mean KE = (3/2)kT (k = 1.38 × 10−23 J K−1).
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Mean KE = (3/2)kT = 1.5 × 1.38 × 10−23 × 300 (1)
Mean KE = 6.21 × 10−21 J (1)State the kinetic theory equation relating pressure, volume and molecular speed.
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pV = ⅓Nm(crms)2 (1), where N is the number of molecules, m the mass of one molecule and crms the root-mean-square speed (1).Smoke particles suspended in air are viewed through a microscope. Describe what is observed, and state what this observation is evidence for.
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The smoke particles move continually in random, jerky paths (Brownian motion) (1). This is evidence that air consists of molecules in rapid random motion, bombarding the smoke particles unevenly (1).State what is meant by the root-mean-square speed of the molecules of a gas.
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The square root of the mean of the squares of the molecular speeds (1).At one instant, four gas molecules have speeds of 300 m s−1, 400 m s−1, 500 m s−1 and 600 m s−1. Calculate their root-mean-square speed.
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Mean square speed = (3002 + 4002 + 5002 + 6002)/4 = 2.15 × 105 m2 s−2 (1)
crms = √(2.15 × 105) = 464 m s−1 (1)Calculate the total kinetic energy of one mole of an ideal gas at 400 K, using R = 8.31 J mol−1 K−1.
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Each molecule carries (3/2)kT on average, so a mole carries (3/2)RT (1)
= 1.5 × 8.31 × 400 (1)
= 4986 J ≈ 5.0 × 103 J (1)A nitrogen molecule has a mass of 4.7 × 10−26 kg. Calculate its root-mean-square speed at 300 K, using ½m(crms)2 = (3/2)kT.
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crms = √(3kT/m) (1)
= √((3 × 1.38 × 10−23 × 300)/(4.7 × 10−26)) (1)
crms = 514 m s−1 (1)A gas of density 1.2 kg m−3 has molecules with an rms speed of 500 m s−1. Calculate its pressure, using p = ⅓ρ(crms)2.
At the same temperature, explain which has the greater root-mean-square speed: a light molecule or a heavy one.
A nitrogen molecule of mass 4.65 × 10−26 kg strikes a container wall head-on at 480 m s−1 and rebounds along the same line at the same speed. Calculate the change in momentum of the molecule. The molecule returns and strikes the same wall once every 1.2 ms. Go on to calculate the average force it exerts on the wall.
A container holds a mixture of helium (molecular mass 6.6 × 10−27 kg) and argon (molecular mass 6.6 × 10−26 kg) at the same temperature. State the ratio of the mean kinetic energy of a helium molecule to that of an argon molecule, and calculate the ratio crms(helium)/crms(argon).
A rigid container of gas is sealed and kept at constant temperature. Explain, with reference to one assumption of the kinetic theory model, why the pressure does not decrease over time.
An oxygen molecule in air has a mass of 5.3 × 10−26 kg. Calculate (a) its mean kinetic energy at 290 K and (b) its root-mean-square speed.
A gas is at 300 K. Calculate the temperature to which it must be raised to double the root-mean-square speed of its molecules.
Explain how the kinetic theory model accounts for the pressure a gas exerts on its container.
A cubical box of side L contains N identical molecules, each of mass m, in random motion. Starting from the change in momentum when one molecule collides elastically with a wall, derive the equation pV = ⅓Nm(crms)2.
The escape speed from the Earth is 11 km s−1. Calculate the temperature at which hydrogen molecules (mass 3.3 × 10−27 kg) would have a root-mean-square speed equal to the escape speed (k = 1.38 × 10−23 J K−1). The temperature of the upper atmosphere is only a few hundred kelvin, yet it contains almost no hydrogen. Suggest why.
A cylinder holds 0.020 m3 of helium, an ideal gas, at a pressure of 2.5 × 105 Pa. Explain why the internal energy of the helium is entirely kinetic, and calculate this internal energy.
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