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We can now gain a qualitative understanding of a puzzle about the composition of Earth’s atmosphere. Hydrogen is by far the most common element in the universe, and helium is by far the second-most common. Moreover, helium is constantly produced on Earth by radioactive decay. Why are those elements so rare in our atmosphere? The answer is that gas molecules that reach speeds above Earth’s escape velocity, about 11 km/s, can escape from the atmosphere into space. Because of the lower mass of hydrogen and helium molecules, they move at higher speeds than other gas molecules, such as nitrogen and oxygen. Only a few exceed escape velocity, but far fewer heavier molecules do. Thus, over the billions of years that Earth has existed, far more hydrogen and helium molecules have escaped from the atmosphere than other molecules, and hardly any of either is now present.

We can also now take another look at evaporative cooling, which we discussed in the chapter on temperature and heat. Liquids, like gases, have a distribution of molecular energies. The highest-energy molecules are those that can escape from the intermolecular attractions of the liquid. Thus, when some liquid evaporates, the molecules left behind have a lower average energy, and the liquid has a lower temperature.

Summary

  • The motion of individual molecules in a gas is random in magnitude and direction. However, a gas of many molecules has a predictable distribution of molecular speeds, known as the Maxwell-Boltzmann distribution.
  • The average and most probable velocities of molecules having the Maxwell-Boltzmann speed distribution, as well as the rms velocity, can be calculated from the temperature and molecular mass.

Key equations

Ideal gas law in terms of molecules p V = N k B T
Ideal gas law ratios if the amount of gas is constant p 1 V 1 T 1 = p 2 V 2 T 2
Ideal gas law in terms of moles p V = n R T
Van der Waals equation [ p + a ( n V ) 2 ] ( V n b ) = n R T
Pressure, volume, and molecular speed p V = 1 3 N m v 2
Root-mean-square speed v rms = 3 R T M = 3 k B T m
Mean free path λ = V 4 2 π r 2 N = k B T 4 2 π r 2 p
Mean free time τ = k B T 4 2 π r 2 p v rms
The following two equations apply only to a monatomic ideal gas:
Average kinetic energy of a molecule K = 3 2 k B T
Internal energy E int = 3 2 N k B T .
Heat in terms of molar heat capacity at constant volume Q = n C V Δ T
Molar heat capacity at constant volume for an ideal gas with d degrees of freedom C V = d 2 R
Maxwell–Boltzmann speed distribution f ( v ) = 4 π ( m 2 k B T ) 3 / 2 v 2 e m v 2 / 2 k B T
Average velocity of a molecule v ¯ = 8 π k B T m = 8 π R T M
Peak velocity of a molecule v p = 2 k B T m = 2 R T M

Conceptual questions

One cylinder contains helium gas and another contains krypton gas at the same temperature. Mark each of these statements true, false, or impossible to determine from the given information. (a) The rms speeds of atoms in the two gases are the same. (b) The average kinetic energies of atoms in the two gases are the same. (c) The internal energies of 1 mole of gas in each cylinder are the same. (d) The pressures in the two cylinders are the same.

a. false; b. true; c. true; d. true

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Repeat the previous question if one gas is still helium but the other is changed to fluorine, F 2 .

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Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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