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By the end of this section, you will be able to:
  • Describe the distribution of molecular speeds in an ideal gas
  • Find the average and most probable molecular speeds in an ideal gas

Particles in an ideal gas all travel at relatively high speeds, but they do not travel at the same speed. The rms speed is one kind of average, but many particles move faster and many move slower. The actual distribution of speeds has several interesting implications for other areas of physics, as we will see in later chapters.

The maxwell-boltzmann distribution

The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This predictable distribution of molecular speeds is known as the Maxwell-Boltzmann distribution    , after its originators, who calculated it based on kinetic theory, and it has since been confirmed experimentally ( [link] ).

To understand this figure, we must define a distribution function of molecular speeds, since with a finite number of molecules, the probability that a molecule will have exactly a given speed is 0.

The figure is a graph of probability versus velocity v in meters per second of oxygen gas at 300 kelvin. The graph has a peak probability at a velocity V p of just under 400 meters per second and a root-mean-square probability at a velocity v r m s of about 500 meters per second. The probability is zero at the origin and tends to zero at infinity. The graph is not symmetric, but rather steeper on the left than on the right of the peak.
The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed v p is less than the rms speed v rms . Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than v rms .

We define the distribution function f ( v ) by saying that the expected number N ( v 1 , v 2 ) of particles with speeds between v 1 and v 2 is given by

N ( v 1 , v 2 ) = N v 1 v 2 f ( v ) d v .

[Since N is dimensionless, the unit of f ( v ) is seconds per meter.] We can write this equation conveniently in differential form:

d N = N f ( v ) d v .

In this form, we can understand the equation as saying that the number of molecules with speeds between v and v + d v is the total number of molecules in the sample times f ( v ) times dv . That is, the probability that a molecule’s speed is between v and v + d v is f ( v ) dv .

We can now quote Maxwell’s result, although the proof is beyond our scope.

Maxwell-boltzmann distribution of speeds

The distribution function for speeds of particles in an ideal gas at temperature T is

f ( v ) = 4 π ( m 2 k B T ) 3 / 2 v 2 e m v 2 / 2 k B T .

The factors before the v 2 are a normalization constant; they make sure that N ( 0 , ) = N by making sure that 0 f ( v ) d v = 1 . Let’s focus on the dependence on v . The factor of v 2 means that f ( 0 ) = 0 and for small v , the curve looks like a parabola. The factor of e m 0 v 2 / 2 k B T means that lim v f ( v ) = 0 and the graph has an exponential tail, which indicates that a few molecules may move at several times the rms speed. The interaction of these factors gives the function the single-peaked shape shown in the figure.

Calculating the ratio of numbers of molecules near given speeds

In a sample of nitrogen ( N 2 , with a molar mass of 28.0 g/mol) at a temperature of 273 °C , find the ratio of the number of molecules with a speed very close to 300 m/s to the number with a speed very close to 100 m/s.

Strategy

Since we’re looking at a small range, we can approximate the number of molecules near 100 m/s as d N 100 = f ( 100 m/s ) d v . Then the ratio we want is

d N 300 d N 100 = f ( 300 m/s ) d v f ( 100 m/s ) d v = f ( 300 m/s ) f ( 100 m/s ) .

All we have to do is take the ratio of the two f values.

Solution

  1. Identify the knowns and convert to SI units if necessary.
    T = 300 K , k B = 1.38 × 10 −23 J/K

    M = 0.0280 kg/mol so m = 4.65 × 10 −26 kg
  2. Substitute the values and solve.
    f ( 300 m/s ) f ( 100 m/s ) = 4 π ( m 2 k B T ) 3 / 2 ( 300 m/s ) 2 exp [ m ( 300 m/s ) 2 / 2 k B T ] 4 π ( m 2 k B T ) 3 / 2 ( 100 m/s ) 2 exp [ m ( 100 m/s ) 2 / 2 k B T ] = ( 300 m/s ) 2 exp [ ( 4.65 × 10 −26 kg ) ( 300 m/s ) 2 / 2 ( 1.38 × 10 −23 J/K ) ( 300 K ) ] ( 100 m/s ) 2 exp [ ( 4.65 × 10 −26 kg ) ( 100 m/s ) 2 / 2 ( 1.38 × 10 −23 J/K ) ( 300 K ) ] = 3 2 exp [ ( 4.65 × 10 −26 kg ) [ ( 300 m/s ) 2 ( 100 ms ) 2 ] 2 ( 1.38 × 10 −23 J/K ) ( 300 K ) ] = 5.74
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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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