The root-mean-square speed vrms gives us a general idea of molecular speeds in a gas at a given temperature. We often want to know more. For example, what fraction of the molecules have speeds greater than the rms value? What fraction have speeds greater than twice the rms value? To answer such questions, we need to know how the possible values of speed are distributed among the molecules. Figure 19-7a shows this distribution for oxygen molecules at room temperature (T = 300 K); Fig. 19-7b compares it with the distribution at T = 80 K.
In 1852, Scottish physicist James Clerk Maxwell first solved the problem of finding the speed distribution of gas molecules. His result, known as Maxwell’s speed distribution law, is
Fig. 19-7 (a) The Maxwell speed distribution for oxygen molecules at T = 300 K. The three characteristic speeds are marked. (b) The curves for 300 K and 80 K. Note that the molecules move more slowly at the lower temperature. Because these are probability distributions, the area under each curve has a numerical value of unity.
Here M is the molar mass of the gas, R is the gas constant, T is the gas temperature, and v is the molecular speed. It is this equation that is plotted in Fig. 19-7a, b. The quantity P(v) in Eq. 19-27 and Fig. 19-7 is a probability distribution function: For any speed v, the product P(v) dv (a dimensionless quantity) is the fraction of molecules with speeds in the interval dv centered on speed v.
As Fig. 19-7a shows, this fraction is equal to the area of a strip with height P(v) and width dv. The total area under the distribution curve corresponds to the fraction of the molecules whose speeds lie between zero and infinity. All molecules fall into this category, so the value of this total area is unity; that is,
The fraction (frac) of molecules with speeds in an interval of, say, v1 to v2 is then
Average, RMS, and Most Probable Speeds
In principle, we can find the average speed vavg of the molecules in a gas with the following procedure: We weight each value of v in the distribution; that is, we multiply it by the fraction P(v) dv of molecules with speeds in a differential interval dv centered on v. Then we add up all these values of v P(v) dv. The result is vavg. In practice, we do all this by evaluating
Substituting for P(v) from Eq. 19-27 and using generic integral 20 from the list of integrals in Appendix E, we find
Similarly, we can find the average of the square of the speeds (v2)avg with
Substituting for P(v) from Eq. 19-27 and using generic integral 16 from the list of integrals in Appendix E, we find
The square root of (v2)avg is the root-mean-square speed vrms. Thus,
which agrees with Eq. 19-22.
The most probable speed vP is the speed at which P(v) is maximum (see Fig. 19-7a). To calculate vP, we set dP/dv = 0 (the slope of the curve in Fig. 19-7a is zero at the maximum of the curve) and then solve for v. Doing so, we find
A molecule is more likely to have speed vP than any other speed, but some molecules will have speeds that are many times vP. These molecules lie in the high-speed tail of a distribution curve like that in Fig. 19-7a. We should be thankful for these few, higher speed molecules because they make possible both rain and sunshine (without which we could not exist). We next see why.
Rain The speed distribution of water molecules in, say, a pond at summertime temperatures can be represented by a curve similar to that of Fig. 19-7a. Most of the molecules do not have nearly enough kinetic energy to escape from the water through its surface. However, small numbers of very fast molecules with speeds far out in the high-speed tail of the curve can do so. It is these water molecules that evaporate, making clouds and rain a possibility.
As the fast water molecules leave the surface, carrying energy with them, the temperature of the remaining water is maintained by heat transfer from the surroundings. Other fast molecules—produced in particularly favorable collisions—quickly take the place of those that have left, and the speed distribution is maintained.
Sunshine Let the distribution curve of Fig. 19-7a now refer to protons in the core of the Sun. The Sun’s energy is supplied by a nuclear fusion process that starts with the merging of two protons. However, protons repel each other because of their electrical charges, and protons of average speed do not have enough kinetic energy to overcome the repulsion and get close enough to merge. Very fast protons with speeds in the high-speed tail of the distribution curve can do so, however, and for that reason the Sun can shine.
A container is filled with oxygen gas maintained at room temperature (300 K). What fraction of the molecules have speeds in the interval 599 to 601 m/s? The molar mass M of oxygen is 0.0320 kg/mol.
Solution: The Key Ideas here are the following:
1. The speeds of the molecules are distributed over a wide range of values, with the distribution P(v) of Eq. 19-27.
2. The fraction of molecules with speeds in a differential interval dv is P(v) dv.
3. For a larger interval, the fraction is found by integrating P(v) over the interval.
4. However, the interval Δv = 2 m/s here is small compared to the speed v = 600 m/s on which it is centered.
Thus, we can avoid the integration by approximating the fraction as
The function P(v) is plotted in Fig. 19-7a. The total area between the curve and the horizontal axis represents the total fraction of molecules (unity). The area of the thin gold strip represents the fraction we seek.
To evaluate frac in parts, we can write
where
Substituting A and B into Eq. 19-36 yields
Thus, at room temperature, 0.262% of the oxygen molecules will have speeds that lie in the narrow range between 599 and 601 m/s. If the gold strip of Fig. 19-7a were drawn to the scale of this problem, it would be a very thin strip indeed.
The molar mass M of oxygen is 0.0320 kg/mol.
(a) What is the average speed vavg of oxygen gas molecules at T = 300 K?
Solution: The Key Idea here is that to find the average speed, we must weight speed v with the distribution function P(v) of Eq. 19-27 and then integrate the resulting expression over the range of possible speeds (0 to ∞). That leads to Eq. 19-31, which gives us
This result, plotted in Fig. 19-7a
(b) What is the root-mean-square speed vrms at 300 K?
Solution: The Key Idea here is that to find vrms, we must first find (v2)avg by weighting v2 with the distribution function P(v) of Eq. 19-27 and then integrating the expression over the range of possible speeds. Then we must take the square root of the result. That leads to Eq. 19-34, which gives us
This result, plotted in Fig. 19-7a, is greater than vavg because the greater speed values influence the calculation more when we integrate the v2 values than when we integrate the v values.
(c) What is the most probable speed vP at 300 K?
Solution: The Key Idea here is that vp corresponds to the maximum of the distribution function P(v), which we obtain by setting the derivative dP/dv = 0 and solving the result for v. That leads to Eq. 19-35, which gives us
This result is also plotted in Fig. 19-7a.
Leave a Reply