Question 18.3: Test assumption 7 at the beginning of this section by comput...
Test assumption 7 at the beginning of this section by computing the fraction of neon molecules at 273 \text{K} whose velocities are faster than (a) V_\text{mp}, (b) V_\text{avg}, (c) V_\text{rms}, and (d) c (the speed of light). Use the molecular data for neon given in Example 18.2
a. The fraction having velocities greater than V_\text{mp} is given by Eq. (18.26) with x = V_\text{mp}/V_\text{mp} = 1.0 as
\frac{N(V →∞)}{N} = 1− \frac{N(0 →V)}{N} = 1− \text{erf} (x) + \frac{2}{\sqrt{π}} xe^{−x^2} (18.26)
\frac{N(V_\text{mp} →∞)}{N} = 1− \text{erf} (1.0) + \frac{2}{\sqrt{π}} (1.0)e^{−1.0}
or
\frac{N(V_\text{mp} →∞)}{N} = 1 − 0.8427 + 0.4151 = 0.5724
Thus, 57.24\% of the molecules have velocities faster than V_\text{mp}.
b. Here, Eq. (18.20) gives
V_\text{avg} = \sqrt{\frac{8kT}{πm}} , V_{mp} = \sqrt{\frac{2kT}{m}}
and
x = \frac{V_\text{avg}}{V_{mp}} = \sqrt{\frac{8}{2π}} = 1.128
Therefore, the fraction of molecules having velocities greater than V_\text{avg} is given by interpolating in Table 18.2 to find
| Table 18.2 Values of the Error Function | |
| x | erf(x) |
| 0.0 | 0.0 |
| 0.1 | 0.1125 |
| 0.2 | 0.2227 |
| 0.3 | 0.3286 |
| 0.4 | 0.4284 |
| 0.5 | 0.5205 |
| 0.6 | 0.6039 |
| 0.7 | 0.6778 |
| 0.8 | 0.7421 |
| 0.9 | 0.7969 |
| 1.0 | 0.8427 |
| 1.2 | 0.9103 |
| 1.4 | 0.9523 |
| 1.6 | 0.9764 |
| 1.8 | 0.9891 |
| 2.0 | 0.9953 |
| 2.2 | 0.9981 |
| 2.4 | 0.9993 |
| 2.6 | 0.9998 |
| 2.8 | 0.9999 |
| ∞ | 1.0 |
| Note: For all \text{x}, \text{erf} (\text{x}) = \frac{2}{\sqrt{π}} (\text{x} – \frac{\text{x}^3}{3(1!)} + \frac{\text{x}^5}{5(2!)} – \frac{\text{x}^7}{7(3!)} + …) and \text{exp} ( – \text{x}^2) = 1 – \text{x}^2 /1! + \text{x}^4 /2! – \text{x}^6 /3! + \text{x}^8 /4! – … | |
\frac{N(V_\text{avg} →∞)}{N} = 1− \text{erf} (1.128) + \frac{2}{\sqrt{π}} (1.128)e^{−1.272}
= 1 − 0.8893 + 0.3566 = 0.4673
Consequently, 46.73\% of the molecules have velocities faster than V_\text{avg}
c. Here, Eq. (18.21) gives V_\text{rms} = \sqrt{3kT/m} ,so
x = V_\text{rms}/V_{mp} = \sqrt{\frac{3}{2}} = 1.225
and
\frac{N(V_\text{rms} →∞)}{N} = 1− \text{erf} (1.225) + \frac{2}{\sqrt{π}} (1.225)e^{−1.501}
= 1 − 0.9168 + 0.3081 = 0.3913
or 39.13\% of the molecules have velocities greater than V_\text{rms}.
d. It can be shown (see Problem 10 at the end of this chapter) that, when V/V_{mp} ≫ 1, the fraction of molecules with velocities in the range from V to ∞ is approximately given by
\frac{N(V →∞)}{N} ≈ \frac{2}{\sqrt{π}} (x + \frac{1}{2x})e^{−x^2} (\text{for} x ≫ 1 \text{only})
Consider x = V/V_{mp} = 10.00 then,
\frac{N(V →∞)}{N} ≈ \frac{2}{\sqrt{π}} (10.05)e^{−100} = 4.22 × 10^{−43}
Thus, only one molecule in about 10^{20} moles of a gas has a velocity ten times greater than V_\text{mp}. Now, the velocity of light c is 3.00 × 10^ 8 \text{m/s}, and for neon at 273 \text{K}. we have m = 3.35 × 10^{−26} \text{kg} (see Example 18.2), and V_{mp} = \sqrt{2kT/m} = 474 \text{m/s}. Thus
x = \frac{V}{V_{mp}} = \frac{c}{V_{mp}} = \frac{3.00 × 10^8 \text{m/s}}{474 \text{m/s}} = 6.33 × 10^5
so that
\frac{N(c →∞)}{N} ≈ \frac{2}{\sqrt{π}} (6.33 × 10^5)e^{−4.0×10^{11}} ≈0
Even though we allow molecules to move faster than the speed of light in our mathematical model, we find that, for all practical purposes, this model predicts that virtually no molecules have velocities this fast at ordinary temperatures.