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## Q. 9.4

What fraction of the molecules in an ideal gas in equilibrium has speeds within ±
1% of v*?

Strategy Recall that F(v) dv is the probability of finding a particle with speed between v and v + dv. In principle we could integrate the distribution F(v) in Equation (9.14) from the limits 0.99v* to 1.01v*.

$F(v) d v=4 \pi C \exp \left(-\frac{1}{2} \beta m v^{2}\right) v^{2} d v$ (9.14)

$P(\pm 1 \%)=\int_{0.99 v^{*}}^{1.01 v^{*}} F(v) d v$ (9.23)

Unfortunately, the indefinite integral cannot be done in closed form. We can obtain an approximate solution by calculating F(v*) and multiplying by $d v \approx \Delta v=0.02 v^{*}$.

## Verified Solution

The product F(v*)(0.02v*) gives the probability

\begin{aligned}P(\pm 1 \%) & \approx F\left(v^{*}\right)\left(0.02 v^{*}\right) \\& \approx 4 \pi C \exp \left(-\frac{1}{2} \beta m v^{* 2}\right) v^{* 2}\left(0.02 v^{*}\right) \\& \approx 4 \pi\left(\frac{\beta m}{2 \pi}\right)^{3 / 2} e^{-1}(0.02)\left(\frac{2}{\beta m}\right)^{3 / 2} \\& \approx \frac{4}{\sqrt{\pi}} e^{-1}(0.02) \approx 0.017\end{aligned}

Students with computer programming experience are encouraged to do the integration in Equation (9.23) numerically and compare the result with this approximation. Your results will be more precise but should agree with this approximation to two significant figures.