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Worked Examples · Example 12.9

Q.Estimate the mean free path for a water molecule in water vapour at 373 K373\ \text{K}. Use information from Example 12.1 and Eq. (12.41) above (mean free path of an air molecule at STP, l=2.9×10−7 ml = 2.9 \times 10^{-7}\ \text{m}).

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The mean free path is l=1/(2 n πd2)l = 1/(\sqrt2\,n\,\pi d^2) with n=P/kTn = P/kT. At fixed pressure l∝Tl \propto T, and taking the effective molecular size for water vapour as roughly that of air, scaling the given air value from 273 K273\ \text{K} to 373 K373\ \text{K} gives l≈4×10−7 ml \approx 4\times10^{-7}\ \text{m}.

Reasoning

The mean free path depends on the number density nn and the molecular diameter dd:

l=12 n πd2,n=PkTl = \frac{1}{\sqrt{2}\,n\,\pi d^2}, \qquad n = \frac{P}{kT}

At the same pressure, n∝1/Tn \propto 1/T, so - treating the effective diameter of a water molecule as roughly the same as that of an air molecule (both a few angstrom, a fair estimate for an order-of-magnitude answer) - the mean free path scales directly with temperature:

lvapourlair=nairnvapour=TvapourTair\frac{l_{\text{vapour}}}{l_{\text{air}}} = \frac{n_{\text{air}}}{n_{\text{vapour}}} = \frac{T_{\text{vapour}}}{T_{\text{air}}} …

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