Chapter 23: Problem 50
The vapor pressure of \(\mathrm{Hg}(1)\) as a function of temperature is \(\log P(\mathrm{mmHg})=(-0.05223 a / T)+b,\) where \(a=61,960\) and \(b=8.118 ; T\) is the Kelvin temperature. Show that at \(25^{\circ} \mathrm{C}\), the concentration of \(\mathrm{Hg}(\mathrm{g})\) in equilibrium with Hg(l) greatly exceeds the maximum permissible level of \(0.05 \mathrm{mg} \mathrm{Hg} / \mathrm{m}^{3}\) air.
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