Mercury has a small but significant vapor pressure, andpoisoning can occur in$20.0\raisebox{1ex}{$\mathrm{mg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3\mathrm{Hg}$}\right.$poorly ventilated rooms wherehas been spilled. What is the partial pressure (in torr) of$\mathrm{Hg}$if the maximum concentration at$25°\mathrm{C}$is?

Mercury has a small but significant vapor pressure, and poisoning can occur in poorly ventilated rooms where has been spilled.

Ideal Gas Equation: $\mathrm{PV}=\mathrm{nRT}$

To get the partial pressure of mercury, the given concentration must first be converted to units that can be used in the ideal gas equation. In this case, $20\frac{\mathrm{mg}}{{\mathrm{m}}^3}$ must be converted to $\frac{\mathrm{mol}}{\mathrm{L}}$

This can be done through using the molar mass of $\mathrm{Hg}$($200.59\times {10}^3\raisebox{1ex}{$mg$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$) and converting ${\mathrm{m}}^3$to $\mathrm{L}$($1{\mathrm{m}}^3=1000\mathrm{L}$)

$\frac{20.0\mathrm{mg}}{{\mathrm{m}}^3}\times \frac{1{\mathrm{m}}^3}{1000\mathrm{L}}\times \frac{1\mathrm{molHg}}{200.59\times {10}^3\mathrm{mgHg}}=\frac{9.97\times {10}^{-8}\mathrm{mol}}{\mathrm{L}}$

Ideal Gas Equation: $PV=nRT$

Upon manipulation, partial pressure can be derived through:

$P=\frac{n}{V}\times RT$

with $R=0.08206\frac{Latm}{molK}$and $T=25°$ or $298.15K.$

$\begin{array}{rcl}P& =& \frac{n}{V}\times RT\\ & =& \frac{n}{V}\times \left(0.08206\frac{Latm}{molK}\right)\left(298.15K\right)\\ & =& \frac{9.97\times {10}^{-8}moL}{L}\times \left(0.08206\frac{Latm}{molK}\right)\left(298.15K\right)\\ & =& 2.44\times {10}^{-6}atm\\ & & \end{array}$

Convert the given pressure in atm to torr through the conversion factor $1atm=760torr$

$2.44\times {10}^{-6}\mathrm{atm}\times \frac{760\mathrm{torr}}{1\mathrm{atm}}=1.85\times {10}^{-3}\mathrm{torr}.$