Question

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?

Short answer

$1.85\times {10}^{-3}\mathrm{torr}$

Step-by-step solution

 Step 1: Given the information

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

Ideal gas equation

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

Convert to units

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 $\mathrm{L}$($1{\mathrm{m}}^3=1000\mathrm{L}$)

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

Derive partial pressure

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 $1atm=760torr$

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