Question

A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2×10^-6mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 ̊C?

Short answer

The partial pressure of gaseous mercuryis1.9×10^-7torr.

Step-by-step solution

Definition of pressure

The force applied perpendicular to an object's surface per unit area across which that force is spread is known as pressure. The pressure gauge in relation to the ambient pressure is known as gauge pressure. Pressure is measured in a variety of ways.

Calculating partial pressure gaseous mercury

We need to calculate the partial pressure of gaseous mercury.

Partial pressure of gaseous mercury,

$$P_{Hg}=X_{Hg}\times P_{Total}$$

To calculate$$\mathrm{P}\mathrm{H}\mathrm{g}$$we need calculate$${\mathrm{X}}_{\mathrm{H}\mathrm{g}}$$and$${\mathrm{P}}_{\mathrm{H}\mathrm{g}}$$.

$${\mathrm{X}}_{\mathrm{H}\mathrm{g}}=\frac{{\mathrm{n}}_{\mathrm{H}\mathrm{g}}}{{\mathrm{n}}_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}$$

Concentration of gaseous Hg atoms =2×10^-6mg/L=2×10^-9g/L

Converting concentration to moles, we have

$$\begin{array}{r}{\mathrm{n}}_{\mathrm{H}\mathrm{g}}=\frac{2\times 10^{-9}\mathrm{g}/\mathrm{L}}{200.59\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}\\ =9.97\times 10^{-12}\mathrm{m}\mathrm{o}\mathrm{l}/\mathrm{L}\end{array}$$

$$\mathrm{n}\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}$$are the moles of all the gases in air required to bring the atmospheric prasine to $$733$$torr at$$26^{{}^{\circ }}\mathrm{C}$$.

If we choose to use

$$\mathrm{R}=0.08206\mathrm{L}$$atm$$\mathrm{m}\mathrm{o}{\mathrm{l}}^1{\mathrm{K}}^{-1}$$,

then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

$$\begin{array}{r}\mathrm{P}\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}=733\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\times \frac{1\mathrm{a}\mathrm{t}\mathrm{m}}{760\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}\\ =0.964\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{T}}^{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}=26^{{}^{\circ }}\mathrm{C}+273\\ =299\mathrm{K},\mathrm{V}=1\\ \mathrm{L}{\mathrm{n}}_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{\mathrm{P}\mathrm{V}}{\mathrm{R}\mathrm{T}}\\ =\frac{0.964\mathrm{a}\mathrm{t}\mathrm{m}\times 1\mathrm{L}}{\left(0.08206\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}\right)\times 299\mathrm{K}}\\ =0.0393\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$$

Now, we can calculate partial pressure of$$\mathrm{H}\mathrm{g}$$:

$$\begin{array}{r}l{P}_{Hg}=X_{Hg}\times P\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\\ =\frac{n_{Hg}}{n_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}\times P_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}\\ =\frac{9.97\times {10}^{12}\mathrm{m}\mathrm{o}\mathrm{l}}{0.0393\mathrm{m}\mathrm{o}\mathrm{l}}\times 0.964\mathrm{a}\mathrm{t}\mathrm{m}\\ =2.45\times {10}^{-10}\mathrm{a}\mathrm{t}\mathrm{m}\times \frac{760\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}{1\mathrm{a}\mathrm{t}\mathrm{m}}\\ =1.9\times {10}^{-7}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\end{array}$$

Therefore, the partial pressure ofgaseous mercuryis $$1.9\times 10^{-7}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$$.