Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): $$\text { ppmv of } X=\frac{\text { vol of } X \text { at } \mathrm{STP}}{\text { total vol of air at } \mathrm{STP}} \times 10^{6}$$ On a certain November day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was \(628\) torr and the temperature was \(0^{\circ} \mathrm{C}\) a. What was the partial pressure of \(\mathrm{CO}\)? b. What was the concentration of \(\mathrm{CO}\) in molecules per cubic meter? c. What was the concentration of \(\mathrm{CO}\) in molecules per cubic centimeter?

We can calculate the partial pressure of CO by first converting the atmospheric pressure (in torr) to atm. 1 atm = 760 torr So, $$ P_\text{total} = \frac{628 \ \text{torr}}{760 \ \text{torr/atm}} = 0.8263 \ \text{atm} $$ Now to find the partial pressure of CO in the air, we use the mixing ratio formula given: $$ \text{ppmv of } X = \frac{\text { vol of } X \text { at STP}}{\text { total vol of air at STP}} \times 10^{6} $$ Rearrange the formula to isolate the partial pressure of CO: $$ P_\text{CO} = \frac{\text{ppmv of CO}}{10^6} \times P_\text{total} $$ Plug in the provided ppmv value and the total pressure: $$ P_\text{CO} = \frac{3.0 \times 10^2}{10^6} \times 0.8263 \ \text{atm} $$ Calculate the partial pressure of CO: $$ P_\text{CO} = 2.479 \times 10^{-4} \ \text{atm} $$ #a. The partial pressure of CO is \(2.479 \times 10^{-4}\) atm.

In order to find the concentration of CO in molecules per cubic meter and per cubic centimeter, we need to convert the pressure to pascals and find the moles of CO per unit volume. 1 atm = 101325 Pa So, $$ P_\text{CO} = 2.479 \times 10^{-4} \ \text{atm} \times \frac{101325 \ \text{Pa}}{1 \ \text{atm}} = 25.11 \ \text{Pa} $$ Use the Ideal Gas Law, \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature. We are given the temperature in Celsius; we need to convert it to Kelvin: $$ T = 0^{\circ} \text{C} + 273.15 = 273.15 \ \text{K} $$ Rearrange the Ideal Gas Law formula to find moles per unit volume, \(n/V\): $$ \frac{n}{V} = \frac{P}{RT} $$ Use the ideal gas constant, \(R = 8.314\) J/(mol K), and plug in the values: $$ \frac{n}{V} = \frac{25.11 \ \text{Pa}}{8.314 \text{J/(mol K)} \times 273.15 \ \text{K}} $$ Calculate the moles per cubic meter: $$ \frac{n}{V} = 1.089 \times 10^{-5} \ \text{mol/m}^3 $$

We will now find the concentration of CO in molecules per cubic meter and per cubic centimeter using the moles per cubic meter value and Avogadro's number. Avogadro's number \(N_\text{A} = 6.022 \times 10^{23}\) molecules/mol Multiply moles per cubic meter by Avogadro's number to find molecules per cubic meter: $$ \text{molecules/m}^3 = \left( 1.089 \times 10^{-5} \ \text{mol/m}^3 \right) \times \left( 6.022 \times 10^{23} \ \text{molecules/mol} \right) = 6.558 \times 10^{18} \ \text{molecules/m}^3 $$ Now convert molecules per cubic meter to molecules per cubic centimeter by using: 1 m³ = 1,000,000 cm³ $$ \text{molecules/cm}^3 = \frac{6.558 \times 10^{18} \ \text{molecules/m}^3}{1,000,000 \ \text{cm}^3/\text{m}^3} = 6.558 \times 10^{12} \ \text{molecules/cm}^3 $$ #b. The concentration of CO in molecules per cubic meter is \(6.558 \times 10^{18}\) molecules/m³. #c. The concentration of CO in molecules per cubic centimeter is \(6.558 \times 10^{12}\) molecules/cm³.