Question

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): ppmv of X 5 vol of X at STP 3 106 total vol of air at STP On a certain November day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached 3.0 3 102 ppmv. The atmospheric pressure at that time was 628 torr and the temperature was 08C.
a. What was the partial pressure of CO?
b. What was the concentration of CO in molecules per cubic meter?
c. What was the concentration of CO in molecules per cubic centimeter?

Answer 1

Answer:

a) $P_{CO_2} =0.188 torr$

b) $6.71\times 10^{21} \frac{molecules}{m^3}$

c) $6.71 \times 10^{15} \frac{molecules}{cm^3}$

Explanation:

a) The partial pressure of CO₂ is given by:

$P_{CO_2} = x_{CO_2} \times P_{T}$

Since the molar fraction of the gas is directly proportional to the volume of the gas, the molar fraction is given by:

$x_{CO_2} = \frac{moles \ CO_2}{total \ moles} =\frac{volume\ of\ CO_2}{total\ volume}$

$x_{CO_2}=\frac{3.0\times 10^2 ppmv}{10^6} =3.0\times 10^{-4}$

Hence,

$P_{CO_2} =3.0\times 10^{-4} \times 628 torr=0.188 torr$

b) Using the Ideal Gas Law, we can find the number of moles of CO₂:

$P_{CO_2}V=n_{CO_2}RT$

where:

$P_{CO_2}=0.19torr\frac{1 atm}{760torr} =2.5\times10^{-4} atm$

$T=0 + 273K=273K$

$R=0.0822\frac{l.atm}{mol.K}$

The concentration of CO in molecules per cubic meter is:

$\frac{n_{CO_2}}{V} = \frac{ P_{CO_2}}{RT}$

$\frac{n_{CO_2}}{V} = \frac{ 2.5\times10^{-4} atm}{0.0822\frac{l.atm}{mol.K}\times 273K}\times \frac{1000l}{1m^3} \times \frac{6.022\times 10^{23} molecules}{1 mole}=6.71\times 10^{21} \frac{molecules}{m^3}$

c) The concentration of CO in molecules per cubic centimeter is:

$6.71\times 10^{21} \frac{molecules}{m^3}\times\frac{1 m^3}{(100cm)^3}= 6.71 \times 10^{15} \frac{molecules}{cm^3}$