Question

The mixing ratio of CO₂ at a pressure of 1 atm and at 300 K is reported as 340 ppmv.

Considering ideal gas conditions, the equivalent concentration of CO₂ in air is ______ mg/m³ (rounded off to two decimal places).

The molecular weight of CO₂ = 44 g/mol

Universal gas constant = 8.314 J/mol-K

Hint:

Find the molar volume of air from the ideal gas law at 1 atm and 300 K, then convert the ppmv mixing ratio into moles of CO2 per cubic metre before multiplying by the molecular weight.

Answer 1

Correct Answer: 607.74

This problem can be solved directly with a single conversion formula instead of working out the molar volume separately.

The total molar concentration of air at the given pressure and temperature is $\dfrac{P}{RT}$. Substituting the numbers with $P$ in pascals: $$\frac{P}{RT} = \frac{101325}{8.314 \times 300} = 40.62\ mol/m^3$$

This tells us that one cubic metre of air contains about 40.62 moles of gas in total, regardless of composition. Since the mixing ratio of $CO_2$ is 340 ppmv (a volume, and hence mole, fraction of $340\times10^{-6}$), the molar concentration of $CO_2$ itself is $$n_{CO_2} = 340\times10^{-6} \times 40.62 = 0.013811\ mol/m^3$$

Converting this to a mass concentration using the molecular weight of $CO_2$ (44 g/mol) and expressing the result in mg/m^3: $$C = 0.013811 \times 44 \times 1000 = 607.7\ mg/m^3$$

Both routes agree because they are really the same ideal-gas relation rearranged differently, which is a good check on the arithmetic. \[\boxed{C \approx 607.74\ mg/m^3}\]