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

Step 1: Write down the known quantities.
The mixing ratio is 340 ppmv, meaning 340 parts of \(CO_2\) by volume per million parts of air, at \(P = 1\ atm = 101325\ Pa\) and \(T = 300\ K\). The molecular weight is \(M = 44\ g/mol\) and \(R = 8.314\ J/mol\text{-}K\).

Step 2: Find the molar volume of air at the given conditions.
From the ideal gas law, \(PV = nRT\), so the volume occupied by one mole is \[V_m = \frac{RT}{P} = \frac{8.314 \times 300}{101325} = 0.024618\ m^3/mol\]

Step 3: Convert the volume ratio into a molar concentration.
A mixing ratio of 340 ppmv means that in every cubic metre of air, \(340\times10^{-6}\ m^3\) is \(CO_2\) (same T and P). The number of moles of \(CO_2\) per cubic metre of air is \[n = \frac{340\times10^{-6}}{0.024618} = 0.013812\ mol/m^3\]

Step 4: Convert moles to mass concentration.
Multiplying by the molecular weight gives the mass concentration: \[C = 0.013812 \times 44 = 0.60774\ g/m^3 = 607.74\ mg/m^3\] This value lies comfortably in the expected 590-615 mg/m³ band for a few-hundred ppmv \(CO_2\) reading, so the working checks out.