# Why are these two methods of converting from ppm to g/m3 equivalent?

I've found two methods to convert from $$ppm$$ to $$g/m^3$$. They both appear to give equivalent results, but I don't understand why. Can anyone join the dots for me and explain why they are equivalent methods, or show a derivation for Method 1? I've found a derivation for Method 2.

Method 1 - Definition

$$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$

Where $$C[ppm]$$ and $$C[g/m^3]$$ represent gas concentration in units of $$ppm$$ and $$g/m^3$$ respectively and $$\rho$$ represents gas density.

Method 2 - Definition (As derived here)

$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$

Where $$M$$, $$P$$, $$R$$ and $$T$$ represent molar mass, pressure, ideal gas constant and temperature respectively.

Method 1 - Example - Carbon Monoxide (CO)

Assuming $$C[ppm]$$ as $$100ppm$$ and CO density as $$1.15 kg/m^3$$ at $$20^{\circ}C$$ and $$100000Pa$$ $$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$ $$C[g/m^3] = \frac{100 \cdot 1.15}{1000}$$ $$C[g/m^3] = 0.115$$

Method 2 - Example - Carbon Monoxide (CO)

Assuming $$C[ppm]$$ as $$100ppm$$ at $$20^{\circ}C$$ and $$100000Pa$$.

$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$ $$C[g/m^3] = \frac{100 \cdot 28 \cdot 100000}{8.3145 \cdot (20+273.15) \cdot 10^6}$$ $$C[g/m^3] = 0.115$$

Using the ideal gas law

$$pV = nRT = \frac{m}{M}RT$$

you can express the density $$\rho = \frac m V$$ in terms of molar mass, pressure, ideal gas constant, and temperature:

$$\rho = \frac{pM}{RT}$$

With the units you use in your second example, this yields a density in $$\mathrm{g\,m^{-3}}$$, whereas you insert a density in $$\mathrm{kg\,m^{-3}}$$ into your first equation. Thus, I will write your first equation as

$$C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6}$$

where the density is in $$\mathrm{g\,m^{-3}}$$.

Inserting the expression for $$\rho$$ derived above into this equation, we get

$$C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6} = \frac{C[\mathrm{ppm}]}{10^6}\frac{pM}{RT}$$

which is identical to your second equation.