<p>The ppm designation means parts in mass per million parts in total mass.</p>
<p>Consider component A in a gas mixture of A + B. Assume the gas is ideal (it is at high enough temperature or low enough pressure such that non-ideal effects are insignificant). Define molar masses <span class="math-container">$M_A$</span>, <span class="math-container">$M_B$</span>, and <span class="math-container">$M_G$</span> (for the total gas). Under these conditions, 1 ppm A in the gas means by definition 1 mg A per 1 kg gas total. The molar mass of an ideal gas mixture is obtained by the mole fraction of A <span class="math-container">$y_A$</span> as</p>
<p><span class="math-container">$$ M_G = y_A\ M_A + (1 - y_A)\ M_B $$</span></p>
<p>Conversion of any <span class="math-container">$ppm$</span> to moles A per total moles is</p>
<p><span class="math-container">$$y_A = ppm\ \left(\frac{10^{-3}}{10^3}\right)\left(\frac{M_G}{M_A}\right)$$</span></p>
<p>leading to</p>
<p><span class="math-container">$$M_G = \frac{M_B}{1 - ppm\ 10^{-6}\ (M_B - M_A)} \approx M_B$$</span></p>
<p>The density of an ideal gas depends <span class="math-container">$\rho_G^\star$</span> depends only on pressure, temperature, and molar mass as</p>
<p><span class="math-container">$$\rho_G^\star = \frac{p\ M_G}{RT}$$</span></p>
<p>By example, for 2 ppm NO<span class="math-container">$_2$</span> in air, where <span class="math-container">$M_{air} \approx 28$</span> g/mol and <span class="math-container">$M_{NO2} \approx 46$</span> g/mol.</p>
<p><span class="math-container">$$M_G = \frac{46}{1 - 2\times 10^{-6}(46 - 28)} = 46.0016... $$</span>
<span class="math-container">$$ y_{NO2} = 2\times 10^{-6}\ \frac{46.0016...}{46} = 2.00007.. \times 10^{-6} $$</span></p>
<p>The exact conversion <span class="math-container">$y = ppm \times 10^{-6}$</span> is approached in an ideal gas at low ppm and when no significant difference occurs between the molar masses of the added component and the gas to which the component is added.</p>

CC BY-SA 4.0
2023-12-19T23:00:07.403Z
@JeopardyTempest fixed thanks
Jeffrey J Weimer
CC BY-SA 4.0
2023-12-19T16:27:49.633Z
2023-12-19T22:59:42.150Z
2023-12-19T22:59:42.150Z
Jeffrey J Weimer
<p>Consider a measurement of the concentration of a gas such as ozone in Earth's atmosphere in pseudo-units of parts-per-million (ppm). What is needed to convert such a measurement to density of the gas? Do we just need to multiply by the density of air at the altitude that the measurement was taken? If so, this can be done using using the ideal gas law, at least for low altitudes. I'm not sure at higher altitudes where compressibility matters more.</p>
<p>If is is easier, assume that this measurement is constant within a small box.</p>

<meteorology><atmosphere><climate><atmospheric-chemistry>
How can gaseous abundance in ppm be converted to density?
user900940
# How can gaseous abundance in ppm be converted to density?

(Asked by: username_0 on 2023-12-09)

Consider a measurement of the concentration of a gas such as ozone in Earth's atmosphere in pseudo-units of parts-per-million (ppm). What is needed to convert such a measurement to density of the gas? Do we just need to multiply by the density of air at the altitude that the measurement was taken? If so, this can be done using using the ideal gas law, at least for low altitudes. I'm not sure at higher altitudes where compressibility matters more.

If is is easier, assume that this measurement is constant within a small box.

**Tags:** meteorology, atmosphere, climate, atmospheric-chemistry


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## 1 answers

### Answer by username_1 on 2023-12-19. Score: 0

The ppm designation means parts in mass per million parts in total mass.

Consider component A in a gas mixture of A + B. Assume the gas is ideal (it is at high enough temperature or low enough pressure such that non-ideal effects are insignificant). Define molar masses $M\_A$, $M\_B$, and $M\_G$ (for the total gas). Under these conditions, 1 ppm A in the gas means by definition 1 mg A per 1 kg gas total. The molar mass of an ideal gas mixture is obtained by the mole fraction of A $y\_A$ as

$$ M\_G = y\_A\ M\_A + (1 - y\_A)\ M\_B $$

Conversion of any $ppm$ to moles A per total moles is

$$y\_A = ppm\ \left(\frac{10^{-3}}{10^3}\right)\left(\frac{M\_G}{M\_A}\right)$$

leading to

$$M\_G = \frac{M\_B}{1 - ppm\ 10^{-6}\ (M\_B - M\_A)} \approx M\_B$$

The density of an ideal gas depends $\rho\_G^\star$ depends only on pressure, temperature, and molar mass as

$$\rho\_G^\star = \frac{p\ M\_G}{RT}$$

By example, for 2 ppm NO$\_2$ in air, where $M\_{air} \approx 28$ g/mol and $M\_{NO2} \approx 46$ g/mol.

$$M\_G = \frac{46}{1 - 2\times 10^{-6}(46 - 28)} = 46.0016... $$
$$ y\_{NO2} = 2\times 10^{-6}\ \frac{46.0016...}{46} = 2.00007.. \times 10^{-6} $$

The exact conversion $y = ppm \times 10^{-6}$ is approached in an ideal gas at low ppm and when no significant difference occurs between the molar masses of the added component and the gas to which the component is added.

> **Comments:**
> **username_1**: @JeopardyTempest fixed thanks  

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