I want to investigate how temperature affect the concentration of oxygen in air as it will affect how much oxygen we inhale per breathe. Oxygen by percent composition in air is 20.95%. However, the oxygen concentration/density in air varies due to the expansion of gas particles and varies inversely wth temperature according to P/RT=n/V (where n/V is the density of gas).

I did an experiment that involves the rusting of steel wool to determine the percent oxygen in air by volume (determined by volume of water rose in a test tube divided by the total volume of air in the test tube). However, I'm not sure if the % oxygen I determined is a measure of the oxygen density (if so I can convert the percentage to ppm).

My initial hypothesis is that as there are less oxygen molecules in hot air, water level will rise less since there is less oxygen reacting with the steel wool (compared to cold air). Hence lower % oxygen concentration. Vice versa for cold air.

Then later, I'm thinking that as oxygen molecules (or any gas particles) are heated, they move quicker so each oxygen molecule will occupy more volume. Vice versa if the gas particles are colder. However, since there are more oxygen molecules in cold air than warm air, the increase in number of oxygen molecules will occupy more volume and make up for the smaller volume each oxygen molecule occupy by slower movement.

Therefore, the % oxygen in air by volume I determined by the volume of rising water level over the volume of air, will theoretically be the same and hence it can't be used to determine the oxygen density in air.

But then I also saw a picture of a vernier O2 probe with percentage O2 reading (which im not sure if it's percent by volume or percent composition) that can be converted to concentration in ppm by a click of a button. I then did some research and found out percentage concentration (by mass or by volume) can be converted to ppm (which will be a measure of oxygen density).


3 Answers 3


No, the concentration does not depend on density in a well-mixed ideal gas.

The key principle here is that the atmosphere is close to being an ideal gas. In an ideal gas the components are well mixed and the composition will not change unless there is a chemical change that acts selectively on one of the components.

The ideal gas equation applies equally to each component of the mixture of gases that is air. If you warm a vessel of fixed size, the pressure goes up. If you warm the mixture in a variable volume, the volume increases. But nothing in either process causes the composition of the gas to change, so the concentrations of the components will remain the same.

You seem to be applying the gas equation to just the oxygen and ignoring the other components (Nitrogen, argon etc.). Yes the warmer air will have a lower density, but the composition will be unaltered. So the relative amount of oxygen in the air remains the same. Warm air has an overall lower density (which is why hot air balloons work) and the volume density of all the components will decrease. So the percentage of oxygen per litre will be lower but so will the percentage of Nitrogen. Mostly when measuring gas composition chemists talk about the concentration relative to all the components in the mixture not to the volume (and in an ideal gas, the volume/pressure/temperature relationship is the same for all components). So, trivially, the volume density of all components is lower when the gas is warmer at constant pressure. There is no selective change for just one component.


Below 90.19 K there is definitely an effect of temperature on the % oxygen in air, as it condensates out. So as general as the question is stated in the header, the answer is 'yes'.


If we observe a closed air parcel so that no gas can go in or out, the amount-of-substance fraction $x$ of each component is given as $$x_{\ce{N2}}=\frac{n_{\ce{N2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.7812=78.12\ \%$$ $$x_{\ce{O2}}=\frac{n_{\ce{O2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.2096=20.96\ \%$$ $$x_{\ce{Ar}}=\frac{n_{\ce{Ar}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.0092=0.92\ \%$$ Since molar mass is defined as $$M=\frac mn$$ we can directly obtain the corresponding mass fraction $w$. $$w_{\ce{N2}}=\frac{m_{\ce{N2}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{N2}}\cdot M_{\ce{N2}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.7557=75.57\ \%$$ $$w_{\ce{O2}}=\frac{m_{\ce{O2}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{O2}}\cdot M_{\ce{O2}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.2316=23.16\ \%$$ $$w_{\ce{Ar}}=\frac{m_{\ce{Ar}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{Ar}}\cdot M_{\ce{Ar}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.0127=1.27\ \%$$ Assuming ideal gas behaviour, i.e. $$pV=nRT$$ we can also calculate the corresponding volume fraction $\varphi$. $$\begin{align}\varphi_{\ce{N2}}=\frac{V_{\ce{N2}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{N2}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt] &=\frac{n_{\ce{N2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{N2}}=0.7812=78.12\ \%\end{align}$$ $$\begin{align}\varphi_{\ce{O2}}=\frac{V_{\ce{O2}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{O2}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt] &=\frac{n_{\ce{O2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{O2}}=0.2096=20.96\ \%\end{align}$$ $$\begin{align}\varphi_{\ce{Ar}}=\frac{V_{\ce{Ar}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{Ar}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt] &=\frac{n_{\ce{Ar}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{N2}}=0.0092=0.92\ \%\end{align}$$ Note how $R$, $T$, and $p$ cancel since they are they same for all involved gases. Thus, volume fraction $\varphi$ is equal to amount-of-substance fraction $x$, but only for ideal gas.

Temperature $T$ and pressure $p$ of the air parcel may change, and the volumes $V_{\ce{N2}}$, $V_{\ce{O2}}$, $V_{\ce{Ar}}$, and $V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}$ may change accordingly; however, the volume fraction $\varphi$ remains constant since the volumes are proportional to each other.


The following values were obtained from the GERG-2008 mixture model, i.e. without using the ideal gas equation.

Assuming dry air with the following initial conditions:

$T_1=10.000\ \mathrm{^\circ C}$
$p_1=1.0000\ \mathrm{bar}$
$V_{\text{air},1}=1.0000\ \mathrm{m^3}$
$\rho_{\text{air},1}=1.2306\ \mathrm{kg\ m^{-3}}$

Thus, the mass of $\ce{O2}$ is

$$m_{\ce{O2}}=V_1\cdot\rho_{\text{air},1}\cdot w_{\ce{O2},1}=1.0000\ \mathrm{m^3}\times1.2306\ \mathrm{kg\ m^{-3}}\times0.23160=0.28501\ \mathrm{kg}$$

The density of $\ce{O2}$ at the initial temperature and pressure is

$\rho_{\ce{O2},1}=1.3603\ \mathrm{kg\ m^{-3}}$

Thus, the initial volume of the $\ce{O2}$ alone is

$$V_{\ce{O2},1}=\frac{m_{\ce{O2}}}{\rho_{\ce{O2},1}}=\frac{0.28501\ \mathrm{kg}}{1.3603\ \mathrm{kg\ m^{-3}}}=0.20952\ \mathrm{m^3}$$

Therefore, the initial volume fraction of $\ce{O2}$ is

$$\varphi_{\ce{O2},1}=\frac{V_{\ce{O2},1}}{V_{\text{air},1}}=\frac{0.20952\ \mathrm{m^3}}{1.0000\ \mathrm{m^3}}=0.20952=20.952\ \%$$

If we now heat the air to $30\ \mathrm{^\circ C}$ at a constant pressure, the volume expands, so that the new parameter values are as follows.

$T_2=30.000\ \mathrm{^\circ C}$
$p_2=1.0000\ \mathrm{bar}$
$V_{\text{air},2}=1.0708\ \mathrm{m^3}$
$\rho_{\text{air},2}=1.1492\ \mathrm{kg\ m^{-3}}$

The density of $\ce{O2}$ at the new temperature and pressure is

$\rho_{\ce{O2},2}=1.2703\ \mathrm{kg\ m^{-3}}$

Thus, the new volume of the $\ce{O2}$ alone is

$$V_{\ce{O2},2}=\frac{m_{\ce{O2}}}{\rho_{\ce{O2},2}}=\frac{0.28501\ \mathrm{kg}}{1.2703\ \mathrm{kg\ m^{-3}}}=0.22436\ \mathrm{m^3}$$

Therefore, the new volume fraction of $\ce{O2}$ is

$$\varphi_{\ce{O2},2}=\frac{V_{\ce{O2},2}}{V_{\text{air},2}}=\frac{0.22436\ \mathrm{m^3}}{1.0708\ \mathrm{m^3}}=0.20953=20.953\ \%$$

We can see that, although the volume increased by about $7\ \%$, the volume fraction of $\ce{O2}$ did not significantly change.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.