## Number of Molecules of Oxygen in 1 cm^{3} of Air

### Task number: 2171

Determine the number of oxygen molecules in 1 cm^{3} of air at normal air pressure and room temperature.

#### Notation

*V*= 1 cm^{3}= 10^{−6}m^{3}volume of the air *p*= 101 325 Panormal air pressure *T*= 20 °C = 293 Kroom air temperature *N*(O_{2}) = ?number of oxygen molecules in 1 cm ^{3}From the Tables:

*k*= 1.38·10^{−23}J·K^{−1}Boltzmann constant *R*= 8.314 J·K^{−1}·mol^{−1}molar gas constant *N*_{A}= 6.022·10^{23}mol^{−1}Avogadro constant #### Analysis

We can determine the number of oxygen molecules in the air in two ways. In both of them we assume that under the given conditions we can consider the air to be ideal gas and use the equation of state.

In the first case we will use the form of the equation of state which uses the number of molecules. The gas in which we are interested in this problem, oxygen, takes up 21 % of the given volume of the air. So, we obtain the number of oxygen molecules by multiplying the determined number of molecules in 1 cm

^{3}of air by the ratio 21/100.In the second way of solving we will start from the equation of state again. This time we will use the form containing the amount of substance and calculate the volume taken up by 1 mole of the air under the given conditions. Then, we will use the Avogadro constant, which states the number of particles in 1 mole of substance. We will determine the number of particles in 1 l of air and therefrom, in 1 cm

^{3}of the air. After that we will determine the respective part of oxygen molecules.#### Hint 1: Equation of State for the Number of Molecules

Write the equation of state for ideal gas. We need it in such a form in which the number of molecules appears.

#### Hint 2: Equation of State for the Amount of Substance

Write the equation of state in the form in which the amount of substance appears. Using this equation, determine the volume taken by 1 mole of air under the given conditions. Note, how many particles are contained in 1 mole of substance, thus, in 1 mole of air also.

#### Solution based on the equation of state for the number of molecules

The equation of state of an ideal gas can be expressed using the number of molecules as follows

\[pV = NkT.\tag{1}\]We express the number of molecules

\[N = \frac{pV}{kT}.\tag{2}\]*N*from equation (1)Values of

\[N(O_{2}) = \frac{21}{100}\frac{pV}{kT}.\tag{3}\]*N*represents the number of all molecules of the gas molecules in the given volume. Air is a mixture of different gases including oxygen, which constitutes 21 % of the air volume. Thus, we can obtain the number of oxygen molecules by multiplying relation (2) by the respective percentage of oxygen.Then we substitute for the known values into relation (3) and calculate the number of oxygen molecules in 1 cm

\[N(O_{2}) = \frac{21}{100}\frac{101325·10^{−6}}{1.38·10^{−23}·293}\] \[N(O_{2}) \dot{=}\, 5.3·10^{18}.\]^{3}of air#### Solution based on the equation of state for the amount of substance

The equation of state of an ideal gas can be expressed through the amount of substance as follows

\[pV = nRT.\tag{4}\]We express the volume

\[V = \frac{nRT}{p}.\]*V*from equation (4)Let’s assume that the amount of substance is 1 mole and then calculate the corresponding volume of air under the given conditions.

\[V = \frac{1·8.314·293}{101325} \mathrm{m^{3}} \dot{=}\, 0.024 \mathrm{m^{3}} \dot{=}\, 24 \mathrm{l}\]The Avogadro constant

*N*_{A}indicates the number of particles in 1 mole of substance. Its value can be found in the tables.Then it holds true that 24 l of air contain

*N*_{A}particles.Number of particles in 1 l of air can be calculated this way:

\[N_{1 l} = \frac{N_{\mathrm{A}}}{24}.\tag{5}\]Since we are looking for the number of particles in 1 cm

\[N_{1 cm^{3}} = \frac{N_{\mathrm{A}}}{24·10^{3}}.\tag{6}\]^{3}, we can obtain it by dividing equation (5) by 10^{3}(because of the conversion from dm^{3}to cm^{3})Oxygen accounts for approximately 21 % of air. Thus, the number of oxygen molecules in 1 cm

\[N(\mathrm{O_{2}}) = \frac{21}{100}\frac{N_{\mathrm{A}}}{24·10^{3}}.\tag{7}\]^{3}of air is obtained by multiplying relation (6) by 21/100Then we substitute for the value of the Avogadro constant into relation (7) and calculate the number of molecules of oxygen in 1 cm

\[N(\mathrm{O_{2}}) = \frac{21}{100}\frac{6.022·10^{23}}{24·10^{3}}\] \[N(\mathrm{O_{2}}) \dot{=}\, 5.3·10^{18}.\]^{3}of air#### Answer

In 1 cm

^{3}of air at the normal pressure and room temperature, there is approximately 5.3·10^{18}oxygen molecules.