The Gas Phase

To study the gas phase, we use a simplified model called the ideal gas (perfect gas). You can study more about it in our lesson Ideal Gas Law

Using this equation, we can calculate, for example, the density of a gas, knowing its molar mass, temperature, and pressure:

The relative density of a gas with another in the same conditions of temperature and pressure is the ratio between their densities:

\[\frac{\rho_1}{\rho_2} = \frac{\frac{M_1}{Vm}}{\frac{M_2}{Vm}} = \frac{M_1}{M_2}\]

The composition of a mixture of substances can be expressed using molar fractions. The molar fraction (\(X_i\)) of a component (\(N_i\)) of a mixture \(N_1, N_2, N_3, ... ,N_i, ...\) is defined by the ratio of the number of moles of \(N_i\) (\(n_i\)) and the total number of moles of the mixture (\(n_1 + n_2 + n_3 + ...\)):

We can also calculate it using \(X_i = \frac{pr}{100}\), where pr is the molar percent of the component. For a mixture, the sum of molar fractions is 1.

The partial pressure of a gas from a mixture is, basically, the pressure it would have if it occupied that volume by itself. We calculate it using the relation \(p_i = X_i \times P\) where \(X_i\) is the molar fraction and P is the total pressure of the mixture.

John Dalton's rule: the pressure of a mixture of gases is the sum of their partial pressures.

In a vessel with the volume 80 dm³, there is a mixture of \(\ce{CO2}\), \(\ce{O2}\), and \(\ce{N2}\). The mixture has a temperature of 52.2°C and a pressure of 2 atm. Knowing that in the vessel are \(18.066 \times 10^{23}\) molecules of \(\ce{CO2}\), 64 g of \(\ce{O2}\), and the rest is \(\ce{N2}\), calculate:

the quantity of \(\ce{N2}\) in the vessel
the partial pressures of the gases
the mean molar mass of the mixture

Solution:

First of all, we transform the temperature from Celsius to Kelvin: \(T = 52.2+273=325.2 \: K\)

We use the ideal gas law to find the number of moles of the mixture:

\(pV = nRT => 2 \times 80 = n \times 0.082 \times 325.2\)\(=> n = \frac{160}{26.67} => n = 6 \:mol\)

We have 64 g of \(\ce{O2}\) which we transform into mol: \(n \ce{O2} = \frac{64}{32} = 2 \: mol\).

We also transform the \(\ce{CO2}\) molecules into mol: \(n \ce{N2} = \frac{18.066 \times 10^{23}}{N_A} = 3 \: mol\).

That means that we have \(6-2-3=1 \: mol\) of \(\ce{N2}\).
\(X \ce{O2} = \frac{2}{6} = 0.33\)

\(X \ce{CO2} = \frac{3}{6} = 0.5\)

\(X \ce{N2} = \frac{1}{6} = 0.167\)

\(p \ce{O2} = X \ce{O2} \times p = 0.33 \times 2 = 0.66 \: \text{atm}\)

\(p \ce{N2} = X \ce{N2} \times p = 0.5 \times 2 = 1 \: \text{atm}\)

\(p \ce{CO2} = X \ce{CO2} \times p = 0.167 \times 2 \)\(= 0.33 \: \text{atm}\)
\(M = X \ce{O2} \times M \ce{O2} + X \ce{N2} \times M \ce{N2}\)\( + X \ce{CO2} \times M \ce{CO2} \)\(= 0.33 \times 32 + 0.167 \times 28 \)\(+ 0.5 \times 44 = 37.236 \: \text{g/mol}\)