Gas Mixture Condensation

A mixture of m_N = 100 g of nitrogen and some oxygen is isothermally compressed at T = 77.4 K. The result of this experiment is plotted as the pressure dependence of the mixture versus volume in arbitrary units (see Figure 1.1). Find the mass of oxygen and the oxygen saturation vapor pressure at this temperature.

Figure 1.1

Hint: T = 77.4 K is the boiling temperature of liquid nitrogen at atmospheric pressure. Oxygen boils at a higher temperature.

SOLUTION

Consider three parts of the plot (see Figure 1.2). At V > V₂ there is a regular gas mixture (no condensation). At V₁ < V < V₂, one of the gases is condensing; let us assume for now it is oxygen (it happens to be true). At V < V₁ they are both condensing, and there is no pressure change. Let us write

(1)

(2)

Here P_N is the partial nitrogen pressure at (V₂, P₂), P_O is the saturation vapor pressure of oxygen, and P_A is the saturated vapor pressure of nitrogen (1 atm) at T = 77.4 K. Between V₂ and V₁, nitrogen is a gas, and since the temperature is constant,

(3)

Using (3) and dividing (1) by (2), we have

(4)

yielding

Figure 1.2

Had we assumed that oxygen is condensing at (V₁, P₁) we would get  This contradicts the fact that oxygen boils at a higher temperature. The saturated vapor pressure at T = 77.4 K should be less than P_A. To find the oxygen mass, we use the ideal gas law at (V₂, P₂)  where the oxygen is just starting to condense (i.e., its pressure is P_O and it is all gas). So

(5)

where μ_O is the oxygen molar mass. For nitrogen a similar equation can be written for (V₁, P₁):

(6)

where μ_N is the molar mass of nitrogen. Dividing (5) by (6), we obtain