# Vapour pressure and Raoult's Law

The following question from my textbook assignment reads

A mixture contains $$1$$ mole volatile liquid $$A$$ ($$P_\mathrm A^\circ = \pu{100 mmHg}$$) and $$3$$ moles of volatile liquid $$B$$ ($$P_\mathrm B^\circ = \pu{80 mmHg}$$). If the solution behaves ideally, the total vapour pressure of the distillate is?

Using Raoult's Law: \begin{align} P_\text{total} &= P_\mathrm A^\circ \times \chi_\mathrm A + P_\mathrm B^\circ \times \chi_\mathrm B\\ P_\text{total} &= \pu{100 mmHg} \times 0.25 + \pu{80 mmHg} \times 0.75\\ P_\text{total} &= \pu{85 mmHg} \end{align}

But the answer is $$\pu{85.88 mmHg}$$.

I tried searching online for an explanatory solution but all of them seem to use the mole fraction of solute in vapour phase without any explanation. I figured it was due to the word distillate. Can someone explain why do we need to use mole fraction of the vapour phase and not of the solute in solution?

• The distillate is the condensed form of the vapor. – Safdar Aug 3 '20 at 9:10
• @Safdar You mean take out all the vapours of the solution separately and then condense them into a solution? – Rew Aug 3 '20 at 9:11
• Exactly.. Then attempt to use Dalton's law of partial pressures to find the mole fractions in distillate/ – Safdar Aug 3 '20 at 9:12
• @Safdar hey doesn't that mean the (condensed) distillate will form its own vapours? – Rew Aug 3 '20 at 9:14
• It would but they have asked for the distillate of the given mixture and not the distillate of the distillate ... – Safdar Aug 3 '20 at 9:15

During the process of distillation, we condense the vapours (B) of the mixture (A) to give a mixture (C) that has the same mole fractions as that of the vapour (B) of mixture (A).

In order to solve this question there would be three steps.

• Firstly, finding the vapour pressure of the initial mixture.
• Second, we find the mole fraction of components in the vapour phase of this mixture.
• The third and final part would be to find the vapour pressure of the distillate leading to our answer.

However, finding the actual composition of the vapours here, would be the biggest challenge in this question.

### Step $$1$$

In order to find the vapour pressure, we use Raoult's Law which mathematically states:

$$P_T = \sum{P^\circ_i\chi_i}$$

Since this is a two-component system, the total pressure would be expressed as a sum of two terms for each component A and B.

$$P_T = P^\circ_\mathrm a \chi_\mathrm a +P^\circ_\mathrm b \chi_b$$

Substituting the values given in the question, we get $$P_T= \pu{85 mmHg}$$

### Step $$2$$

The next step is to find the composition in the vapour phase. For this, we use the property that the partial pressure of a component in the vapour phase is equal to the contribution of the same component in the liquid phase. Mathematically, it is equivalent to stating:

$$P_Ty_\mathrm a= P^\circ_\mathrm a \chi_a$$

Here, $$y_a$$ is the mole fraction of component A in vapour phase. Solving for $$y_\mathrm a$$, we get that $$y_\mathrm a = \frac{5}{17}$$, which implies $$y_\mathrm b = \frac{12}{17}$$.

### Step $$3$$

The final is exactly the same as the first step, wherein you use Raoult's law to find the final vapour pressure of the distillate (mole fraction of components equal to that mole fraction of vapour in initial mixture).

Solving for the vapour pressure, you get $$P_T^{'} = \pu{85.88 mmHg}$$, which is the answer provided.