It is the pressure exerted by the vapors’ on the liquid surface when the rate of evaporation and the rate of condensation is same.
$\text{Boiling Point} \alpha \frac {1}{\text{Vapour pressure}}$
$\text{Boiling point} \alpha \text{Atm. Pressure (ext. Pressure)}$
$P_A^o$ = Vapour pressure of pure liquid
$P_A$ = Partial Vapour pressure of component in a solution
Raoult’s Law:
According to Raoult’s law;
Partial vapour pressure of a component in a solution is directly proportional to it’s mole fraction.
$\text{Partial V. P} \alpha \text{Mole fraction}$
$P_A \alpha \chi _A$
$P_A = P_A^0 \chi _A$
Let us take two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually equilibrium would be established between Vapour phase and the liquid phase. Let the total vapour pressure at this stage be p total and $p_1$ and $p_2$ be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions $\chi _1$ and $\chi _2$ of the two components 1 and 2 respectively
Now
$p_1=p_1^0 x_1$
$p_2=p_2^0x_2$
According to Dalton’s law of partial pressures, the total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as:
$p= p_1 + p_2$
$p= p_1^0 \chi _1 + p_2^0 \chi _2$
$p=p_1^0(1-\chi _2) + p_2^0 chi _2$
$p=p_1^0 + \chi _2(p_2^0 - p_1^0)$
The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components.
If $y_1$ and $y_2$ are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures
$p_1 = y_1 p_{total}$
$p_1 = y_1 p_{total}$
Comparison of Raoult's law and Henry's law
It is observed that the partial pressure of volatile component or gas is directly proportional to its mole fraction in solution. In case of Henry’s Law the proportionality constant is $K_H$ and it is different from $p_1^0$ which is partial pressure of pure component. Raoult’s Law becomes a special case of Henry’s Law when $K_H$ becomes equal to $p_1^0$ in Henry’s law.
Solved Example
Example 1
Vapour pressure of chloroform ($CHCl_3$) and dichloromethane($CH_2Cl_2$) at 298 K is 200 mm of Hg and 415 mm of Hg respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of $CHCl_3$ and 40 g of $CH_2Cl_2$ at 298 K
(ii) mole fractions of each component in vapour phase. Solution
(i) for $CHCl_3$
$P_A^O=200$ mm of Hg
$W_A = 25.5$ gm , $M_A = 119.5$ gm/mol
For $CH_2Cl_2$
$P_B^O = 415$ mm of Hg
$W_B = 40$ g,$M_B=85$ g/mol
$n_A = W_A/M_A =25.5/119.5 = 0.215$
$n_B = W_B/M_B =40/85=0.47$
Now,
$\chi _A = n_A/(n_A+ n_B )= 0.213/(0.213+0.47)=0.213/0.683=0.31$
$\chi _B = 1 – chi _A = 1 – 0.31 = 0.69$
$P_A = P_A^O \chi _A = 200 \times 0.31=62.0$ mm of hg
$P_B = P_B^O \chi _B = 415 \times 0.69=286.35$ mm of hg
$P_S = 62 + 286.35=348.35$ mm of Hg
(ii)
Now,
$Y_A = P_A/P_S = 62/348.35=0.179$
$Y_B = 1 – 0.179= 0.821$
Example 2
heptanes($C_7H_{16}$) and octane($C_8H_{18}$) forms ideal soln at 373 k. The vapour pressure of the two liquid components is 105.2 k Pa and 46.8k Pa respectively. Find the vapour pressure of the mix obtain by mixing 26 gm of heptanes & 35 gm of octane
Solution
fot Heptanes
$P_A^O= 105.2$ k Pa
$W_A = 26$ g
$M_A = 100$ g
$n_A = W_A/M_A =26/100=.26$
for octane
$P_B^O= 46.8$ k Pa
$W_B = 35$ g
$M_B =114$ g
$n_B = W_B/M_B =35/114=.31$
Now,
$\chi _A = n_A/(n_A+ n_B )= .456$
$\chi _B = 1 – chi _A = .544$
$P_A = P_A^O \chi _A = 200 \times 0.31=47.97$ kPabr>
$P_B = P_B^O \chi _B = 415 \times 0.69=25.46$ kPa
$P_S = 47.97 + 25.46=73.43$ kPa