Question

4. Two liquids, A and B, obey Raoul's law at all concentration. That is PA = DAPA and PB = xpPg where r, is the mole fraction of species i in the liquid mixture, P. is the partial vapor pressure of species i in the mixture and P* is the vapor pressure of pure i. (a) (5 points) Express the total vapor pressure as a function of x. (b) (5 points Express the total vapor pressure as a function of ya, the mole fraction of A in the vapor phase. (HINT: Recall Dalton's law.) (c) (10 points) Prove that the difference between the mole fraction of the vapor and that of the liquid (i.e. DA-YA) is maximum when

Answer 1

a)

Total vapour pressure will be sum of two vapour pressures,

$P=P_A+P_B\\ \Rightarrow P=x_AP_A^*+x_BP_B^*\\ \Rightarrow P = x_AP_A^*+(1-x_A)P_B^*\qquad[x_B=1-x_A]\qquad(1)$

We have obtained P in terms of x_A

b)

In vapour Phase, the Mole fraction is given as,

$y_A=\frac{P_A}{P}\\ \Rightarrow y_A=\frac{x_AP_A^*}{P}\\ \Rightarrow y_A= \frac{x_AP_A^*}{x_AP_A^*+(1-x_A)P_B^*}\qquad(2)\\ \Rightarrow \frac{1}{y_A}=\frac{x_AP_A^*+(1-x_A)P_B^*}{x_AP_A^*}\\ \Rightarrow \frac{1}{y_A}= 1+\left ( \frac{P_B^*}{x_AP_A^*}-\frac{P_B^*} {P_A^*} \right )\\\Rightarrow x_A=\frac{P_B^*y_A}{P_A^*+(P_B^*-P_A^*)y_A}\qquad\textrm{[After rearranging terms]}\qquad(3)$

Again putting (3) in the below equation,

$y_A=\frac{x_AP_A^*}{P}$

we will obtain,

$y_A=\frac{P_B^*y_A}{P_A^*+(P_B^*-P_A^*)y_A}\frac{P_A^*}{P}\\ \Rightarrow P=\frac{P_A^*P_B^*}{P_A^*+(P_B^*-P_A^*)y_A}$

Thus, we have obtained P in terms of y_A

(c)

From (2), we have,

$x_A-y_A= x_A-\frac{x_AP_A^*}{x_AP_A^*+(1-x_A)P_B^*}$

For this part, for our convenience, we will use x_A = x, P_A* =a and P_B* = b

Thus, we will have,

$x_A-y_A= x-\frac{xa}{xa+(1-x)b}$

For maximum, we will differentiate the above equation with respect to x and equate it to zero. Thus, we will have,

$\frac{d}{dx}\left (x-\frac{xa}{xa+(1-x)b} \right )=0\\ \Rightarrow 1-\frac{[xa+(1-x)b]\cdot a-ax\cdot[a-b]}{[xa+(1-x)b]^2}=0\\ \Rightarrow [xa+(1-x)b]^2=[xa+(1-x)b]\cdot a-ax\cdot[a-b]\\ \Rightarrow x^2a^2+x^2b^2+b^2-2x^2ab-2b^2x+2abx=xa^2+ab-abx-xa^2+abx\\ \Rightarrow x^2a^2-abx^2+2abx-ab+x^2b^2-x^2ab-2b^2x+b^2=0\\ \Rightarrow a(ax^2-bx^2+2bx-b)+x^2b^2-x^2ab-2b^2x+b^2=0 \\ \Rightarrow a(ax^2-bx^2+2bx-b)-b(ax^2-bx^2+2bx-b)=0\\ \Rightarrow (a-b)(ax^2-bx^2+2bx-b)=0$

Assuming 'a' and 'b' are different, we have,

$(ax^2-bx^2+2bx-b)=0\\ \Rightarrow ax^2=b(x^2-2x+1)\\ \Rightarrow ax^2=b(1-x)^2\\ \Rightarrow \frac{(1-x)^2}{x^2}=\frac{a}{b}$

Now back substituting, x_A = x, P_A* =a and P_B* = b, we have,

$\frac{(1-x_A)^2}{x_A^2}=\frac{P^*_A}{P_B^*}\\ \Rightarrow \frac{x_B}{x_A}=\sqrt{\frac{P^*_A}{P_B^*}}\qquad[\textrm{Taking only positive root}]$

Thus, shown.