0
$\begingroup$

I have been studying about liquid solutions and came across these graphs.

enter image description here enter image description here

From the first graph it is evident that the solution considered has been assumed ideal as there is no deviation from ideal behavior and the graph is straight. However, I wonder what causes the curvature in the second graph. Both Raoult's law and Dalton's law describe the proportionality between the mole fraction and the vapour pressure. So, why does the graph deviate from its proportionality in graph 2 despite the ideal nature of the vapours?

$\endgroup$

1 Answer 1

3
$\begingroup$

So, why does the graph deviate from its proportionality in graph 2 despite the ideal nature of the vapours?

It does not deviate, both curves are the result of the ideal behavior. Lets do some math.

For an ideal binary mixture where: (1) the gas phase behaves as an ideal gas, and (2) the liquid phase behaves as an ideal solution; the LV equilibrium reduces to Raoult's law $$ y_j P = x_j P_j^\pu{sat} \tag{1} $$ where $P_j^\pu{sat}$ is the saturation pressure of chemical species $j$ at a temperature $T$.

Bubble Curve Summing twice Eq. (1) for both components yields \begin{align} &y_1 P + y_2 P = x_1 P_1^\pu{sat} + x_2 P_2^\pu{sat} \\ &(y_1 + y_2) P = x_1 P_1^\pu{sat} + (1 - x_1) P_2^\pu{sat} \\ &\boxed{P(x_1, T) = [P_1^\pu{sat}(T) - P_2^\pu{sat}(T)]x_1 + P_2^\pu{sat}(T)} \tag{2} \end{align} Thus, at a fixed temperature, the bubble curve is a linear function of the liquid composition $x_1$. This is the scenario in the left-hand side.

Dew Curve First we put Eq. (1) in a more suitable form to eliminate $x_j$ $$ x_j = \frac{y_jP}{P_j^\pu{sat}} \tag{3} $$ Summing twice Eq. (3) for both components yields \begin{align} &x_1 + x_2 = \frac{y_1P}{P_1^\pu{sat}} + \frac{y_2P}{P_2^\pu{sat}} \\ &1 = P\left(\frac{y_1}{P_1^\pu{sat}} + \frac{1 - y_1}{P_2^\pu{sat}}\right) \\ &1 = P\left(\frac{y_1P_2^\pu{sat} + (1 - y_1)P_1^\pu{sat}} {P_1^\pu{sat}P_2^\pu{sat}}\right) \\ &\boxed{P(y_1,T) = \frac{P_1^\pu{sat}(T)P_2^\pu{sat}(T)}{y_1[P_2^\pu{sat}(T) - P_1^\pu{sat}(T)] + P_1^\pu{sat}(T)}} \tag{4} \end{align} Thus, at a fixed temperature, Eq. (3) has a mathematical equation in the form of $a/(bx + c)$. This is the scenario in the right-hand side.

Therefore in LV equilibrium, as indicated by Raoult's law, the two-phase region is confined between an upper straight line and a lower hyperbolic curve for the $Pxy$ diagram.

$\endgroup$
1
  • $\begingroup$ Thank You for your explanation. It is very clear to me. I have one more question, it might sound silly, but why do we converge the two graphs to show that there exists equilibrium between the straight line and the hyperbolic curve? How is it possible to converge both graphs if the x-axis represents completely different things? In one graph it represents the mole fraction component B in the solution and in the other it represents mole fraction of it in the vapor phase. xB and yB aren't equal. $\endgroup$ Apr 27, 2023 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.