# Cases when Raoult's law is unsolvable

### Setup

Consider a closed binary mixture of known total molar composition $$n_i$$, held at volume $$V$$ and temperature $$T$$. The equilibrium phase composition is determined by $$\mu_i^{vap} = \mu_i^{liq}$$. Assume the vapor is an ideal gas mixture, the liquid is an ideal mixture, Poynting's correction is negligible, and $$V_{liq} \ll V$$. Then Raoult's law holds:

$$y_i P = x_i P_i^{sat}(T)\quad \tag{1}\label{1}$$

### Simplify

For convenience, denote vapor quantity = $$v_i$$, liquid quantity = $$l_i$$, total quantity = $$n_i=v_i+l_i$$, and "saturation quantity" = $$s_i=\frac{P_i^{sat}(T)\,V}{R T}$$. A closed system means $$n_i$$ are constant, and $$s_i$$ are constant at constant $$T$$. For simplicity, choose units such that $$R T = V = \sum_i n_i=1$$, so that Raoult's law becomes

$$n_i - l_i = s_i \frac {l_i}{\sum_i l_i} \tag{2}\label{2}$$

where $$l_i$$ are the only independent variables. (The solution/result is independent of these choices. It holds for any Raoult's law system at constant $$N, V, T$$. It generalizes to infinite systems with fixed $$\sum n_i/V$$.)

### Question

Eq. ($$\ref{2}$$) can be algebraically solved. But it has no solution when $$s_1>1$$ and $$s_1 - n_1 ≥ n_2 \frac{s_1}{s_2}$$. For example, if $$s_1=2$$ and $$s_2=1/2$$, then no solution exists when $$n_2≤1/3$$ ($$x_2=y_1$$ and $$y_2=x_1$$ at $$\lim n_2\to \frac{1}{3}^+$$).

This seems to predict that any generic species can form an uncondensable mixture if one is uncondensable (as a pure component), and there are few enough of the condensable species. Is this correct?

• Hi there, Alex, and welcome to the Chemistry StackExchange. It seems that you are trying to solve Raoult's Law in a manner that I have never seen before, and with quite a few non-standard variables. Might I suggest that you look into using the Antoine Equation to solve for the Saturation Pressure, of each species, as a function of Temperature. With this you can solve the Raoult's law equation as a function of Temperature, Pressure and either the gas or liquid fraction of species i. As for the prediction, I would not draw any conclusion about species based off of so many assumptions. – Taylor Scott Oct 19 '19 at 21:09
• Thanks. My only assumptions are ideality and constant T, V, and N. That's a vessel with no reactions. Everything else is just a choice of units to simplify Eq.1 into Eq.2. – alexchandel Oct 19 '19 at 21:38
• How do you justify this inequality: $s_1 - n_1 ≥ n_2 \frac{s_1}{s_2}$? What does it mean in plain words? – Buck Thorn Oct 19 '19 at 22:01
• Perhaps towards this point it would be better to define your non-standard variables of $l_i$ and $v_i$ in more explicit terms. As you have it written, the only sensible unit for these would be in terms of moles, and they could be re-written as $l_i = n_{liq}x_i = n_{liq,i}$, and $v_i = n_{vap}y_i = n_{vap,i}$, where $\sum_i x_i = 1$, $\sum_i y_i = 1$, and $n_{total} = n_{vap} + n_{liq}$. Using this method, you can keep track of both the phase and the component of each variable – Taylor Scott Oct 20 '19 at 13:53
• I think that a large amount of my confusion about your approach is in how and why you are defining a "saturation quantity". As per your definition from the ideal gas law, and Rault's law, you can find that $s_i=\frac{n_{vap}y_i}{x_i}=\frac{n_{vap,i}}{x_i}$. This has no intuitive meaning to me, and as I have never seen such an approach, I am unsure as to why you are using it. – Taylor Scott Oct 20 '19 at 14:09

As you have not provided a source for your definition of $$s_i$$, I shall redefine it in terms of Raoult's Law and the Ideal Gas Law. To do this, we shall first take Raoult's Law and solve for $$P$$.

\begin{align}y_iP&=x_iP_i^{sat}(T)\\P&=\frac{x_iP_i^{sat}(T)}{y_i}\end{align}

we shall then take this form, and plug it into the ideal gas law, considering only the gas phase of the equilibrium system. Note that because the system is at equilibrium, $$P_{bulk}=P_{liq}=P_{vap}$$ and $$T_{bulk}=T_{liq}=T_{vap}$$. If the volume of the liquid phase is assumed to be significantly lower than that of the vapor, then $$V_{bluk}=V_{vap}$$. After plugging in this, we will then work towards the form of $$s_i$$ as shown.

\begin{align}P_{vap}V_{vap}&=n_{vap}RT_{vap}\\PV&=n_{vap}RT\\n_{vap}&=\frac{PV}{RT}\\n_{vap}&=\frac{x_iP_i^{sat}(T)}{y_i}\frac{V}{RT}\\\frac{n_{vap}y_i}{x_i}&=\frac{P_i^{sat}(T)V}{RT}\end{align}

From this, and the definition of $$s_i$$ as you stated, we can show that

$$s_i=\frac{P_i^{sat}(T)V}{RT}=\frac{n_{vap}y_i}{x_i}$$

which is subject only to the reasonable assumptions of ideality as stated above, and the bounds per the definition of each term. Namely, $$\left\{P_i^{sat}(T),V,T\right\}$$ must be positive, real numbers and that $$\left\{x_i,y_i\right\}$$ must be positive, real numbers with bounds of $$\left\lbrack0,1\right\rbrack$$.

No further inequalities are required to describe the system.

To solve such a system, you would then set up a system of equations and numerically backsolve for an answer, in terms $$n_{vap},x_{i},y_{i}$$. For example you might feed a set of equations the seed values of $$n_{vap}=1$$ and $$x_i=.5=y_i$$ such as seen below, then optimize your values to minimize the error between the calculated and target values. The above seed values could then optimize to a particular solution that looks like. Note that this does not represent the whole of the solution space. If you wish to further control where your particular solution ends up, you can restrict the solution space by hold one of the values to be a known constant. Below is one such example as described per your comments. In this case, it is not possible to find a reasonable solution, as $$n_{vap}=0.25~\text{mol}$$ lies outside the solution space for this system of equations. Without going too heavily into the derivation, it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution. This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at in order to reach equilibrium than it says the system could not exist, though yours is not an unreasonable reading of the result.

A good rule of thumb for this would be that you can expect to find a reasonable solution for $$\left\{x_i,y_i\right\}$$ for any system in which $$s_1 \leq n_{vap} \leq s_2$$ holds, however this is by far the most complicated way to solve out this system of equations. A simpler approach to this would be to hold either $$x_i$$ or $$y_i$$ constant to restrict the solution space, as any constant value on the bounds of $$\left\lbrack0,1\right\rbrack$$, as already assumed will garuntee a unique solution to the system of equations.

• My question isn't how to solve Raoult's law; for < 5 components, at constant $V, T$, & $V_{liq} \ll V$, Raoult's law has (if any exists) an exact algebraic solution for a given $N_i$, which numerical methods approximate. I'm talking about equilibrium behavior when Raoult's law is unsolvable. "This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at equilibrium" — then that isn't equilibrium. – alexchandel Oct 23 '19 at 3:26
• "…it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution." — The proof of that is the premise of the question. "…you can expect to find a reasonable solution for $\left\{x_i,y_i\right\}$ for any system in which $s_1\leq n_{vap}\leq s_2$ holds" — Not so: take $s_1=0.5$ mol, $s_2=2$ mol, & begin with $n_1 = 0.25$ mol, $n_2 = 0.75$ mol ($\to n_{vap}≤1$ mol). No solution exists for any $n_{liq}>0$, and limiting $f_i^{liq}>f_i^{vap}$ imply $n_{liq}=0,n_{vap}=1$ mol at equilibrium. So $s_1\leq n_{vap}\leq s_2$ with no solution. – alexchandel Oct 23 '19 at 3:33
• "this is by far the most complicated way to solve out this system of equations." — How else would you determine the equilibrium phase compositions of a given closed mixture at a constant $V, T$? "hold either $x_i$ or $y_i$ constant to restrict the solution space…, as already assumed will garuntee a unique solution to the system of equations." — Fixing $y_1$ and $y_2$ (or $x_1$ and $x_2$) is incompatible with finding the equilibrium of a given closed binary mixture at constant $V, T$, as solution space is already zero-dimensional. – alexchandel Oct 23 '19 at 3:42
• In regards to the second comment, note that $n_{vap} \neq n_{vap,i}$, and is the total number of moles in the gas phase from all $i$ species, and therefore must be the same of both species. If you hold $n_{vap}=0.75$ a solution exists at $x_1=0.167$ and $y_1=0.444$. – Taylor Scott Oct 23 '19 at 15:38
• In regards to the third comment - I was assuming that $n_{vap}$ would be able to be modulated by changing the pressure of the system, but after looking over your most recent revision, I see that this is not the case, which means that $n_{vap}$ would be held constant under the ideal gas law and the definition of the system. – Taylor Scott Oct 23 '19 at 15:59