Why does decreasing the pressure of the system increase the relative volatility of a binary solution?

Question

In our teaching lab, we were posed with the following question as an exersize:

If the boiling points of two compounds differ by $\pu{50 ^\circ C}$ at atmospheric pressure, what will be the effect on the relationship by decreasing the pressure to $\pu{1 mm}$ (no number, just a general process)?

I did not find much information on the subject in our lab manual or book but, but based off of some reading I've done around the internet (see below), it seems that for most binary solutions, decreasing the pressure increases the relative volatility. However, asking some of my peers, they seems to believe that while both boiling points would decrease, the difference in their new boiling points would remain about the same (i.e. $\pu{\approx 50 ^\circ C}$), To make matters slightly more confusing, using a pressure-temperature nomograph, picking arbitrary numbers, I found that the relative volatility decreased (see Figure 1).

1. Who is more correct? Does decreasing the system pressure increase the relative volatility or does it stay the same?

2. If the relative volatility increases when system pressure is decreased, why is this the case? I would actually guess the opposite, because as you decrease the pressure towards $0$ (a vacuum), it would seem that all boiling points would approach $\pu{0 K}$, thus be getting closer together.

3. If the relative volatility increases when system pressure is decreased, does that mean it is always more efficient to perform fractional distillation under a partial vacuum?

Figure 1 Two nomographs measured via Pressure-Temperature Nomograph Interactive Tool by Sigma-Aldrich decreasing the system pressure to $\pu{1 mm}$ for two arbitrary substances with $\pu{50 ^\circ C}$ boiling point disparity at $\pu{1 atm}$.

References:

• Effect of temperature or pressure on relative volatility chemengineering.wikispaces.com via the Internet Archive (under "Effect of temperature or pressure on relative volatility")
• Industrial-scale applications of vacuum distillation at Wikipedia

Answer 1

Let's assume that the vapor pressure of both components can be modeled with the Antoine equation. This gives us a convenient way to address the question algebraically. The vapor pressure of most chemicals is excellently modeled by the correct Antoine equation for that chemical. The equation is: $$\log_{10}{p} = A-\frac{B}{C+T}$$

where the the parameters $A$, $B$, and $C$ are compound specific parameters. The equation can be solved for temperature explicitly yielding $T = \frac{B}{A-\log_{10}\, p} - C$.

So at $\pu{1 atm}$, the equation for pure species $i$ and also a different pure species $j$ that is higher boiling by $\pu{50 ^\circ C}$ is:

\begin{align} T_1 &= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\ T_1 &= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C} \end{align}

where $T_1$ is the boiling point of $i$ at $\pu{1 atm}$. To make the math easier, let's suppose the equation is in units of atmospheres, in which case $\log_{10}\, p_{\pu{1 atm}} = 0$.

\begin{align} T_1 &= \frac{B_i}{A_i} - C_i\\ T_1 &= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C} \end{align}

The question is, what happens when $p$ changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.

\begin{align} T_{0,i} &= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\ T_{0,j} &= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j \end{align}

Now we have four equations but six unknown parameters, $A_i$, $A_j$, $B_i$, $B_j$, $C_i$, and $C_j$. So in general, there is no generally valid constraint for the temperature difference $T_{0,i}-T_{0,j}$: It could be higher, lower, or about the same as the difference at room temperature.

However, we could make some further assumptions. A good one here could be that both species in question have a constant heat of vaporization, which means that $C_i=C_j=0$. If that assumption is true, then the four equations become:

\begin{align} T_1 &= \frac{B_i}{A_i}\\ T_1 &= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\ T_{0,i} &= \frac{B_i}{A_i+3} =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}} =\frac{T_1}{1+\frac{3}{A_i}}\\ T_{0,j} &= \frac{B_j}{A_j+3} =\frac{T_1+50 °C}{1+\frac{3}{A_j}} \end{align}

Now, the temperature difference of interest $T_{0,j}-T_{0,i}$ is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$

To keep the formulas easy let me define new parameters $a_i$ and $a_j$ such that $a_x=1+\frac{3}{A_x}$. Then the temperature difference is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$

Now we must introduce further assumptions about the Antoine $A$ parameters. They are positive and thus the $a$ parameters must also be positive, and further positive $A$ implies $a>1$. If we further assume $a_j=a_i$, then

$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$

Since we know already that $a_j>1$, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is less than the temperature difference of $\pu{50 ^\circ C}$ at \pu{1 atm}. Now, even if $a_i$ is not precisely equal to $a_j$, then the temperature difference will still be less than $\pu{50 ^\circ C}$ as long as the $a_i$ and $a_j$ parameters are similar enough to keep $(a_i-a_j)T_1 << a_i(\pu{50 ^\circ C})$. It would take a very unusual choice of chemicals $i$ and $j$ for that to happen. I don't think I could identify such a pair.

So, in summary, to get to the result that I think your instructor was asking for, we required many assumptions:

• Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption).
• Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine $C$ parameters.
• The Antoine $A$ parameters for both chemicals must be positive. (A good assumption but it still must be made.)
• The chemicals have Antoine $A$ parameters (and thus also $a$ parameters as I defined them here) that are not too different from each other.

It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than $\pu{50 ^\circ C}$, but I'm confident that the result is not necessarily generally true. It isn't a thermodynamic law or anything. It definitely depends on several key assumptions about how vapor pressure "usually" or "typically" behaves for most chemicals.

Answer 2

My first answer focused on the boiling point difference. But that's not what's practically important when separating compounds. The concentration ratio in the vapor phase of the two compounds to be separated is what is important.

Let's define some variables. The total overall pressure is $P$ and the temperature is $T$. The partial pressures of components $i$ and $j$ are $p_i$ and $p_j$. The mole fractions of $i$ and $j$ in the liquid phase are $x_i$ and $x_j$ and in the vapor phase are $y_i$ and $y_j$. Lastly, the vapor pressure of pure $i$ is $p_i^*(T)$ and of pure $j$ is $p_j^*(T)$. Those vapor pressures are functions of temperature, we could choose the Antoine equation for example. I won't write the $(T)$ but keep in mind vapor pressures of pure species are functions of temperature (only).

For simplicity let's focus on ideal solutions, which mean's the simplest form of Raoult's law applies:

\begin{align} \tag1 p_i&=p_i^*x_i\\ \tag2 p_j&=p_j^*x_j \end{align}

The sum of mole fractions must be $1$ in both vapor and liquid:

\begin{align} \tag3 1&=x_i+x_j\\ \tag4 1&=y_i+y_j \end{align}

There is also the definition of partial pressure:

\begin{align} \tag5 p_i &= y_i P\\ \tag6 p_j &= y_j P \end{align}

And there are vapor pressure relationships, e.g. Antoine equations for each of $i$ and $j$:

\begin{align} \tag7 p_i^* &= p_i^*(T) = \text{some equation with parameters and T only}\\ \tag8 p_j^* &= p_j^*(T) = \text{some equation with diff. parameters and T only} \end{align}

That is eight independent equations but there are ten variables. So there are two degrees of freedom. Let's assume that we have an equimolar liquid, i.e. that $x_i=0.5$. From equation 3 that means $x_j=0.5$ too. Now there is only one degree of freedom left. The general question is what pressure $P$ maximizes the relative volatility $\frac{y_i}{y_j}$. The math is tough because you have to substitute one Antoine equation into the other to eliminate temperature. I assumed $C_i=C_j=0$ again for Antoine parameters, and got that $p_j^*=K (p_i^*)^{\frac{B_j}{B_i}}$, where K is a proportionality constant that depends in a complex way on Antoine $A_i$, $A_j$, $B_i$, and $B_j$ parameters. The other non-Antoine equations reduce to:

\begin{align} y_i P &= p_i^*/2\\ (1-y_i)P &= p_j^*/2 \end{align}

Subbing the Antoine result in for $p_j^*$ and then subbing the first equation into the result gives

$$2^{d-1}P^{d-1} = \frac{K y_i^d}{1-y_i} $$

That expression doesn't have an analytical solution for $y_i$ as a function of $P$ but in theory you could find the solution numerically if you knew all the Antoine parameters. I graphed the equation for a few different values of $d$ and y_i was always increasing as P got lower and lower.

So that's a long-winded way of saying that "relative volatility" is different than the difference in boiling point temperatures.