Why does decreasing the pressure of the system increase the relative volatility of a binary solution?
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 &amp;= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\
T_1 &amp;= \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 &amp;= \frac{B_i}{A_i} - C_i\\
T_1 &amp;= \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} &amp;= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\
T_{0,j} &amp;= \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 &amp;= \frac{B_i}{A_i}\\
T_1 &amp;= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\
T_{0,i} &amp;= \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} &amp;= \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  $a&gt;1$.  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 $a_j&gt;1$, 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 $(a_i-a_j)T_1 &lt;&lt; a_i(\pu{50 ^\circ C})$.  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.