Chemical Equilibrium - Le Chatelier's Principle, Change in Volume Question: Consider the following reaction at equilibrium at a total pressure that we will call $P_1$. 
  
  $$\ce{2 SO2(g) + O2(g) &lt;=&gt; 2 SO3(g)}$$
  
  Suppose the volume of the system is compressed to $\frac 12$ its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be: 
  
  a. Twice $P_1$
  
  b. Three times $P_1$
  
  c. 3.5 times $P_1$
  
  d. less than twice $P_1$
  
  e. unchanged


My thought process: Since I knew the volume had decreased, that meant the pressure had increased and so the reaction will head towards the side with less moles, which would be the products. I thought that since the volume halved, that meant the pressure should increase two times.

However, the answer is D. Could someone please explain this to me?
Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.

The partial pressure would change as follows:
$$\begin{array}{c|c|c|c}
&amp; \ce{SO2} &amp; \ce{O2} &amp; \ce{SO3} \\
\hline
\text{Initial state} &amp; P_A &amp; P_B &amp; P_C \\
\text{On halving volume} &amp; 2P_A &amp; 2P_B &amp; 2P_c\\
\text{Final state} &amp; 2P_A-2x &amp; 2P_B-x &amp; 2P_c + 2x\\
\end{array}$$

where $P_A + P_B + P_C = P_1$

Since initial and final states are in equilibrium,
$$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$

We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.

Adding the partial pressures at the final state, we have
$$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x &lt; 2P_1$$

Which corresponds to option (d)