The reaction for the formation of sulfur trioxide from sulfur dioxide and oxygen is:

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$

According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas. In this case, the product, sulfur trioxide, has two moles of gas, while the reactants, sulfur dioxide and oxygen, have three moles of gas. Therefore, increasing the pressure will favor the formation of sulfur trioxide, resulting in a higher yield.

Conversely, decreasing the pressure will shift the equilibrium towards the side with more moles of gas, which is the reactants' side. This will decrease the yield of sulfur trioxide.

Therefore, increasing the pressure increases the yield of sulfur trioxide, while decreasing the pressure decreases the yield.