For discussion purposes, we will assume the cylinder is 4 inches in diameter, the rod is 2 inches in diameter, and all math figured here is approximate. This means the delta cross-sectional area would be near 10 square inches. At the first de-stroke of the piston, the part of the cylinder above the piston would be partially evacuated. If the vacuum attained was 20 inches of mercury, the upper surface of the piston would then feel an atmospheric pressure of about 5 psia, or about 10 psia less than if it were filled with air instead of partial vacuum. Ten pounds per square inch times (in this example) about 10 square inches would imply that the next and every subsequent lifting stroke would have about 100 pounds less to lift, depending on where the piston was within the cylinder. Not a great savings, but measurable at any rate. There may be issues with seals that might need to be overcome.