Suppose there is no air trapped in the syringe. When you put small amount of weight on the pan, the piston stays in its position. If you pull the piston down and release, it will be pushed up because the force by the atmosphere is more than the weight you put. But if the weight is equal to the force by the atmosphere, and you pull the piston little bit and release, the piston will stay wherever it is released. By measuring the inner area of cross section of the syringe, you can get the atmospheric pressure by the formula Atmospheric Pressure = Weight/Area.
You need a syringe with one end closed and very little air trapped in it, an identical extra syringe, support system for keeping the syringe fixed in vertical position, a pan suspended from the lower end of the syringe barrel, known weights.
Find the inner cross sectional area of the syringe. The weight required is definitely more than 1/2 kg-wt. So put 1/2 kg weight on the pan. Because of little air trapped, the piston might come down little bit. From here pull the barrel little bit and release. If it goes up, you need to put more weight. Use smaller weights to find the position where the piston stays wherever it is left. Try to get this situation correctly.
You might find that even if you make small variation in weight the piston behaves similarly and stays wherever it is left. This is due to friction. Find the range of weights for which this situation occurs. Use the mean weight and calculate the atmospheric pressure.
Questions: Due to the trapped air in the syringe do you expect the measured atmospheric pressure to be more or less than the actual value? Can you estimate this error in percentage.