2. Torricelli Barometer The mercury in the tube pushes down with its weight. The bottom of the tube is open to the atmosphere. The air pushes on the open surface of the mercury. On an average day, the pressure of the air equals the pressure exerted by a column of mercury 760 mm high. Above 760 mm, there is a vacuum in the tube. Weight of mercury

3. Barometer Detail Why doesn’t the diameter of the column of Hg make a difference? Recall that Pressure = force/area. The “force” is the weight of the mercury, but the pressure that results is that weight divided by the area of the column. So … a bigger column weighs more but also has a proportionally bigger area, and the two factors cancel one another out. The pressure caused by the column of mercury pressing down is independent of the diameter of the column.

4. Manometer A manometer is comprised of a bulb containing a gas and a U-shaped tube. The U-shaped tube is partially filled with mercury. The weight of the mercury puts pressure on the gas. If the U-tube is OPEN there is also air pressure acting on the gas. The gas molecules put pressure on the mercury. P Hg

5. Closed Manometers There is a balance between the weight of the mercury on the left (P Hg ) and the pressure of the gas on the right (P gas ). The difference between the heights of the mercury on each side of the tube is a measure of the pressure of the gas. P gas =  h vacuum P Hg

6. Open Manometers When gas pressure is greater than atmospheric pressure, the mercury is pushed toward the open end. The balance is between the gas on the right, and the air plus mercury on the left. P air + P Hg = P gas The weight of the mercury is measured as the height difference: P Hg =  h So P gas = P air +  h P air P Hg

7. Open Manometers When gas pressure is less than atmospheric pressure, the mercury is pushed toward the gas reservoir. The balance is between the air on the left and the gas plus mercury on the right: P air = P gas + P Hg The weight of the mercury is measured as the height difference: P Hg =  h So P air = P gas +  h Or P gas = P air -  h P Hg P air

8. Sample Problems P AIR = 765 mm  h = 27 mm  h = 13 mm P AIR = 753 mm  h = 20 mm P air = 790 mm Find the pressure of the gas in each manometer. Pay attention to whether the manometer is open or closed! 1.2.3.

9. Sample Problem Answers 1. P gas +  h = P air P gas = 790 mm - 20 mm = 770 mm Hg 2. P gas = vacuum + 13 mm = 13 mm Hg 3. P gas = P air +  h P gas = 765 + 27 = 792 mm Hg