Recall Pascal’s principle: An object within a fluid experiences pressure The figure to the right shows a cylinder of liquid of height h and area A. The weight of the cylinder is its mass m times the acceleration due to gravity g. This is the force exerted by the cylinder of liquid on whatever is just below it: F = m g The pressure p is this force divided by the area A of the face of the cylinder. p = F/A The mass of the cylinder is the density of the liquid times the volume V. m = V The volume is the area A of the face of the cylinder times its height h. V = A h

So, the pressure p is: p   = F / A   = ρmg / A   = ρV g / A   = ρAh g / A   = ρhg Thus, if , h and g are measured in SI units, the pressure p will be in pascals. Note that the value is independent of the area of the cylinder.

The three basic states of matter are Solids Liquids and Gases (for your personal knowledge know there are also Bose-Einstein Condensates Plasmas

The U-Tube In the figure to the right we show such a U shaped tube filled with a liquid. Note that both ends of the tube are open to the atmosphere. Thus both points A and B are at atmospheric pressure. The two points also have the same vertical height.

Now the top of the tube on the left has been closed Now the top of the tube on the left has been closed. We imagine that there is a sample of gas in the closed end of the tube. The right side of the tube remains open to the atmosphere. The point A, then, is at atmospheric pressure. The point C is at the pressure of the gas in the closed end of the tube. The point B has a pressure greater than atmospheric pressure due to the weight of the column of liquid of height h. The point C is at the same height as B, so it has the same pressure as B. And we have already seen that this is equal to the pressure of the gas in the closed end of the tube. Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h.

The point A is at atmospheric pressure. The point C is at whatever pressure the gas in the closed end of the tube has, or if the closed end contains a vacuum the pressure is zero. Since the point B is at the same height as point A, it must be at atmospheric pressure too. But the pressure at B is also the sum of the pressure at C plus the pressure exerted by the weight of the column of liquid of height h in the tube. We conclude that pressure at C, then, is less than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h. If the closed end of the tube contains a vacuum, then the pressure at point C is zero, and atmospheric pressure is equal to the pressure exerted by the weight of the column of liquid of height h. In this case, the manometer can be used as a barometer to measure atmospheric pressure.

For example, atmospheric pressure varies with the weather and is usually about 100 kilopascals (KPa). Another common unit for measuring atmospheric pressure is mm of mercury, whose value is usually about 760 mm. Put another way, if the closed end of the tube in Case 3 above contains a vacuum, the height h is about 760 mm. 1atmosphere = 760mmHg = 101.325KPa