 For an example of a situation where we could use the Baugh equation to describe a pressure difference, let's consider a barometer. Old-fashioned barometers used to be made by taking a glass cylinder, dipping it into a pool of mercury, turning it upside down, and then lifting it back up again. What you end up with is a pressure at the top that is pretty close to zero. Because that vial was allowed to fill up with mercury entirely before you turned it over and lifted it up, the pressure at the top is the vapor pressure of mercury which is very, very, very small. The pressure at the bottom, the pressure at state point B here, is the same as atmospheric pressure because B and the surface between the mercury and the atmosphere are at the same height. Therefore, Pb is pretty close to atmospheric pressure. Then you could describe the pressure difference between A and B as being the density of mercury times gravity times the height of the column of mercury. Since the pressure at the top is effectively zero, that means Pb minus Pc is really just atmospheric pressure. So if you know the height of a column of mercury, you can calculate what the atmospheric pressure is. Let's try an example. If you used a mercury barometer to measure atmospheric pressure and that pressure was one atmosphere, how tall would the column of mercury have to be? Well, if that delta P really is simplifying to atmospheric pressure minus zero and that's equal to the density of mercury times gravity times height, then I could say the height of that column is going to be atmospheric pressure divided by the density of mercury times gravity. If atmospheric pressure is one atmosphere and the density of mercury I retrieved from table A19 of my textbook where density is the second column here being 13,529 and if we assume standard gravity, we could calculate the height. It doesn't ask for a specific unit here, but let's go with, oh I don't know, millimeters as our goal. I will write that out in the form of 1000 millimeters is one meter and then I have to break the atmosphere part into its component pieces. For that I will go to the conversion factor sheet on the inside of the front cover of my textbook that one atmosphere is equal to 1.01325 bar. A bar is 10 to the fifth Newtons per square meter and I know a Newton is a kilogram meter per second squared at which point Newton cancels Newtons, bars cancels bars, atmosphere cancels atmospheres, kilograms cancels kilograms, second squared cancels second squared square meters and meters and meters cancels cubic meters and meters leaving me with an answer of millimeters. So, with a little help from my calculator, I can say 1 times 1000 times 1.01325 times 10 to the fifth divided by 13,529 times 9.81 will give me the height of the column of mercury in millimeters. So it would be about 760 millimeters tall that is so ubiquitous or at least it was decades ago that millimeters of mercury is still used as a pressure unit. What if that were made out of water instead of mercury? And let's ignore the issues with the water evaporating over time. How tall would the barometer have to be? Why don't you pause this video and try that on your own? Well, the only thing that's different here is the density of water. Instead of 13,529, we are going to grab the density of water at about standard temperature and pressure which from table A19 is 996.5. We would need a column of water 10.4 meters tall. How many feet is that? Well, if we jump back to our conversion factor sheet we can see that one meter is 3.2808 feet. So if I multiply here this number divided by 1000 multiplied by 3.2808 I need to be 34 feet tall. So like two or three storey building. That's the power of mercury when you're reading off pressure.