>> Hello class, Professor Anderson here. Let's take a look at a pressure problem with something called a U-tube and a U-tube is just that. It's something that looks like a u but lets attach this end of it to a box with a pressurized gas in it. So this will have p gas. This end is open to the atmosphere. And now lets give you some facts. We're going to put mercury in there and the mercury comes to equilibrium when it's in this position. It's lower level on this side than it is on that side. Everything above the mercury over there is gas and lets say that that difference in height is given by d. And now we want to calculate, based on this information, what is the pressure in the gas. All right, so how do we do that? One thing we realize is if we look horizontally, the pressures have to be the same. So if we draw a dash line straight across there, the pressure on the left side tube has to be the same as the pressure on the right side tube. So if we call this left side position a and the right side position b, those pressures have to be equal. P sub a = p sub b. But p sub b is at the top of the mercury on this right side and so there's nothing above it but air that's open to the atmosphere and so that pressure is p naught, which is one atmosphere. All right, so now let's think about the pressure on the left side of the tube. What do we know? We know that p a = the pressure above it, which is the pressure that's in that gas chamber, plus any weight of fluid that's above it, the pressure due to that, which is rho g d. D is the height of the mercury that is above our point a. All right, and so now we can solve this for pressure in the gas chamber. Pressure in the gas chamber is p a - rho g d, p a we just said was p naught. And now let's plug in some realistic numbers, okay? So what is rho for liquid mercury? Rho is equal to 13,600 kilograms per cubic meter and let's take a height difference of 10 centimeters; 10 centimeters in SI units is 0.1 meters and now we should have everything we need to plug into this equation. P naught is 1 atmosphere, 1 atmosphere is 1 times 10 to the 5 pascals. Rho, we just said, was 13,600, g is, of course, 9.8, d is 0.1, and now you can plug in all these numbers and in SI units you should get around 8.7 times 10 to the 4 pascals. All right? Hopefully that one is clear, if not come see me in my office. Cheers!