 Problem number two. A monometer is used to measure the pressure inside of a gas tank. The atmospheric pressure is reported to be 0.92 atmospheres. The height of the monometer fluid is measured as being 25 centimeters, and you know that the monometer contains a fluid with a specific gravity of 1.75. So let's think about what those numbers mean in regards to my diagram down here. I know that I was given an atmospheric pressure. So I'm going to assume that the top of this monometer, because it's open, is exposed to the atmospheric pressure. So the pressure out here is what was given, and that was that 0.92 atmospheres. I'm also going to assume that the pressure changes across the height difference in the atmosphere are negligible. Therefore, regardless of where we look, the pressure outside in this region is always 0.92. Then because the top of this monometer is exposed to atmosphere, that means that the pressure at this interface must also be the atmospheric pressure. So I'm going to call this, I guess we have a 1 and a 2, I'll call this 3. So I know that P3 is the atmospheric pressure. Now, I was also given the height difference, yes, 25 centimeters. So that's referring to this H here. And it's significant because I know that this height of my fluid is going to contribute a pressure difference. So the pressure at state 3 is not going to be the same as state 2. But I can use that height of that fluid along with the density of the fluid to figure out what its contribution to the pressure at state 2 is. Now, I can also see that states 1 and 2 here are at the same height, because they're at the same height and there's the same fluid here. This fluid is constrained therefore due to Pascal's law, I know that P1 is going to equal P2. So I know P3, I can step down to P2 because I know the height and I can figure out the density and then P2 will give me P1. Now, the problem is actually asking us for the absolute pressure and relative pressure of the gas inside the tank. So what I'm trying to find is the pressure out here. But again, I generally assume that the change in pressure caused by the height difference of a gas because there's so little density there, I assume that that change in pressure is negligible. Therefore the pressure at any point in this gas is assumed to be the same. So that means that if this is P, the pressure at state 1 is also P, so let's call this P gas. Then I know that P1 must also be P gas. So I can use the atmospheric pressure to get down to P2 and I know that P2 is equal to P1, therefore P2 is also P gas. So what I'm really concerned with is this thing here. So the question becomes, how can I use the specific gravity of the fluid, which is 1.75, and the height difference across that fluid to figure out the difference in pressure between 3 and 2? Well, I know that. I can calculate that because I know that the changed pressure caused by the height of a fluid is the density times the acceleration that it's experiencing times the height. So I know the density can be calculated from specific gravity because I know what specific gravity is. Specific gravity is the proportion of the density of some fluid to the density of a known fluid. In this case, I generally deal with the known fluid as being water. Most of the time in thermo, that known fluid is water. So we could figure out the density of our fluid by taking the specific gravity of our fluid multiplied by the density of water. Let me write that h a little better. Density of water. So what is the density of water? Well, I could just assume that the density of water is about a thousand kilograms per cubic meter here. But the better practice would be to flip over to my textbook. So if I open my textbook here, let's open it to the appendix. I can start to look through, and I have properties for a variety of substances back here. I have a large table here in A2, which is properties of saturated water, but that's not particularly helpful. I'm going to assume that I am dealing with liquid water at standard temperature and pressure. So if I flip back here, eventually I should get to properties of various liquids. I could have used the table of contents at the beginning of the appendix, but that's way too easy. We have properties of selected solids and liquids here on table A19. So down here at the bottom of our table, we see that we have properties of water at a variety of temperatures. So it's safest for us to use about 25 degrees Celsius, which is about 300 kelvin. And at 300 kelvin, the density of water is 996.5. So at this point, I'm going to be assuming that my density of water is 996.5. Because let's assume that standard temperature and pressure apply to the density of water here. So density of H2O is about 996.5 kilograms per cubic meter. Jump back to mine. There we go. Now you can see what I'm writing. So 1.75 times 996.5 kilograms per cubic meter is going to give me the density of my substance. So let me get on my calculator. And this would be 1.75 times 996.5, which is 1,743.88. 1743.88. So this is a little bit more precise than it had to be. I could have used 1,000 kilograms per cubic meter. I could have assumed that and it would have worked out just fine. But we have all the time in the world right now. So I can take this density and plug that in to this equation. I know my density now. That's a known. I know the height difference because that's given to me. And I'm going to be assuming that the acceleration that this fluid is experiencing is just standard gravitational acceleration. So I'm going to use gravity for that acceleration. So this is now going to be 1,743.88 kilograms per cubic meter multiplied by 9.81 meters per second squared times 25 centimeters. But I'm going to want to get that centimeters in meters because everything else is in meters here. So there are 100 centimeters in a meter. That's going to give me an answer in units of the meters cubed are going to cancel partially. I should have done a better job. The cube here is actually going to cancel this meter and this meter. Centimeters canceled. So I'm left with kilograms per meter second squared. It isn't a very common unit of pressure. So let's convert it to something more useful. Let's say kilopascals. So I know that a kilopascal is a thousand Pascals. And I know that a Pascale is defined as being a Newton per square meter. And I know that a Newton is a kilogram meter per second squared. So now my Newton's Newtons cancel. My meter and meter are going to cancel my square meters. My second squared are going to cancel second squared. And let's see here Pascals cancels Pascals. So I could take 1743.88 times 9.81 times 25 divided by 100 divided by 1000. And that's going to give me an answer in kilopascals. Should have also mentioned kilograms and kilograms cancel. So let's see if I can type that in. 1743.88 times 9.81 times 25 times. It's easiest for me to just write this as 1 over 100 times 1 over 1000. So that's going to give me my answer which is 4.27. Therefore the pressure difference between 3 and 2 is going to be 4.27. Let's call it 7 kilopascals. Now I know that difference in pressure and I can figure out P2 because I have P3 and the difference in pressure. So this is going to be the bigger number minus the smaller number. In this case the height is pushing down. So my pressure at state 2 is going to equal the difference in pressure plus P3. So my pressure at state 2 is going to be 4.277 kilopascals plus P3 which is 0.92 atmospheres because that was given. So 0.92 atmospheres. But I can't just add these numbers together. I have to convert one of them to the other. So I can either convert kilopascals to atmospheres and then add them together or I can convert atmospheres to kilopascals and then add those together. I prefer to work in kilopascals so I'm going to convert atmospheres to kilopascals. So one atmosphere is 101.325 kilopascals. And if you didn't know that off the top of your head we could flip back over to our textbook and we could look that up. If we go to the, I think it's the inside front cover, yeah here we go. So on the inside of the front cover of our textbook we have a variety of conversions and in this case what I'm using is the pressure conversion from atmospheres to kilopascals. So I have to combine things here. I know that one atmosphere is 1.01325 bars and I know that one bar is 10 to the fifth Newton's per square meter and I know that a Newton per square meter is a Pascal. Therefore I know that one atmosphere is equal to 1.01325 Pascal's times 10 to the fifth. So that would be 1, 2, 3, 4, 5, 101,325 Pascal's. Therefore one atmosphere is 101.325 kilopascals. But you do it enough times and you just eventually commit it to memory. So this case, let's get the calculator again, 0.92 times 101.325 is 93.21 kilopascals. So I have 4.277 plus 93.219 kilopascals. So that would be this number plus this number which is 97.4959 kilopascals. Now the question is, is that an absolute or a relative pressure? Well remember that when we talk about relative pressure what we're actually talking about is what a gauge would be measuring. So if we had a gauge on this tank, a state-of-the-art gauge, it's probably going to give us the difference in pressure between the gas and the atmosphere. That's what we mean when we talk about relative pressure or gauge pressure. So the relative pressure is going to give us the difference between the actual tank pressure which is P gas minus my atmospheric pressure. So the question is, does this pressure that I calculated contain the atmospheric pressure already or not? And it does. That was this term over here. Therefore, this answer that I came up with is an absolute pressure. This would be the answer to part A. So to answer part B, I have to convert from my absolute pressure into my relative pressure, which I can do using this relationship. So this would be my pressure of my gas, which is 97.4959 kilopascals minus the atmospheric pressure, which I already came up with, which was 93.219 kilopascals, which is going to give me 4.277 kilopascals. And I know that because I got this number, 97.5, by adding together 4.277 plus 93.219. So if I take this number and then subtract this number, I'm going to get this number. Therefore, the relative pressure, which is the answer to part B, is going to be 4.277 kilopascals. That's problem number two solved.