 A manometer is used to indicate the pressure inside of a suspiciously cubicle gas tank. Note that that's gases in vapor, not gases in gasoline. The manometer is filled with a gauge fluid which claims a specific gravity of 0.85 and the fluid is currently 55 centimeters tall. The ambient atmospheric pressure is 96 kilopascals. Determine the absolute and gauge pressure inside the gas tank. Well, Pascal's law tells us that the pressure at state 1 is going to be the same as the pressure at state 2. Furthermore, because there's a vapor inside of this box, I'm going to assume that the density of that vapor is so low that the forces are not affected by the weight of the fluid. Therefore, the pressure at all of these little red dots, including all the way down here, is the same. Because of that assumption, I can write the pressure of our gas is going to be equal to the pressure at state 1, which is equal to the pressure at state 2. So this problem really just becomes, how do I get from an atmospheric pressure up here down to the pressure at state 2? Well, remember that this pressure difference caused by the height of that column of fluid can be described using the PAH equation. The density of that fluid times the acceleration it's experiencing times height. So let me write that out a little bit more clearly. The pressure difference across the height of the column of fluid is going to be the density of that fluid multiplied by the acceleration it's experiencing multiplied by its height. In this problem, we know the specific gravity of the fluid that will allow us to calculate its density. We know that the height of the fluid is 55 centimeters and we know the atmospheric pressure. So we can write P2, which is equal to P1 minus atmospheric pressure is equal to the density of the fluid. I'm going to abbreviate that GF for gauge fluid times the acceleration it's experiencing, which I'm assuming is standard gravity times the height, which is 55 centimeters. I'm going to make the substitution from specific gravity. Remember that specific gravity is going to be the density of our gauge fluid divided by the density of water at standard temperature and pressure. Therefore, the density of our gauge fluid is going to be the specific gravity of our gauge fluid multiplied by the density of water at standard temperature and pressure. Now, where do we get the density of water at standard temperature and pressure? I hear you thinking to yourself. Well, remember that anytime we need a material property, we go to our appendices. I want a metric property, which means I'm going to go to our first appendix and I want the properties of some common liquids, which means that I'm going to go to table A19. So, flipping on down through the tables breaking the potato. Table A19 going to be at a number higher than 12. We are making progress. Superheated propane. Propane accessories, not quite 19. Haha, table A19. The density of water is going to be this column right here and if we scroll on down to our water, we have a variety of different densities. Ah, that's because the density of water is going to vary as a function of temperature. So, which temperature do we want? Well, we want standard temperature, which for our purposes is assumed to be about room temperature. Room temperature is about 21 or 22 degrees Celsius, which when you add 273 to that, you get about 295. For our purposes, anytime that you are grabbing a property at standard temperature, it is perfectly adequate to use 300 Kelvin instead of trying to interpolate for a value at 295. Does that make sense? So, we are using the property at 300 Kelvin, 996.5. So, from table A19, the density of water at 300 Kelvin is 996.5 kilograms per cubic meter. Then when we multiply by specific gravity, that is 0.85, we would come up with a density of the gauge fluid. Now, personally, I would prefer to just keep that within the other calculation that we're going to be performing, but just for fun here, let's calculate an actual number. We get 996.5, we'll go by 0.85. We have 847.025 kilograms per cubic meter. That's the density of our gauge fluid. How neat is that? I'm solving this equation for P2. So, I will rewrite this as P2 is equal to the density of our gauge fluid multiplied by gravity, multiplied by height, and then we are adding atmospheric pressure. Interesting that. So, I will write this as specific gravity of our gauge fluid times the density of water at standard temperature and pressure times gravity times height plus P atmosphere. If you don't have a gravity, I'm going to assume standard gravity. I'll write that under assumptions. The pressure change of the gas has a function of height. That was our first assumption, and I will dutifully note it. Then I'm also going to say gravity is about 9.81 meters per second squared. So, we have... Come on, we can do a better horizontal line than that. There we go. 0.85 multiplied by 996.5 kilograms per cubic meter, multiplied by 9.81 meters per second squared. And then I think our height was 55 centimeters. Yeah, 55 centimeters. And we are going to be adding that result to our ambient atmospheric pressure, which is 96 kilopascals. Next, let's recognize that it wants kilopascals as an answer. So, I'm going to want to convert my first term into kilopascals so that I can add it to 96. I will write out 1 kilopascal is 1,000 pascals, and a pascale is Newton per square meter. And then I will make a little bit more space. Newton is a kilogram meter per second squared. Newton cancels Newton's, pascals cancels pascals. Second squared cancels second squared. Kilograms cancels kilograms. Meters and meters squared. Cancel cubic meters. Which means that I just need to convert between centimeters and meters. I know that there are 100 centimeters in one meter. That allows me to cancel everything, meaning I'm left with kilopascals. So, I will summon our calculator. It says C, means that it's ready to help us out. By 100 centimeters, not 1,000, yeah. And we get 100.57 as our answer. Now, the next question I have to pose is, is that an absolute pressure or a gauge pressure? To answer that question, just ask yourself, is atmospheric pressure being considered? Yes, it is. We are adding it to our quantity here. Gauge pressure just would give us the pressure difference between our actual pressure and our atmospheric pressure, which would just be this term. Another way you could think about it if you wanted would be to consider gauge pressure as what would happen if you used an atmospheric pressure of zero. Anyway, since this is absolute, we are going to convert it to a gauge pressure as well because that the problem wants. Again, I will give you a couple of seconds to think about what you want to do to convert it from absolute pressure to gauge pressure. That's right, we subtract atmospheric pressure, which is 96 kilopascals. When we get rid of 96 here, we have 4.57. That's the pressure at state 2, which is the same as the pressure at state 1. And then we are assuming the pressure of the gas is the same everywhere because the density of the vapor is so low that it doesn't increase much as you drop down. You can think of that like the PAH equation with a sufficiently low density compared to a thousand isn't going to be enough to matter. Just for good measure here, I'll write this as P gas instead of P2.