Intro to Pressure: Study with Video Lessons, Practice Problems & Examples

Pressure is defined as force divided by area, measured in pascals (Pa). Atmospheric pressure, the weight of air above, is approximately 1.01 x 10⁵ Pa at sea level. As altitude increases, atmospheric pressure decreases due to less air above. In liquids, pressure increases with depth, calculated using the equation: P_bottom = P_top + ρgh. Understanding these concepts is crucial for applications in fluid dynamics and atmospheric science.

Pressure and Atmospheric Pressure

Hey guys, so in this video we're gonna start talking about pressure and atmospheric pressure, both of which are huge topics in this chapter. Let's check it out. Alright, so pressure is defined as force divided by area. Force divided by area. So it's a measurement of how much a particular force is spread out over a surface area, and it has units of Newton per square meter, and that's because force is measured in Newtons, and area is measured in square meters. Now they got tired of writing Newtons per square meter over and over again, so they decided to call this something, and this is called a Pascal. Pascal, named after Mr. Pascal, abbreviated PA, and it just means that if you have 1 Pascal, you have 1 Newton per square meter. Let's look at a quick example here. So, two identical wood blocks, these two guys here, 12, and it specifies here that they have 800 kilograms per cubic meter. Hopefully, right away, you identify that this is density because of the units. It doesn't say that, but you need to know that. So the density, which is ρ, the Greek letter rho, is 800 kilograms per cubic meter. This is why we covered density earlier. And these are the dimensions of the blocks. So 0.2 by 0.2 by 1. So if you notice, this is a long side here. This must be 1 meter, and these guys here will be the point twos meters. And here, it's just oriented in a different direction. This is the long side so this is going to be 1 meter, and this must be 0.2. Heights. And the depth here must be 0.2 as well. Okay. So, they're placed outdoors, meaning that there's a bunch of air around them and horizontal surface, on horizontal surfaces. So the idea is that it's placed on a sort of surface here, the floor, something like that. We want to know the pressure of each block on the surfaces that they sit on. So the idea is that if you have a block and it sits on a surface, it is pushing against the surface, and it's applying a pressure. Why? Because there's a force over an area, and then whenever you have a force over an area, you have a pressure. So I want to know how much pressure is this block over here applying on the surface right underneath it. So you might imagine that it looks kind of like that, right? If you draw sort of the 3D version here. And you might imagine that there's the bottom here of this guy is also pushing against the surface, against the floor. And I want to know the pressure. So we're calculating pressure. Pressure against the floor, let's call that How do we find pressure? Well, the equation for pressure is force over area. So let's write that. It's the amount of force that the block applies on the floor, divided by the area, the area that they're touching. How much area is that? It's just this area down here, the area of interaction. Okay, so what is the force? This block pushes against the floor because it has weight because of gravity, right? So gravity pulls on the block down, the Earth pulls down the block, the block pulls on the table, or on the surface, on the floor. So the force that's causing the block to push against the surface is mg. So I'm just gonna rename this to mg, and this happens a lot, by the way, that the force on a pressure problem is the weight force divided by the area. And I can just sort of start plugging in that the area here is going to be 0.2 times 0.2. Okay? Obviously, we know gravity is 9.8. For the sake of this problem, to keep it simple, we're going to use that gravity is approximately 10 meters per second squared to make our lives easier. But I still have to find the mass. And once I find the mass, I plug it in, and we're done. How do we find mass? You may remember that if you have density, which you do, and if you have volume, which you do, you can find mass. Because density is mass over volume, therefore, mass is density times volume. Now, please don't get the little p, the little curvy Greek p, which is rho, confused, that's density. Don't get that confused with big P, pressure. Okay. Those are different things. It's unfortunate that they look so similar. So, do I have pressure and volume? Yes. So that I can find I'm sorry. Do I have density? See, I just did it. Do I have density and volume? We do. So we're gonna be able to just plug all this stuff in and figure out the mass. So let's do that real quick. Mass is gonna be density, which is 800 kilograms per cubic meter. Remember to always put units like this, it's easier to cancel, times the volume. The volume is just the 3 sides multiplied. So 0.2 times 0.2 times 1.0. And because this is meter meter meter, this is cubic meter. And this is nice because cubic meter cancels here and we end up with the mass. The mass will be, I have it here, 32 kilograms. Now that I have the mass, I can plug it in here, 32. Gravity is 10. This is 0.04. And if you do this entire thing, you get that the pressure is 8000. Now the question is, what are the units here? Well, because I'm using the standard units, this is just going to be Pascal. Now if you don't see that, just keep in mind that mg, because this is in kilograms and this is in meters per second squared, this mg here is in Newtons. And this was meter and meter, so this is meter squared. So Newtons per meter squared gives you a Pascal. So that is the answer to this one. Okay. Now I'm gonna do the second one in a different way so you can see another way that you could have done this. That is gonna be a little bit easier, and it's gonna be helpful later on. But this is sort of the most straightforward way. You could have done it without anything fancy. Okay. So let's do this a little bit differently. And the first thing you might be wondering is, isn't it just the same thing because it's the same block? Well, pressure is force over area. And while the force is the same because the mass is the same, because it's the same block, right? The area is different. The floor is touching, is interacting with the surface underneath it, via a much larger area. So the area that they are touching against each other is much bigger. And if the area is bigger, you might imagine that the pressure will be smaller. Okay. The pressure will be smaller. But we're gonna calculate this differently. So the pressure with the floor is still gonna be the force against the floor, divided by the area. And the force against the floor, by the way, it's still mg, divided by the area. But I'm going to show you something a little bit different now. So what is the area? The area is these two dimensions here, right? These two dimensions here, not all 3 of them but just 2, which is the width times the depth. Okay? So let's leave it there. And in mass, remember we just did this here, mass is right here. Mass is density times volume. But what is volume? Volume is width times depth times height. Okay? Width, depth, and height. So I can plug in this stuff in here, and I can say the m is going to become row wdh, don't forget the g over here, divided by w times d. Okay? And this is the only time I'm going to do this, just to show you this is actually very helpful for you. W cancel, d cancels, and you're left with that the pressure against the floor is gonna be ρhg (row h g) . So this is interesting because the pressure actually does not depend on the area. It only depends on how high this thing is. Why doesn't pressure

A large warehouse is 100 m wide, 100 m deep, 10 m high:
a) What is the total weight of the air inside the warehouse?
b) How much pressure does the weight of the air apply on the floor?

• A. (a) 1.2×10⁴ N;
  (b) 1200 Pa
• B. (a) 1.2×10⁵ N;
  (b) 120 Pa
• C. (a) 1.2×10⁵ N;
  (b) 1200 Pa
• D. (a) 1.2×10⁶ N;
  (b) 120 Pa

Pressure In Air and In Liquids

Hey guys. So in this video, we're going to look into how pressure works. If you have an object that is surrounded by air, as opposed to if you have an object surrounded by a liquid such as water. Let's check it out. Alright. So imagine that you are next to the ocean. Okay. So you're next to the ocean. And if you're next to the ocean, remember that the pressure of the air molecules around you, let's make air molecules green, the pressure that the air is going to exert on you at this level, next to the ocean, is going to be 1ATM, which is the standard atmospheric pressure at sea level. Okay. So if you're out in the open, you have air molecules around you and that's what happens. Now, as you go up in heights, the pressure will change. The easiest way I think to remember what happens to the air pressure, whether it increases or decreases, is just to think that if you go up in air, there's going to be less air above you. So if you are here, you can imagine there's a column of air molecules, on top of you. But if you're here, there's a smaller column of air molecules on top of you. So because there's less air on top of you, the air pressure will be lower. It will decrease. Okay. Remember air pressure decreases. Air pressure comes from the weight of air molecules on top of you. So if there's less air on top of you, there's less weight. So the air pressure decreases. Okay, because there's less weight pushing down on you. Now, the air density is going to decrease as well. And number 2 follows from number 1. So here, you have a ton of air molecules on top of you and the air molecules up here push down against the air molecules down here. So the air molecules down here are more squished together because there's all this weight on top of them. Okay? So you can think that the air density here is lower and here it is higher of molecules. The molecules are more spread out up there because they're not being squished by the weight of the molecules on top of them. If you remember that, you don't even have to memorize that as you go up in height, your pressure goes down and then your density goes down as you go up in height. Okay? Now, here's what's even more important for you to remember: the density of air is very low as it is. So both of these changes are very insignificant. In fact, most of the time, we're going to ignore changes in pressure and density of air. Okay. So this first example deals with that. Which of the following is the best approximation for the atmospheric pressure, Pair at 100 meters above sea level? So remember, changes are only significant over large distances. And I should say for very, very large distances, such as how high an airplane is flying. So 100 meters is not a very large distance, even though it'd be pretty tall, but it's not significant. Therefore, the answer is that the atmospheric pressure here is basically going to be the same as it is at sea level, okay. So it's the same because there's very little difference. It's approximately the same. Cool. So if you're not sure which pressure to use, you should be using 1ATM, which of course is this number right here. It's 1.01. I made 1.00 just because I was rounding. And if they don't tell you, you can use that number. Cool. So it's a little bit different if you have liquids however. So if you were an object or under a liquid, submerged in a liquid, the pressure differences will be much more pronounced. They're going to be much bigger differences in pressure even for a little bit of a distance because liquids have much higher density than air. Okay? So but now in air, we moved up and our pressure changed. But if you are in water, you're going to move down. Okay? So here the pressure depends on your height. And here, it depends on your depth. Okay? Now we just used h for both of those, but the idea is that the pressure will increase as you go down here. And everyone knows that if you start swimming, if you start going underwater, the deeper you go, your ears start to feel a lot of pressure. And that's because the water pressure increases as you go down in height or depth, okay? It increases. And it increases because there's more liquid above you. So before, if you went up, it would go down because there's less air. Now if you go down, the pressure will go up because there's more stuff on top of you. There's more liquid on top of you. So there's more weight pushing down. It's the same logic as before. The difference here is that changes are significant even for small distances, right? And if you're swimming and you just go a little bit lower underwater, you can tell those differences are pretty significant. Water density does not change much. So we're always going to assume that water density is constant because the changes are very insignificant even for very large distances. So you can pretty much assume, you could even assume that I never even mentioned this and just pretend that water density is always the same always. Cool. And then the last point here is that the pressure in the liquid, such as water but really in any liquid, depends on this equation or can be calculated according to this equation. This is a very important equation and it tells us that the pressure at the bottom of a column, so let's draw a little beaker here and let's say we have, let's say we have some liquid. And there are two lines that are important here. The highest point here, okay, and the lowest point of the liquid here. So the pressure at the bottom, right here, at the bottom, is going to be equal to the pressure at the top, which is this, plus ρ, this is the density of the liquid, g, gravity, and h, which is the height difference or the depth of the liquid. Okay? So I can calculate the pressure at the bottom if I know the pressure at the top and if I know the h. Okay. We're going to use this equation quite a bit. Now, you should know that the pressure at the bottom is called the absolute pressure. The pressure at the top is called the relative pressure, and the pressure difference between these two is called the gauge pressure, okay. Gauge pressure is the difference between the two pressures, how much greater one is than the other. And the idea is that this pressure here is relative to the top pressure. The pressure at the bottom depends on the pressure at the top. That's why this one's called relative. So sometimes you see questions that will throw these terms at you. So you should know what they are. Let's do an example real quick and then we'll be done with this. So it says, suppose you are 1.8 meters tall and your heart is located 1.4 meters from your feet. So I'm going to draw, I'm going to draw a person here, pretty big. So that we can do this and your heart, let's say is over here and your total height is 1.8 meters and your heart is 1.4 meters away from your feet. So it follows that your, so if this is 1.4 and this is 1.8, This gap here, heart to top of your head, must be the difference between those 2 which is 0.4 meters. So that's you. It says the blood pressure near your heart is 1.3×104 Pascal. And we want to know the, we want to calculate the blood pressure at the top of your head. So we want to know the blood pressure here, pressure of head, and we want to know the pressure at the bottom of your feet. Pressure feet. And guess what? We're going to use this equation highlighted in green right here to figure this out. One at a time. So the first one we wanna know, what is the pressure of your head? Okay. Blood pressure of your head. This here by the way, is the density Alright. So Alright. So check this out. We know this here, this is our known and these are our unknowns. So for both of these questions, we're going to do the same thing. We're going to set up an equation be with a known pressure and an unknown pressure. Between these two guys. Which means we can set up an equation between these two guys. So if you set up an equation between these two guys, it's always going to be that pbottom=ptop+ρgh. And the h is the gap between them which is 0.4. So I know pbottom, I'm looking for ptop and this is just because this is at the bottom, this is at the top. It's that simple, right? This is the guy at the bottom, that's at the top. So I know, I want ptop. I know the density, 1060. I know gravity. We're going to use 10. Actually for gravity, we're gonna use 9.8 because I wanna be more accurate since we're dealing with the human body here. And h, h is going to be the distance between top and bottom. So this is very important. h is the distance between top and bottom, which in this case is 0.4. So let's add sub. I can write that pbottom=1.3×104=ptop+ρ×1060×9.8×h×0.4. For the sake of time, I'm not including units here, but all the units are standard, which means my pressure will have standard units at the end Pascal. So if I move this around, you end up with ptop=1.3×104-right,thisgoestotheotherside,-andIhaveithere,1060,9.8×0.4.Andthisisgoingto,I'mroundinghere8800pa. So, let me write this here. This went from 1.3×104. If I want to rewrite this with a 104, I'm going to do this kind of quickly, but it would look like this. 0.088×104. Okay. You can validate that if you would like. Let me get out of the way. Alright. And I want to do that so that we can write all of these answers with a power of 4. Let's do part b. So for part b, we want to know what is the pressure of blood or blood pressure on your feet. So again, we're going to set up an equation pbottom=ptop+ρgh. But now we are talking about this interval here. Okay? So this green height was this height right here, but now the blue height has to do with this height right here. And perhaps obviously, this is and now for this equation, bottom is the feet and top is the heart. Okay? Just to be very careful here, when I did this, the bottom was the heart and the top was the head. But this is all sort of relative, right? So now that I'm writing another equation for a different interval for this height here, bottom and top change, okay? So be careful there. And we are looking for pbottom whereas before we were looking for ptop, okay? Be careful. If you're careful, it's going to be easy. So let me write this over here. pbottom=ptop+ρ×1060×g×9.8×h×1.4, 1.4 meters. All the units are standard so I'm going to get the answer in Pascal. And if you do all of this, you get 275100, 275100. Or if you want to write it in terms of a power of 104, right, you can write this as 2.75×104<

Calculating Pressure in Liquids

Hey, everyone. So now that we've seen the basics of pressure in air and in liquids, in this video, I'm going to show you how to calculate different kinds of pressures at different points throughout a liquid, and we'll do an example. Alright? So let's check it out. So remember that pressure in a liquid changes with depth. As you go down in a liquid, the pressure increases according to this equation here.

P = P bottom = P top + ρ g h

Now there's one super important thing about this equation that I really want you guys to understand. It has to do with this h term here. Normally, when we've used h, we use it as a height from the bottom. Right? We use it sort of like to represent heights, but this is actually wrong. Alright? So I want you to write this. This is wrong. You should never use height this way. And instead, height is always going to be a depth that is measured from the top down. Alright? So the correct way to use and represent your h in your diagrams is basically go from the top of whatever boundary it is that you're looking for and go down to whatever point that it is that you're interested in. This is going to be your H.

H, a sort of like an upside-down height. Alright. Now a lot of problems will look like this where you'll have multiple objects or multiple materials that are mixing together or sort of settling in layers like the problem they're going to work out down below. And in this case, the point or the boundary where two materials touch has a special name, it's called an Interface. So hopefully, you'll see here in this diagram that I've got a couple of interfaces. We've got one right here, which is where the yellow or the orange and the white liquid meet. There's another one where the white and the yellow liquid meet, and there's actually a third one which is where the yellow liquid will touch air. So interfaces are always between two kinds of fluids, whether it's liquid-liquid or liquid-air, but it's never going to be where some liquid hits something solid like glass or something like that. Right? So that's not really interface. You can forget about that. Alright?

Now at these points, what's really important is that the pressure of both materials is the same. So what do I mean by this? Well, if you had a molecule like a little tiny molecule, it was at the bottom of the glass, and you had one that was a little bit higher up, then the pressure of this bottom molecule would be higher than the pressure at the top. Right? That makes sense. We've learned that so far. But what this is saying is that if you grab the molecule that's sort of at the very, very top here and you grab the bottommost molecule of the layer on top of it, then these two things would basically have the same pressure. Alright? So, all this point is saying here is that in these kinds of problems or these kinds of situations, you're never going to have like crazy jumps in pressure from one boundary to the next. It's sort of going to be smooth, and the boundary is where they're equal. Alright?

Now one important consequence of this is that everywhere you have a liquid that's touching air like you have in this topmost boundary over here, what that means is that the pressures are equal. So in other words, the pressure of the liquid is equal to air. So this topmost molecule that's sitting right underneath the surface of the liquid is going to have the same pressure as standard atmospheric pressure. Right? And that's just a number that's the, you know, 101,000 pascals. All right. So let's just go ahead and jump straight into our problem here.

We're going to talk about something else called the gauge pressure, but I'm going to save it for later on. Alright. So in this problem here, we have these 2 liquids that are going to combine in this beaker, but they're not going to mix, and basically, so let's get started. So we have a 6 centimeter column of blue liquid, and we're told what the density of it is. So you pour some liquid into a cup, and it's going to be 0.06 meters. Right? That's a column like this. The density of this blue liquid is 1200. Then you're going to pour a 4 centimeter column on top of it. So this is obviously not to scale, this is going to be 0.04 and the density of this yellow liquid is 800. Now, we're told is that the entire beaker is 12 centimeters tall, so this whole entire thing here is 0.12. And what that means is if you kind of notice here, this is 46, which makes 10. All it just means here is this little tiny gap like this, which is 2, is going to be 2 centimeters, so 0.02 meters. Alright? Might need it. Not not not. No, I don't know. So let's go ahead and get started here.

For part a, we're going to calculate the absolute pressure at the blue-yellow interface. What does that mean? Well, if we're using the absolute pressure, right, we've got some densities, we're going to have to use our P bottom = P top + ρ g h equation. You're usually going to start off with that. Right? So what we have here is that we have that P bottom = P top + ρ g h . Now in these kinds of problems where you have lots of you know, we have different layers and you have different liquids that are mixing, it's very easy to get confused between what's the bottom and what's the top of what you're looking at. Alright? So the first thing you need to do is define what's the bottom and what's the top. Now in this problem here, what I'm asked for is I'm asked to calculate what is the pressure at this point over here. So this is the pressure at the blue-yellow boundary. Now to do that, I could set up an equation. I could set up this equation here and either use this interval from here down to the bottom, or I can use the top down to that boundary there.

Now what happens here is if I want to calculate this pressure, but I also don't know what that pressure is, then I can't use this interval because I have 2 unknown pressures. So you always want to combine a known pressure with an unknown pressure so that you can use this equation. Alright. So we're not going to use this bottom interval. Instead, we're going to use this interval over here because hopefully you guys realize that the pressure at the top right at this point over here is really just going to be the pressure of air because it touches air. Alright. So this is just going to be air like this. And this bottom, what I'm going to call the bottom, is really just going to be the blue-yellow interval. Alright? So even though it's not the real bottom of the cup, that's what I'm just calling my bottom for this interval. Okay? And so therefore, what you're going to use is you're going to use the pressure or the density of the yellow liquid and then the height of the yellow liquid. Alright? So in other words, that your P , blue-yellow interface is just going to be this is going to be 101,000 Pascals plus the density of the yellow, which is 800. Then we're going to use g for 10 to make things a little bit simpler.

Now what about the depth? What about h y ? It's basically just the height down and measured from the top down to wherever it is that you're measuring. So I'm measuring from here all the way down to the boundary. So this distance right here, which is just 0.04 , that's going to be my H y . Alright, that's going to be 0.04 . Now this just combines, if you multiply this out, it's going to be 320. So your answer is 101,320 Pascals. Alright? That's the first part. Now let's move on to the second part here.

Now they're not asking for the absolute pressure. They're asking for something else which is called the gauge pressure also at the blue-yellow interface. So this is going to be P gauge, at the blue-yellow interface . Alright. Now what is this? What's the gauge pressure? Well, basically, the gauge pressure is really just the difference between the bottom, the pressure, the absolute pressure you're trying to measure, and the atmospheric pressure. The gauge pressure is really just the pressure that you're measuring relative to the atmospheric pressure. So, for example, in this problem in part a, we were calculating the pressure at the bottom, and we used p , top , which is the air pressure. Basically, this term right here, the Rho g h term, is the difference between these two terms, right? If you move this p , top over, then all you're left with is this, and that's what the gauge pressure is. Your gauge pressure is really just the difference between these two pressures. Alright? It's the absolute pressure and atmospheric. Alright? So in other words, it's just going to be Rho g h , and we actually don't know what that is. That's just 320 Pascals. The reason it's called the gauge pressure is that if you actually took like an instrument and went down to this level here, you would not see this number. You would probably see 320 Pascals. It's a measurement done relative to the atmospheric pressure outside of the cup. Alright? Okay.

Now I got one last point, which is over here, but I'm actually just going to go ahead and continue on with part c. Alright. So part c, now we're going to calculate another absolute pressure, but now it's going to be at the very bottom of the blue liquid. Alright, so let's set up our equation again. So we got P bottom = P top + ρ g h . Alright, so what's the bottom? I'm trying to figure out this pressure all the way down here. So this is going to be P , blue . So in other words, this is going to be P , blue . Now, what's the top? Well, in this case, I also have 2 intervals where I can go from here all the way down or I can go from all the way at the top all the way down. Because I actually know both of these pressures now, I know what the blue-yellow boundary is, and I know what the pressure at the top is, I can actually use both intervals and both of them would be fine. But the simpler one is actually just to use this one all the way down, the blue-yellow one because I already have what that number is. It's right here. I calculated it in part a. So basically, what you're going to do is you're going to you're going to say your top is actually just going to equal the blue-yellow boundary. The bottom is going to equal the absolute pressure, and because now we know that this term here is really just this. So in other words, your p bottom is just going to be 101,320 plus, and now we've got the row for the blue, the g times the height of the blue. In other words, this is going to be 1200 times 10.

Now, what do we plug in for H B ? Well, you might be thinking you're going to measure from the top down, so you're going to go from here all the way down, but that's actually not what you're going to do. Your H B is actually just going to be the height of just the blue parts. And the reason for that is you're measuring from your interval, which is, in this case, this is the top, and this is the bottom. Okay? So what happens is, is in this interval, because you've called the top of this thing right here, the blue-yellow interface, then your h is going to be measured from the top down relative to that point. So in other words, this is going to be your H B , which is 0.06. It's not going to be the height all the way to the top of the liquids. Alright? Because you've already included, you've already counted the 0.04 when you calculated this first part here. So if you were to do it again, you'd be double counting. Alright. It's super important here. That's why it's really important to label everything. So it's going to be 0.06, and then, what you're going to end up with is 102,040 Pascals. Alright. Now basically, we're done here. You can go ahead and move on. What I want to do is show you real quick how we can actually calculate this in one quick step because some questions might actually just ask you right off the bat to do this. Alright? So I'm going to show you how to do this in one step. So I'm going to say that or you can also calculate this, your p blue as you can say this is going to be p top, in which case, you're going to go all the way up, and you're basically going to call a top the air. You're going to go all the way and say, well, this is actually going to be my interval now. So if I take this whole entire thing as my interval, the top is going to be air. And then what happens is you've got to add a Rho GH , except you've got to add it for the yellow liquid first. So they're going to be Rho yellow</

The deepest known point on Earth is called the Mariana Trench, at ~11,000 m (~36,000 ft). If the surface area of the average human ear is 20 cm2, how much average force would be exerted on your ear at that depth?

• A. 2.22×10⁵ N
• B. 2.22×10⁷ N
• C. 1.11×10⁸ N
• D. 2.22×10⁹ N

A tall cylindrical beaker 10 cm in radius is placed on a picnic table outside. You pour 5 L of an 8,000 kg/m³ liquid and 10 L of a 6,000 kg/m³ liquid into. Calculate the total pressure at the bottom of the beaker. (Use g=10 m/s².)

• A. 3.20×10⁴
• B. 3.52×10⁴
• C. 1.33×10⁵
• D. 1.36×10⁵

A wooden cube, 1 m on all sides and having density 800 kg/m³ , is held under water in a large container by a string, as shown below. The top of the cube is exactly 2 m below the water line. Calculate the difference between the force applied by water to the top and to the bottom faces of the cube (Hint:calculate the two forces, then subtract. Use g=10 m/s².)

• A. F_bot − F_top = 800 N
• B. F_bot − F_top = 1000 N
• C. F_bot − F_top = 8,000 N
• D. F_bot − F_top = 10,000 N

Here’s what students ask on this topic:

What is the definition of pressure in physics?

Pressure in physics is defined as the force exerted per unit area. Mathematically, it is expressed as:

$P=\frac{F}{A}$

where $P$ is the pressure, $F$ is the force, and $A$ is the area over which the force is distributed. The SI unit of pressure is the pascal (Pa), which is equivalent to one newton per square meter (N/m²).

How does atmospheric pressure change with altitude?

Atmospheric pressure decreases with increasing altitude. This is because as you ascend, there is less air above you, resulting in a lower weight of the air column pressing down. At sea level, the standard atmospheric pressure is approximately 1.01 x 10⁵ pascals (Pa). As you go higher, the air becomes less dense, and the pressure drops. This relationship is crucial for understanding phenomena in meteorology and aviation.

What is the equation for pressure in a liquid and how is it used?

The pressure in a liquid increases with depth and can be calculated using the equation:

$P_{\mathrm{bottom}}=P_{\mathrm{top}}+\rho gh$

where $P_{\mathrm{bottom}}$ is the pressure at the bottom, $P_{\mathrm{top}}$ is the pressure at the top, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the height (or depth) of the liquid column. This equation is essential for solving problems related to fluid statics and understanding pressure variations in liquids.

What is gauge pressure and how is it different from absolute pressure?

Gauge pressure is the pressure relative to the atmospheric pressure. It is calculated by subtracting the atmospheric pressure from the absolute pressure. Mathematically, it is expressed as:

$P{\mathrm{gauge}}_{}=P{\mathrm{absolute}}_{}-P{\mathrm{atmospheric}}_{}$

where $P{\mathrm{gauge}}_{}$ is the gauge pressure, $P{\mathrm{absolute}}_{}$ is the absolute pressure, and $P{\mathrm{atmospheric}}_{}$ is the atmospheric pressure. Gauge pressure is often used in practical applications, such as tire pressure measurements, where the atmospheric pressure is not included in the reading.

How do you calculate the force applied by atmospheric pressure on a surface?

To calculate the force applied by atmospheric pressure on a surface, you can use the equation:

$F=PA$

where $F$ is the force, $P$ is the atmospheric pressure, and $A$ is the area of the surface. For example, at sea level, the atmospheric pressure is approximately 1.01 x 10⁵ pascals (Pa). If you have a surface area of 1 square meter, the force applied by the atmospheric pressure would be:

$F=1.01\times {10}^5\times 1$

This results in a force of 101,000 newtons (N).