Joule and pascal relationship tips

Calculating internal energy and work example (video) | Khan Academy

joule and pascal relationship tips

Option: A. Volume. B. Pressure. C. Density. D. Purity. Answer: B. Pressure. Justification: Joule is the unit of energy and. Pascal is the unit of pressure. You cant since Pa is the unit for pressure and joules is the unit for energy. They are entirely separate. However, Pressure x Volume is energy. A joule is 1N m A pascal is 1N/m^2 So I guess to convert one to the other you'd Basically meaning that a joule = Pa m^3 But they're not equivalent anyway, . Flatland can be interpreted in two ways: A satire of Victorian.

If I could just rewrite this exact equation, I could say-- the input force, and let me just divide it by this area. The input here-- I'm pressing down this piston that's pressing down on this area of water. So this input force-- times the input area. Let's call the input 1, and call the output 2 for simplicity.

joule and pascal relationship tips

Let's say I have a piston on the top here. Let me do this in a good color-- brown is good color. I have another piston here, and there's going to be some outward force F2. The general notion is that I'm pushing on this water, the water can't be compressed, so the water's going to push up on this end. The input force times the input distance is going to be equal to the output force times the output distance right-- this is just the law of conservation of energy and everything we did with work, et cetera.

I'm rewriting this equation, so if I take the input force and divide by the input area-- let me switch back to green-- then I multiply by the area, and then I just multiply times D1.

You see what I did here-- I just multiplied and divided by A1, which you can do. You can multiply and divide by any number, and these two cancel out. It's equal to the same thing on the other side, which is F I'm not good at managing my space on my whiteboard-- over A2 times A2 times D2. Hopefully that makes sense. What's this quantity right here, this F1 divided by A1? Force divided by area, if you haven't been familiar with it already, and if you're just watching my videos there's no reason for you to be, is defined as pressure.

Pressure is force in a given area, so this is pressure-- we'll call this the pressure that I'm inputting into the system. What's area 1 times distance 1? That's the area of the tube at this point, the cross-sectional area, times this distance. That's equal to this volume that I calculated in the previous video-- we could say that's the input volume, or V1.

Pressure times V1 is equal to the output pressure-- force 2 divided by area 2 is the output pressure that the water is exerting on this piston.

Pressure and Pascal's principle (part 2)

So that's the output pressure, P2. And what's area 2 times D2? The cross sectional area, times the height at which how much the water's being displaced upward, that is equal to volume 2. But what do we know about these two volumes?

Gas Pressure Unit Conversions - torr to atm, psi to atm, atm to mm Hg, kpa to mm Hg, psi to torr

I think that's one of the trickiest things in these kind of problems. So here, since our system transferred energy to the surroundings and not the other way around, Q should be negative, because when your system transfers energy to the surroundings, then it's internal energy should go down.

Work on the other hand, since V two is less than V one, the volume of our system went down, which means the surroundings had to do work on the gas to get the volume to decrease.

joule and pascal relationship tips

We would expect if the surroundings did work on our system, that would increase the internal energy. So that means the work done here is positive. We can also calculate work because we know the external pressure, we know it's constant, and work can be calculated as the external pressure times the change in volume, and we know both of those things, we know the external pressure and we know the initial and final volumes.

So if we start plugging that in, we get that delta U is equal to negative joules, so that's our heat, we know it should have a negative sign because the heat was transferred to the surroundings. So negative joules, minus, we should have a negative sign there, minus the external pressure 1. So that's our final pressure, 2.

We could at this point be like," okay, "we figured it all out, "we just have to stick all of these numbers "in a calculator and we're done," and that's probably what my first instinct would often be, but there's one more thing that we should check before we actually plug in the numbers and have a party. And that's our units, we have our heat, in terms of joules, we probably want our change in internal energy in terms of joules too.

On this other side, we're calculating our work here, and we have pascals, so pascals times, so we have joules, and joules, we have pascals times liters, so then the question is, okay, we're doing joules minus pascals, times liters, we need to make sure that whatever we calculate here, in terms of work, also has units of joules. Otherwise, we will be, we'll be subtracting two things that don't have the same unit, and that's bad.

A gas is compressible, which means that I could actually decrease the volume of this container and the gas will just become denser within the container. You can think of it as if I blew air into a balloon-- you could squeeze that balloon a little bit. There's air in there, and at some point the pressure might get high enough to pop the balloon, but you can squeeze it. A liquid is incompressible. How do I know that a liquid is incompressible? Imagine the same balloon filled with water-- completely filled with water.

If you squeezed on that balloon from every side-- let me pick a different color-- I have this balloon, and it was filled with water. If you squeezed on this balloon from every side, you would not be able to change the volume of this balloon. No matter what you do, you would not be able to change the volume of this balloon, no matter how much force or pressure you put from any side on it, while if this was filled with gas-- and magenta, blue in for gas-- you actually could decrease the volume by just increasing the pressure on all sides of the balloon.

You can actually squeeze it, and make the entire volume smaller. That's the difference between a liquid and a gas-- gas is compressible, liquid isn't, and we'll learn later that you can turn a liquid into a gas, gas into a liquid, and turn liquids into solids, but we'll learn all about that later. This is a pretty good working definition of that.

Let's use that, and now we're going to actually just focus on the liquids to see if we could learn a little bit about liquid motion, or maybe even fluid motion in general. Let me draw something else-- let's say I had a situation where I have this weird shaped object which tends to show up in a lot of physics books, which I'll draw in yellow.

joule and pascal relationship tips

This weird shaped container where it's relatively narrow there, and then it goes and U-turns into a much larger opening. Let's say that the area of this opening is A1, and the area of this opening is A this one is bigger.

Now let's fill this thing with some liquid, which will be blue-- so that's my liquid. Let me see if they have this tool-- there you go, look at that. I filled it with liquid so quickly. This was liquid-- it's not just a fluid, and so what's the important thing about liquid?

Pressure and Pascal's principle (part 2) (video) | Khan Academy

Let's take what we know about force-- actually about work-- and see if we can come up with any rules about force and pressure with liquids. So what do we know about work?

Work is force times distance, or you can also view it as the energy put into the system-- I'll write it down here. Work is equal to force times distance. We learned in mechanical advantage that the work in-- I'll do it with that I-- is equal to work out.