 A piston with additional weights has been suspended on top of a cylinder containing a gas. The weight of the piston and weights is a combined 200 pounds and the diameter of the cylinder is half a meter. The ambient atmospheric pressure is assumed to be 95 kilopascals. Determine the absolute pressure inside the cylinder. Well, I will start with a poor quality diagram, as usual. I recognize that the gas inside is exerting a force on the piston because the pressure of the gas is exerting a force everywhere that the gas touches. And that includes not the shadowy place, but the piston itself. So I could draw that like a statics problem. There is a force applied along the bottom of the piston. I'm going to write that as F the force of the pressure of the gas. F subscript P subscript gas. That is the force on the piston in the upward direction. The force on the piston in the downward direction includes the weight of the weights and the piston itself, but it also includes the force applied by atmospheric pressure because since we are looking for an absolute pressure, we are including the effects of the atmosphere. Because this problem is in equilibrium, that is the piston isn't moving, we are going to consider this just like we would a statics problem. We are going to say the sum of forces in the y direction are zero. Because for it not to be moving, it must have balanced forces. So I can say the forces in the upward direction are going to counteract the forces in the downward direction. The forces in the upward direction are going to be the forces resulting from a pressure. Remember that pressure is defined as a force distributed over an area, therefore the force exerted by a pressure can be written as pressure times area. In this situation, the area of effect is going to be the area of the piston, which is the area of a circle with the diameter of the piston. So I'm going to write that as pressure times pi over 4 times the diameter of the piston squared. One that was area of a circle is pi over 4 times diameter squared. It's the same as pi r squared, it's just that we happen to have diameter in this problem, therefore I'm substituting in radius is diameter over 2, distributing the square and taking 2 squared and writing it as 4. So the force exerted by the pressure of the gas is going to be pressure of the gas times the area of the piston, which again is pi over 4 times the diameter of the piston squared, and that is equal to the weight of the weights and the piston, which I'm going to call weight I guess, it's a good variable name, plus atmospheric pressure times the area of effect for the atmospheric pressure, which again is just going to be the top circle of the piston. Now I know what you're thinking, you're thinking, but John, what about all this area? Well, remember that from the perspective of directly above, it doesn't matter because it's still going to appear as a circle anyway. The forces on the side here aren't being considered because we're only talking about the forces in the y direction, plus they're going to cancel out because there's going to be the same area on either side. So this is still going to be pi over 4 times diameter of the piston. In this equation, we know the diameter of the piston, we know atmospheric pressure, we know pi, we know the weight of the weight, therefore the only unknown is P gas. So I will write P gas as being equal to the weight of the weight plus atmospheric pressure times pi over 4 times the diameter of the piston squared and divided by pi over 4 times the diameter of the piston squared. So that simplifies down to weight over pi over 4 times diameter of the piston squared plus the atmosphere, note that that's a P not a D, times pi over 4 times the piston squared over pi over 4 times the piston squared. And pi over 4 times the piston squared cancels in the right term, which means that we're just left with this. And the reason I'm pointing that out is because we are calculating an absolute pressure. We have the relative effects plus atmosphere pressure. If we wanted to calculate a relative pressure, a gauge pressure, we would just be figuring out the pressure incurred by the weight, that is the force of the weight divided by the area of effect. You could also think of it as this equation minus P atmosphere. Anyway, now it's time to do some math. The weight of our weight is 200 pounds, and that's going to be pound force because it's described as a weight, but note that if we assume standard gravity, it could also be pound mass, and the force exerted by 200 pound mass at standard gravity would be 200 pound force. That's the convenience of the pound mass instead of slugs. Hashtag slugs are the worst. So 200 pounds of force divided by pi over 4 times the diameter of the piston, which was weird. Half a meter. Oh man, we're mixing together metric and imperial unit systems. I'm sure that doesn't happen in the United States very often. And then in an effort to add these together, I need them to be dimensionally homogenous. My atmosphere pressure was 95 kilopascals. So I could convert 95 kilopascals into pound force per square meter, or I could convert pound force per square meter into kilopascals. Since the problem is asking for the answer in kilopascals, it's going to be convenient for me to take this and get it into kilopascals. So as per usual, I will write my destination and work backwards. That's a thousand pascals. A pascale is a Newton per square meter, and a Newton is a number of pound force that is available to me somewhere on the conversion factor sheet. I don't know why I'm so zoomed in. Come on, potato, you can do it. Hey, there it is, force. So one Newton is 0.22481 pound force, but I already wrote one Newton, so I'm kind of stuck. 0.22481. 0.22481. Generally speaking, I don't like to write decimals because I have a habit of misreading my hand or anyone I do. I'm sure that won't ever happen in one of these example problems. Pound force cancels pound force, Newton cancels Newton's, Pascal's cancels Pascal's, square meters cancel square meters, leaving me with kilopascals, a kilopascal plus a kilopascal will yield an answer in kilopascals. So let's do this calculator, if you would please, 200 times 4 divided by, I'm going to use the fancy pi button on this calculator, multiplied by 0.5 squared times 1000, that's 100 John. One more zero, there you go, times 0.22481. Look guys, I remembered the decimal, aren't you proud of me? Plus 95, probably going to be a syntax error because I forgot my opening parentheses. So we'll jump all the way back to the left and then a step to the right and I get 99.53 one inches for good measure, P gas is 99.53 kilopascals. So I know that I just asked for absolute pressure here, but I made the problem so let's modify it. If I wanted gauge pressure in addition to absolute pressure, what would you do now? I'll give you a second. In order to get gauge pressure from absolute pressure I need to subtract atmospheric pressure. So we are going to subtract 95. Why 95? Because that was the explicitly given atmospheric pressure. A very common mistake at this point would be to just subtract one atmosphere, but in the previous example problem, we didn't have an atmospheric pressure so we assumed it was one atmosphere. If we have an actual number, we don't need to make that assumption. If in fact, we were to use one atmosphere and look at the conversion from atmospheres into kilopascals in the conversion factors of the textbook, I can see one atmosphere is equal to 1.01325 bar, a bar is 10 to the fifth Newton's per square meter, which means it's 10 to the fifth pascals, which means one atmosphere is equal to 101,325 pascals, which is 101.325 kilopascals. Again, one atmosphere is 101.325 kilopascals. That means if I were to subtract 101.325 kilopascals, I would end up with a negative number, which I don't want. That would imply that the pressure of the gas was actually less than atmosphere pressure, which doesn't make sense. It would not be able to hold up the weight here. So 99.53 minus 95 gives us 4.531. If I had one at the gauge pressure, this would be the answer. But I don't, so I won't write it down.