 problem number five 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 0.5 meters the ambient atmospheric pressure is assumed to be a hundred kilopascals determine the absolute pressure inside the cylinder in kilopascals so this is similar to the problem we answered earlier we are looking for an absolute pressure and remember that absolute pressure is going to contain the atmospheric pressure's contribution that absolute pressure is the actual pressure the relative pressure would be what a gauge would read it would be the gauge pressure which is the difference between the actual pressure the absolute pressure and the atmospheric pressure so we're not looking for a gauge pressure let's see here we know that the combined weight is 200 pounds that's pushing down on the piston and the diameter of the piston is 0.5 meters the ambient atmospheric pressure is assumed to be 100 kilopascals let's okay let's look at it this way let's say this was a statics problem if this is an equilibrium the piston isn't moving so in order for it to not be moving I know that the some of the forces in the y-direction would have to equal zero so I have some force pushing down in the form of the weight of the piston and cylinder combined and or the piston and weights on top combined excuse me and I know that I also have some force pushing down from the atmospheric pressure so the force of the atmospheric pressure if that makes sense and then this is held up by the force of the pressure inside so this is going to be the force of the pressure inside the cylinder let's call it yeah that that's a logical name for a variable so if the some of the forces in the y-direction are equal to zero that means that the weight pushing down plus the force contributed by the atmospheric pressure are equal to the force contributed by the pressure inside the cylinder so the weight is known that's 200 pounds awesome and then the force of the atmospheric pressure could be calculated because I know that a pressure is defined as being a force applied over an area so I could calculate the force being applied by taking the pressure times the area so this force being applied by the atmospheric pressure is just going to be the actual atmospheric pressure multiplied by the area over which it's applied so this is a cylinder therefore the top looking down from the top that would be a circle so this would be the area of the cylinder and I could calculate that area because I know the diameter so I could calculate the area of a circle by taking the pi times the diameter of the cylinder squared and then dividing that by 4 because area of a circle is pi r squared diameter is 2 times pi excuse me 2 times r this would be atmospheric pressure times pi over 4 times the diameter my cylinder squared this is a known this is a known and then the force contributed by the cylinder the pressure inside the cylinder is just going to be the pressure inside the cylinder times the same area because it's still going to be a circle so power 4 times the diameter of the cylinder squared and again I know the diameter of my cylinder but I don't know the pressure inside the cylinder that's actually what I'm looking for however when I substitute these back into my equation that I came up with earlier this is weight plus pressure of the atmosphere times pi over 4 times diameter of the cylinder squared is equal to the pressure inside the cylinder times pi over 4 times the diameter of the cylinder squared now in this equation I know everything except for the pressure inside the cylinder so I can solve for the pressure inside the cylinder so this would be pressure inside the cylinder is equal to the weight plus the atmospheric pressure times power 4 times the diameter of the cylinder squared all divided by power 4 times the diameter of the cylinder squared so I'm going to distribute my denominator here just to make it a little bit easier to do the math so this is going to be the weight over power 4 times diameter of the cylinder squared plus atmospheric pressure okay now I'm running out of room so I'm going to write this over here so the pressure inside the cylinder is going to be atmospheric pressure let's do that one first so my atmospheric pressure was given it's assumed to be a hundred kilopascals so hundred kilopascals plus weight which is 200 pounds divided by power over 4 times the diameter of my cylinder which is 0.5 meters like the earlier problem it is a large piston cylinder arrangement so 0.5 meters squared this is a force it's a weight therefore this is pound force by the way so I have a hundred kilopascals plus 200 pound force divided by power over 4 times is a half a meter squared so in order to get this into an answer that I could add together I would have to convert kilopascals to pound force per square meter or pound force per square meter into kilopascals I'm going to use kilopascals partially because the problem is asking for an answer in kilopascals but also because that that unit makes a lot more sense than pound force per square meter so in order to get into kilopascals I'm gonna have to recognize that a kilopascal is a thousand pascals and then a pascale is a Newton per square meter and let's see I'm just becoming a bit of a mess let's just erase this line hopefully this will make sense and then I can convert from Newton's into pound force so let's jump back to the textbook in unit conversions for force over here I have one Newton is equal to 0.22481 pound force 22481 22481 0.22481 pound force per Newton right that's correct I think yes 0.22481 pound force per Newton cool let's make this a little bit easier to read so much easier to read now my Pascal's are gonna cancel Pascal's Newton's cancels Newton's square meters cancels my square meters some left with an answer that is going to be in just kilopascals cool so what is 200 divided by power over 4 times 0.5 squared times a thousand times 0.22481 let's get rid of the BH again and we have 200 divided by the quantity see that be pi times 0.25 times 0.5 squared times a thousand times 0.22481 so this is equal to 4.5309 kilopascals so 100 plus 4.5309 is going to be 104.53 let's just call it 53 kilopascals so the absolute pressure inside this piston cylinder arrangement is about 104.5 kilopascals that's a relatively low number right I mean we didn't we already had a hundred kilopascals so the weight this 200 pounds of weight is only contributing four and a half more kilopascals that seems like a lot of weight for not much increase in pressure right that's because this cylinder is so large it's half a meter in diameter so that weight has a lot of area to be distributed over but regardless that concludes problem five which is the last problem on this exam congratulations we've completed exam number one from the spring 2015 semester of thermal one