 A football official inflates a football to the required gauge pressure of 13 psi prior to a game. The football has an internal volume of 160 cubic inches and the air is at a temperature of 75 degrees Fahrenheit when the football is first inflated. The ball is taken onto a cold playing field and by the time it is put into play, the air temperature inside the ball has decreased to 30 degrees Fahrenheit. Assuming that the football doesn't leak and that the volume of the football doesn't change significantly during the cooling process, calculate the following. A. The pressure of the air in the ball when play begins and B. The initial pressure to which the ball must be inflated so that it will be at the proper and required 13 psi gauge pressure when the temperature reaches 30 degrees Fahrenheit. So I will begin by parsing the problem statement out into some given properties and then begin listing my assumptions. So ignoring part B for a moment, we are considering a cooling process. The football is initially at 13 psi and 75 degrees Fahrenheit and is cooled to 30 degrees Fahrenheit. I'm going to describe that as a process between two state points. State one, the beginning of the cooling process, wherein the pressure inside the ball is 13 psi. The temperature inside the ball is 75 degrees Fahrenheit. The volume of the ball is 160 cubic inches. At state two, once it has cooled, the pressure has dropped to something less than one. T2 is going to be equal to 30 degrees Fahrenheit. And because the problem told us the volume of the football doesn't change significantly, v2 is going to equal v1. At state one, I have two independent intensive properties, which means that I have fully defined the state point. From those two independent intensive properties, I can come up with whatever else I need using the ideal gas law. At state two, I know one independent intensive property and I know that the volume doesn't change. Now volume itself is not intensive, so that doesn't give us enough information to fully define state two. However, I know that the volume doesn't change and I know that the mass doesn't change. The fact that the volume doesn't change and the mass doesn't change means that our specific volume, which we define as volume over mass, specific volume is a specific property and therefore intensive. Because the volume doesn't change, the mass doesn't change, the specific volume will also not change. So the two independent intensive properties I know for state two are temperature and specific volume. So my approach is going to be take p1 and t1, use those two independent intensive properties to come up with another independent intensive property, v1, and then use t2 and v2 to come up with p2. For that, I'm going to be using the ideal gas law. In order to use the ideal gas law, I'm going to be assuming the air in this football is ideal. Before I move on, I should also point out that my system here is defined as the air inside the football. So the boundary of my system is going to be the inside surface of the football. Now normally, I have to identify a control mass or control volume as my system, but here they are one and the same. Because the mass doesn't change, the system is a control mass and because the volume doesn't change, it is also a control volume. When I write out my ideal gas law, I'm going to be writing pressure times volume is equal to mass times specific gas constant times temperature. Remember that when we normally write pressure times volume is equal to number of moles times universal gas constant times temperature. This form is most useful in chemical applications where we're keeping track of say reactants and some sort of chemical reaction or in situations where the number of moles is the relevant parameter by which to keep track of the amount of stuff in your system. In our case, mass is a little bit more useful as what we're doing is primarily a graphometric analysis. And so the mass form is the more convenient way of writing the ideal gas law in most of our circumstances. That being said, they are equivalent because remember our specific gas constant for any specific substance is the universal gas constant divided by the molar mass for that substance. For example, the specific gas constant for air would be the universal gas constant divided by the molar mass of air. So if I were to make that substitution in this equation, what I'm doing is writing pressure times volume is equal to number of moles times specific gas constant for air times the molar mass of air times temperature. The number of moles times the molar mass can be rewritten as the mass of air. So even though we are using PV is equal to MRT, it is not any different equation than the ideal gas law you may know and love. All that we have to do is use mass instead of number of moles and calculate our specific gas constant for the substance at hand. For example, if I wanted to calculate the specific gas constant for this particular example problem, I would need to determine the universal gas constant, which I know is 8.314 kilojoules per kilomole kelvin. But if I did not know that off the top of my head, I can also refer to my conversion factor sheet at the inside of my front cover of my textbook. On my conversion factor sheet at the bottom left, I have a variety of different ways of writing the universal gas constant, depending on what units I have for this particular problem. In our case, kilojoules per kilomole kelvin is perfectly valid, but it isn't going to be as convenient as perhaps one of the imperial forms because everything else in the problem is imperial. So it might be more convenient for me to grab say 1545 feet times pound force per pound mole rankine. If I were to grab that property and plug it into the top of my equation here, 1545 pound of force, rankine, and divide by the molar mass of air, which I'm going to get from the appendix of my textbook, specifically appendix A1. Note that the first two appendices in our textbook are the same information, except one is an imperial and one is in metric. So if I were to jump to table A1 itself in the first appendix, I would have the metric form of the molar mass of air, which is going to be 28.97 kilograms per kilomole. If I were to jump back an appendix to A1e, this is the same information in imperial units, the air's molar mass here is 28.97 pound mass per pound mole. Because it's a proportion and because of the way that we define a pound mole relative to a kilogram and a kilomole, it's going to be the same proportion of mass to number of moles anyway. So 28.97 pound mass, which I will add in an M for good measure, per pound mole. Now pound mole cancels pound mole, degree rankine is still there, and pound force and pound mass aren't going to cancel because one is a representation of mass, one is a representation of force. If I were trying to solve for specific volume specifically at state one, I could write this as V1 over M1. And then if I'm doing the math in my head correctly, that would be R of air times T1 over P1. The specific gas constant for air is going to be 1545 divided by 28.97 pound force per pound mass degree rankine. We could plug that in symbolically, or we could calculate a number if we wanted to. Generally speaking, I prefer to calculate as few numbers as possible. We are multiplying by T1, which we know is 70 degrees Fahrenheit, excuse me 75 degrees Fahrenheit. But we're not going to plug in 75 degrees Fahrenheit, because we are performing a calculation, and we need to use an absolute property for that calculation. Degrees Fahrenheit is a relative temperature, which means that we have to convert it to absolute temperature before we plug it in. So T1 being 75 degrees Fahrenheit would be 75 plus 459.67 rankine. If I pop up my calculator, I can calculate what that number actually is, despite the fact that I'm going to plug it in symbolically, 75 plus 459.67, 534.67. Then we're going to be dividing by the pressure inside the ball. That pressure was described to us as 13 psi, but we're not going to plug in 13 psi either, because just like with temperature, this is a relative pressure. The problem describes the football official inflating the ball to a required gauge pressure. Gauge pressure here indicates that the official is reading the pressure off of a gauge attached to the football. That gauge does not report actual pressure. That gauge reports pressure difference between the inside and outside of the football, which means if we were to just plug in 13 psi, we'd be using a relative pressure. In order to convert it to an absolute pressure, we have to add atmospheric pressure. However, I don't know an atmospheric pressure, so I will have to assume one. It's probably reasonable to assume this football official is inflating the ball relatively close to sea level elevation on earth under relatively consistent conditions. So I think one atmosphere would be a reasonable atmosphere pressure to assume. Then P one absolute, which would be our actual pressure is going to be P one gauge plus atmospheric pressure. So that's 13 psi plus one atmosphere. But I can't add them together because I don't have dimensional homogeneity. The result is not 14 psi atmospheres. I have to convert them to the same units in order to be able to actually add them together. So would you rather convert atmospheres to psi and add them together or convert psi to atmospheres and then add them together? I think converting atmospheres to psi probably makes more sense, but it's an arbitrary distinction. So I'm going to jump into my conversion factor sheet. I'm going to find pressure and I'm going to read that one atmosphere is equal to 14.696 pounds of force per square inch. That is a psi. So I'll write this as 13 psi plus one atmosphere divided by one atmosphere is equal to 13.696 psi. Excuse me, 14.696. So the result would be 27.696 psi. That's the pressure that I actually plug in. So I'm going to write 27.696 psi. And then just for convenience, I'm going to split the psi into its components. So I can keep track of what's cancelling a little bit more conveniently. Okay, so so far pound force is going to cancel pound force. Rankine is going to cancel Rankine. I have square inches in the numerator. I appear to be missing a dimension. I should have one more unit of length. I'm going to double check that I grabbed my universal gas constant in the correct units. That would be the dimension I missed this foot here. So I will go back and add that here. I know my specific volume should be in the dimensions of volume per mass. So I should have three length units in the numerator. The fact that I didn't was an indication that I forgot something. That's an easy way of catching yourself, adding in the units and keeping track of them. You are much more likely to overlook things like that if you just ignore all the units and jump ahead hoping that the end result is the unit that you think it is. So I want probably cubic feet per pound mass even though it's also arbitrary as my end result, which means that I will convert 12 inches to one foot square everything. One squared is boring square inches cancel square inches and I am left with cubic feet divided by pound mass. Then that result is going to be something we can calculate come on calculator you can do it. So I'm going to take 1545 times 75 plus 459.67 divided by 28.97 times 27.696 times 12 squared. My result is 7.14967. 1497 let's call it cubic feet per pound mass. And that specific volume of state one is also equal to specific volume of state two, which means that I now have two independent intensive properties at state two, which means that I can calculate whatever it is that I want at state two. For example, I could calculate p2 using the ideal gas login. This time I'm writing v2 is equal to volume two over mass two, which means that I'm substituting and writing p2 times little v2 is equal to r of air times t2. Therefore p2 is going to equal r of air times t2 divided by little v2. For r of air, I will grab the same quantities as earlier. At this point, it maybe would have been faster to just calculate a number, but you know, in for a penny and for a pound, then we're multiplying by t2, which is given as 30 degrees Fahrenheit. Quick question, are we plugging in 30? No, we are plugging in 30 plus 459.67. And we are dividing by that shiny new specific volume we just calculated, which was 7.1497. And our pressure here is going to be, let's go with psi. It doesn't actually explicitly ask for a unit in its answer. But generally speaking, if you don't know, if you aren't required to calculate a result in the specific unit, it's best usually to default to the units that you are given that allows a reader or a viewer to compare and contrast the given information to the result. So I could say yesterday it rained two tenths of an inch. Today it rained half a tenth of an inch. Tomorrow it should rain five tenths of an inch. That would allow the viewer, the reader, to compare those three numbers together directly. You'll often see that sneaky tactic used to imply that a smaller number is actually bigger, especially in, say, political videos, but it is useful in scientific applications to be as clear as possible. We are not trying to manipulate the reader into trying to construe the result of this football problem to be any different than what it actually is. Anyway, if we want psi, that's a pound of force per square inch. Nope. Inches squared. Nope. That's still 12, John. There you go. Now, how do we get there? Let's see. Pound of force is going to cancel pound of force. Rankine is going to cancel Rankine. Pound of mass is going to cancel pound of mass. In order to get cubic feet to cancel square feet, I have to convert inches squared to feet squared. So there are 12 inches in one feet. Then I square everything. One squared is boring. 12 squared. We leave inches squared, cancels inches squared, square feet, and feet cancel cubic feet. That leaves me with just psi as an answer. So again, I will summon up the mighty calculator 1545 multiplied by the quantity 30 plus 459.67 divided by 28.97 times 7.14967 times 12 squared. And I get 74.6174. Wait, that's not right. Did I miss something? Is it the number? Oh, I wrote 4545 instead of 1545. Try that again. We get 25.365. That's a much more reasonable number. By the way, the thing that clued me off as to having a mistake was the fact that our pressure at the end was higher than it was at the beginning. My temperature is dropping with the same volume, which means that my pressure should also drop. So is 25.365 the answer to part A? Well, it is an answer. But if we're trying to express an answer in the same unit system we were given, we shouldn't express an absolute pressure as the answer. We should express a relative pressure. So how do we get this absolute pressure back to a gauge pressure? You're right. We have to subtract one atmosphere. So P2 gauge would be P2 absolute minus P atmosphere, which is going to be 25.365 psi minus one atmosphere, which was 14.696 psi, I believe. Yep, 14.696. So the better answer for part A would be 10.669 psi. So the official inflates the ball to the correct pressure in the locker room, but once it comes outside and is allowed to cool down, the pressure drops as a result. We performed that calculation by stepping on a stepping stone that was specific volume estate one. We calculated the specific volume estate one and that allowed us to calculate the pressure at state two. We did this a rather long way though. We calculated the actual number. Was it important that we calculate that actual number? Not really. We need the fact that the specific volume estate two is the same as the specific volume estate one, but the number itself doesn't really matter. In this situation, we could have just taken the expression that we came up with for V1 and written that as being equal to the expression we wrote out for V2. I mean if V1 is equal to V2, and I write R of air times T1 divided by P1 is equal to R of air times T2 over P2, I could get to an answer with about six fewer steps. In fact, I can make it even easier by recognizing that the specific gas constant cancels. So I can write P2 is equal to P1 multiplied by T2 over T1. This equation is all I actually need. That's the real advantage of working through problems as symbolically as possible. If you can get as far as you can symbolically before you start punching in numbers, you can often find faster ways to work through a problem, plus you're not wasting time on calculating numbers that you don't actually need. On top of all that, you also don't incur the rounding error inherent to plugging in a number that you calculated in a previous step in your current step. Let's try this all again in an alternate reality where we hadn't calculated all of these intermediate steps. P2 is equal to P1 times T2 over T1. So surely I just plug in 13 psi multiplied by 30 over 75, right? Well no, because anytime I'm doing math with a pressure or temperature, I need an absolute property. That means I have to convert my temperatures to Rankine and I have to convert my pressures to absolute. So I'm going to say P2 gauge plus the atmosphere is equal to P1 gauge multiplied by T2 plus 459.67 divided by T1 plus 459.67. Rearrange this a little bit so that I calculate the actual answer that I actually want. Now we can get to calculating. So I'm going to say 13 plus 14.696 times T2 was 30, I believe. Let me just double check before I go punching in 35. T2 was 30 plus 459.67 divided by 75 plus 459.67. And then at the end of all that I'm going to add 14.696. Excuse me, I'm going to subtract. What algebra was I doing? Ah yes, I was doing the algebra you do when you're too focused on how your formatting sucks right now and you really wish that you had just opened up a new page as opposed to trying to cram this into the side. Hey look, we got the exact same number. That's fun. So we got to the same result in oh about a quarter of the effort. Let's see if we can use that same approach to quickly answer part B, shall we? We are looking for the initial pressure to which the ball must be inflated so that I will reach the required 13 psi gauge pressure when the temperature reaches 30 degrees Fahrenheit. So in this situation the only thing that's different is we are figuring out P1 so that P2 ends up being 13 psi. So we are getting rid of 13 up here. We're adding in some question marks and we're saying P2 is equal to 13 psi. Does that make sense? They were essentially doing part A backwards. Anyway, I'm going to take my newfound approach to the ideal gas on. I'm going to once again recognize that because V1 is equal to V2 I can write T1 over P1 is equal to T2 over P2. And actually while I'm here, let me take a moment to warn you about the dangers of relying on equations like this. It's very common for thermal one students to see equations like this written on an example problem on a previous test solution and then they write it on their equation sheet and then they start deploying it in situations where it's irrelevant. This equation is only applicable in situations where you have a process occurring on an ideal gas within a constant volume process where the mass is also the same. Now you could write that equation down on your equation sheet and also all of the assumptions that are required to be able to use it or you could just build it yourself from the ideal gas law anytime. If you know the ideal gas law and you know how to simplify it based on quantities being constant, if you have a constant temperature process or a constant volume process or a constant mass process or a constant pressure process, you can manipulate the ideal gas law into whatever you need for the moment and not rely on a big equation sheet full of assumptions. It is much better for you to spend a little time on the algebra than to try to jump to these shortcuts. If you had written this equation on a problem where it didn't make sense, you would get no partial credit for the effort. Anyway, P2 over T2 is equal to P1 over T1. Note that I'm actually writing this the opposite of the way that I did before. It doesn't matter, like one half being equal to one half. If I take the reciprocal of both sides, I can write that as 2 is equal to 2. It's still true. I'm just doing this for algebra convenience. P1 is equal to P2 times T1 over T2. And again, I'm in a situation where I need these pressures to be in absolute and I need my temperatures to be in absolute. So I am going to write this as P2 gauge plus atmospheric pressure plus T1 plus 459.67 divided by T2 plus 459.67. And then we are subtracting atmospheric pressure at the end to get our answer into a gauge pressure. Okay. So we're taking 13 plus 14.696 times 75 plus 459.67 divided by 30 plus 459.67. And then we are subtracting 14.696. And we get 15.545 PSI. So in this situation, for the football official to inflate the ball to a pressure so that it's actually the correct pressure during the game, he or she would need to inflate it to 15.545 PSI in the locker room beforehand. And that's what part B is saying.