 A frictionless piston cylinder device contains two kilograms of nitrogen at 100 kilopascals and 300 Kelvin. Nitrogen is now compressed slowly, according to the relation pressure times volume to the 1.4 is constant, until it reaches a final temperature of 360 Kelvin. Calculate the work input during the process and be sure to indicate direction. So first of all, I recognize that I have a compression process, which implies that our work is going to be in the inward direction. Furthermore, I'm going to parse out what I know into state point properties. So I'm going to say mass 1 is 2 kilograms, P1 is 100 kilopascals, T1 is 300 Kelvin. Then I know T2 is 360 Kelvin, and I'm going to assume I have a closed system, i.e. no mass of nitrogen escapes around the piston. Furthermore, I'm going to assume that the nitrogen is ideal. If it's ideal, then I can describe its behavior with the ideal gas law. The last thing we know is that the relation pressure times volume to the 1.4 is constant, describes a polytropic process with an n value of 1.4. That n value of 1.4 is specific to this process, but we had already solved through how the boundary work of a polytropic process simplifies, back when we had worked through our simplifications of boundary work. We say the boundary work of a polytropic process is P2V2-P1V1 divided by negative n plus 1. You probably remember how much fun I had saying those words. So because this is a polytropic process and the polytropic exponent value is 1.4, I can say the boundary work for this process is P2V2-P1V1 divided by negative n plus 1. I know n, I know P1, but I don't know P2 nor do I know either volume. So just like in the previous example problem, we could come up with everything. We could use mass pressure temperature with the ideal gas law to determine the volume at state 0.1, and then we could use the fact that pressure times volume to the 1.4 is constant, along with our ideal gas law for state 2, and use the fact that the mass is constant to say P1 times V1 divided by T1 is equal to P2 times V2 divided by T2, and we would have two equations and two unknowns. Those two unknowns would be volume at 2 and the pressure at 2. We could do that and calculate P2 and V2, and at that point we would have already come up with V1, so we would have everything we need to calculate the boundary work. That would be a perfectly valid way to approach this, but I don't want to do any more algebra than I have to. So let's spend some time thinking about if there's an easier way. I'll give you a hint there is. Do you spot it? Remember that the nitrogen is ideal. So for an ideal gas, the pressure times the volume is equal to the mass times the gas constant times temperature. Therefore, instead of writing P2 V2 here, I could write mass times gas constant times temperature, all of those evaluated at state 2, but mass at 1 is the same as 2, and r at 1 is the same as 2, so I will just write T2. And then P1 V1 would become M1 specific gas constant 1 times T1, which again, mass and gas constant are constant for 1 to 2, so I will just write them as mass times specific gas constant, and then T1 divided by negative n plus 1. Then I will factor out our mass and our gas constant, so mass times gas constant divided by negative n plus 1 times the quantity T2 minus T1. Now, what do we know about that process? Well, we know the mass, we know the temperature, add both state points. We know n. We have broken this from five steps down to two steps. All we need is the specific gas constant. The specific gas constant for nitrogen is going to be the universal gas constant divided by the molar mass for nitrogen. The universal gas constant comes from the inside cover of our textbook, specifically the bottom left-hand corner. We can use whichever form we want. I have a metric unit problem here, and I'm looking for kilojoules. So I would probably be easiest off if I plugged in 8.314 kilojoules per kilomole kelvin and then looked up the molar mass of nitrogen in kilograms per kilomole. For that molar mass, I'm going to be going to table A1 Remember, when you would use your periodic table, you want steak sauce or whatever it is that you use to remember which table is which. On table 1, I can see that the molar mass for nitrogen is 28.01 kilograms per kilomole. So I could take 8.314 divided by 28.01, recognizing that kilojoules per kilomole kelvin divided by kilograms per kilomole are going to simplify down to kilojoules per kilogram kelvin and then plug that into my equation, but you guys know that I like to avoid doing any more calculations that I actually have to. So I will actually write this as mass times universal gas constant divided by molar mass of N2 times negative N plus 1. And yes, I know, I could write that as 1 minus N and write one fewer thing, but negative N plus 1 is just so much easier to commit to memory. And then I have T2 minus T1, and at this point, I know everything. So it's just a matter of computing a result. And let me just remind you, because we know this is a compression process, we're expecting to see a negative boundary work. Okay, here we go. First up, mass. The mass was 2 kilograms. The universal gas constant is 8.314 kilojoules per kilomole kelvin. The molar mass of N2 was 28.01 kilograms per kilomole as per table A1. And then negative N plus 1 is going to be negative 1.4 plus 1 because N was equal to 1.4. So I could write that as negative 0.4, but I'll just write negative 1.4 plus 1, you know, just to try to reduce mental math errors. And then I'm multiplying by T2 minus T1. T2 was 360 minus T1, which is 300. Again, I could write 60, but I don't want to incur any mental math errors if I can avoid it, so I'll just actually write out the subtraction and then we can make the calculator do it for us. So kilomoles is going to cancel kilomoles, kilograms is going to cancel kilograms, and kelvin is going to cancel kelvin leaving me with kilodrules. So calculator, if you would please, what is 2 times 8.314 times 60 divided by 28.01 times negative, which is a different number on my calculator, plus 1. So I get negative 89.0468. So boundary work is equal to negative 89.05, let's call it kilodrules. Is that the answer to my question? No. What I want is a work either in or out. Because it's a negative boundary work that implies it's a compression process and the work associated with a compression process is done on the system. It is a work input. Negative boundary work goes with work in, positive boundary work goes with work out. So this would be a work in of positive 89.05. Because remember, when you switch the sign, you are flipping the positive to a negative or vice versa, and boundary work is always out. That's what the negative, therefore, input means. Note I would also accept this in the form of work is equal to 89.05 kilodrules in the inward direction.