 Problem number one. A well-insulated rigid tank containing two kilograms of oxygen is heated using a fine wire electric resistance heater. The oxygen is initially at a temperature of 300 Kelvin and a pressure of one bar. Electricity is provided to the wire at a constant voltage of 120 volts in one ampere for 10 minutes. Assuming ideal gas behavior, determine the temperature and the pressure of the oxygen at the end of the heating process. Let's begin by drawing out a diagram here. I have a rigid tank containing oxygen. So let's go through and pick out the given information in the problem statement. The first keyword here, or key phrase maybe, is well-insulated. Well-insulated is referring to the fact that I'm assuming no heat transfer is coming from or going into the tank. So I'm going to denote that on my diagram. A well-insulated tank of oxygen. Check. The next keyword or key words is rigid tank here. Rigid tank means that the volume of my tank is not going to be changing. No change in the volume of my tank. So if we add oxygen to it, or we remove oxygen from it, or we add heat to it and the pressure increases, that doesn't change the volume of the tank. It is rigid. Next I'm told that I have two kilograms of oxygen. So I'm going to identify the two kilograms of mass. And this is the mass at the beginning of the process, but I wasn't really told any information about mass entering or exiting. So one of the things I'm going to be assuming is that the mass is constant. This is a closed system. Therefore I'm going to say that this is just the mass. Mass one and mass two are both just M, two kilograms. But that does mean that I'm going to identify my system a little bit more precisely here. I'm going to say that my system, I'm going to declare it as being the mass of the oxygen inside the tank. So I'm ignoring the walls. Two kilograms of oxygen. That's my system. The oxygen is heated using a fine wire electric resistance heater. So I'm going to indicate a heater in my system. Then I know that electricity is flowing through this wire and the wire has some amount of resistance. So that resistance is going to cause the wire to heat up, which is energy added to my system. So I'm actually going to identify this as work, because where this energy crosses the boundary of my system, it is electricity, not heat. It is turned into heat inside my system boundary. Therefore, I'm calling it a work, or rather a power input. Next up, I know my oxygen is initially at a temperature of 300 Kelvin under pressure of one bar. So I'm going to arbitrarily define the beginning of my process as state one and the end of my process as state two. So I know that the temperature at state one is 300 Kelvin, and I know that the pressure at state one is one bar. Then at state two, what I'm actually trying to find is the temperature and pressure. So T2 and P2 are the things I'm trying to find. Next, electricity is provided to the wire at a constant voltage of 120 and a current of one ampere. I know my voltage is 120. So presumably this is connected to a wall outlet, or an American wall outlet, giving it 120 volts AC-ish and a current of one amp. I'm using V to denote voltage. If you see a V without a line that's referring to voltage, I will denote my volume abbreviations with a horizontal line through the V. Okay, so voltage of 120 volts, one ampere for 10 minutes. So I'm going to say that my heating process is applied over the course of 10 minutes. So state two occurs 10 minutes after state one. Now the other pieces that I assumed here, or rather were told here, was that the volume of state two is the same as the volume of state one. Delta V is zero, and that M2 is equal to M1, which is just M, which is two kilograms. And the problem asked me to list my assumptions, so as a good example here, I should list my assumptions. So let's see, the first thing was that the system was well insulated. That's not really an assumption because I was told it was an additional assumption that I had to make, but it's better to err on the side of listing things. Technically, even if it's well insulated, there could be a very, very small amount of heat transfer. So I suppose that the adiabaticness of my system is still technically an assumption. But Delta V being zero definitely isn't. So let's see, what else have I assumed here? Oh, yes. I've assumed that my mass is constant, which gives me a closed system. So I could write that as being, ah, this tank is sealed. So there is no change in mass. Now, as I make any additional assumptions throughout my analysis, I will be sure to go back and write my assumptions over here. And if I don't, you'd be sure to yell at your screen, and maybe I'll hear you through the internet. Let's see here. I have all of my given information written down, so now I can actually begin my analysis. I'm told to find the temperature and pressure of the oxygen at the end of the heating process. And I don't know either one of those yet, but I do know that if I can come up with one of them, I can use the ideal gas law to come up with the other. And that's knowing that the ideal gas law tells me that the pressure times the volume is equal to the mass times the specific gas constant times the temperature. So in my system, the oxygen doesn't change. Therefore, the specific gas constant, which is referring to oxygen specifically here, the way that I wrote out my ideal gas law, the specific gas constant doesn't change. So r is constant. I also know that the tank is rigid, therefore the volume is constant, and I assumed that the tank was sealed, so the mass is also constant. So these three variables are all constant. So if I wanted to rearrange my ideal gas law a little bit here, I could write this as being pressure times temperature, I guess. Excuse me. Pressure divided by temperature, algebra, is equal to the mass times the specific gas constant divided by the volume. So because these three variables are all constant, that means that this whole quantity is constant. So I could write this out as being pressure at state one divided by the temperature at state one is equal to the pressure at state two divided by the temperature at state two. So if I come up with one of these two quantities, that plus the fact that I know P1 and T1 will give me the ability to calculate the other property at state two. And I do have enough information to be able to come up with T2, and that's because I was told everything about the energy occurring in this process. So I was told energy inputs, and I have enough information to determine all of the energy inputs and outputs, that will give me the energy change of my oxygen. And because I know the mass at both the beginning and the end of the process, I can use that to determine the change in temperature. So the key to this is going to be an energy balance. So I'm going to be performing an energy balance, and I need to be performing the energy balance on something. The thing that I'm using for my energy balance is the mass of the oxygen. So all these quantities will refer to the oxygen. So the energy balance is that the change in energy of my oxygen is going to be equal to the energy into my oxygen minus the energy out of my oxygen. And because I assumed that my system was closed here, that reduces the different forms that this energy could take. For example, for the energy change of a closed system, I generally only consider the change in internal energy, the change in kinetic energy, and the change in potential energy. That covers nearly all of my bases. And then energy entering my system, because it's closed, mass can't enter. So the only forms of energy able to cross the boundary are going to be heat and work. So the energy in could exist as heat in and or work in. Same with energy out. I know that the energy out of my system could consist of heat and or work. So running out my full energy balance here, I'm saying that the change in internal energy plus the change in kinetic energy plus the change in potential energy of my oxygen is going to be equal to whatever heat enters plus whatever work enters minus whatever heat leaves plus whatever work leaves. So this is just being an energy accountant. I'm accounting for all of the different forms that energy could take in my analysis. And now I can begin to narrow things down. First and foremost, narrow down. I can make is that the problem told me that it was a well insulated tank. It's adiabatic. Therefore, no heat transfer. Next, I'm going to be neglecting changes in kinetic and potential energy. And this comes from the fact that I wasn't told any significant changes in velocity or height of my system. So I'm assuming that the tank of oxygen is pretty much just hanging out where it is. So I'm going to be assuming that the changes in kinetic and potential energy are about zero. So let me go back up to my assumptions here. I'm going to be assuming that the changes in kinetic and potential energies of my oxygen are about zero. So that leaves me with the change in internal energy of my oxygen is equal to the work in minus the work out. And the only form that I have of work is that electricity crossing the boundary. So now I have to identify is that a work in to my oxygen or a work out. And since I have energy entering and being absorbed by the heater, I'm going to call that a work in. So I'm going to be neglecting this work out. Therefore, the change in internal energy of my oxygen is equal to the entering work. And I was told the voltage and current of electricity running through that wire, which gives me enough information to calculate the power input. Which is not the same as the work input, but I can use the power input to get to work input. So I know that my power input can be calculated by just knowing the voltage and multiplying it by the current. And then I know that my power in general is just work per time. So this would be the work in divided by the change in time. So I could combine these two things together and write, my work in to my system is equal to the voltage times the current times the change in time. So 120 volts times 1 amp times 10 minutes. So multiplying 120 by 10 will give me 1200 volt amp minutes. And that's technically a way to express energy, but it's going to be easier for me if I just bring it back to some standardized units of some sort. So I'm going to use kilojoules. So I'm going to want an answer in kilojoules. So the way I get from volts times current or volts times amps times minutes into kilojoules will just be first converting that I know there are 1000 joules in a kilojoule. Now I know that a volt times an amp times an amp is going to be a watt. So I'm assuming that volt amps are the same as watts. I'm ignoring power factor for the moment. So volts will cancel volts, amps will cancel amps. Now I know that a watt is a joule per second. So watts will cancel watts and joules will cancel joules. So the last thing I have to account for here is that there are 60 seconds in a minute. So minutes and minutes cancel, seconds and seconds cancel. So if I take 120 times 10 divided by 1000 times 60, that'll give me an answer in kilojoules. So let me jump over to my calculator. 120 times 10 times 60, all divided by 1000, gives me an answer of 72. So the work into my system over the 10 minute period is 72 kilojoules. So I could go back over here and then write delta u, the change in total internal energy of my oxygen throughout the 10 minute process is 72 kilojoules. Awesome, now what? How do I use the known change in internal energy of my oxygen to get to change in temperature? Well the first thing I can do is factor out my mass. So this delta u is equal to u2 minus u1 and I can write out u2 as being m2 times specific u2 and I can write out u1 as being m1 times specific u1 and since m1 is equal to m2 this just becomes mass times the quantity specific u2 minus specific u1 and I know mass, that was 2 kilograms I think, I don't want to scroll way up. So that means that now I just have to use these specific internal energies to get to temperature and I could do that, I know of a way to do that. I actually have a couple of options here. The first and most straightforward option, the most accurate option would be to go into my tables here and then look for properties of oxygen. If I have a table of properties for oxygen then using t1 to get to u1 and then using that equation to get to u2 and then working backwards from u2 to t2 is the most accurate albeit longer time consuming option. So the closest thing here, if I scroll through, would probably be ideal gas properties of selected gases. So A23, that's page 965. Awesome, it's sideways. I don't have a way to turn that page so I'm just going to be turning my head. Hopefully I'll remember to rotate that image when I'm editing this video. But if not, sorry about that, you can all just turn your head with me. So I do have properties of oxygen here. As a function of temperature I have specific enthalpy and specific internal energy. Those are both useful to these sorts of problems. Note that this is h-bar and u-bar which means that it's a specific internal energy on a molar basis. So an answer of 4575 would be 4575 kilojoules per kilomole. And that's not a big deal, I'm working in mass but I could use this kilojoules per kilomole value with the molar mass of oxygen to get it to kilojoules per kilogram. So that's option one. So let's first of all zoom out a little bit. If I were a student taking this test I could consider option one. But it would be faster in the long run for me to consider other ways to get to that answer. So let's think about the fact that we're heating oxygen. And we're heating it for 10 minutes with a 120 watt heater. So with a little bit of experience I know that that's not going to add a huge amount of heat. It's going to change the temperature but probably by less than 100 Kelvin. So that's a relatively small temperature change. I also know that the oxygen is starting as a vapor and ending as a vapor. Therefore there's no change in phase. So the other option here would be to assume constant heat capacity. So as a student taking this test I would have to consider both of these options and recognize that option one would give me a more accurate answer with the cost of time. And option two would give me a faster answer but slightly less accurate. So since I have all the time in the world on the internet here I'm going to actually work through both options. So let's start off with option one using the tables. It occurs to me that you can read me writing out option one and option two over here. You were just staring at a list of tables. Sorry about that. Anyway, option one tables I know that I could use T1 being 300 Kelvin to get to U1. So going back to my tables T1 was 300 Kelvin. So if I read across here 300 Kelvin would give me a specific internal energy of 6242. So 6242 kilojoules per kilomole. And then I'm going to use the molar mass of oxygen to get that into an answer of kilojoules per kilogram. So going back to my tables here we go table A1 atomic or molecular weights and critical properties of selected elements and compounds. Diatomic oxygen has a molar mass of 32 kilograms per kilomole. So I could calculate a specific U1 on a mass basis by taking my 6242 kilojoules per kilomole and then dividing it by 32 kilograms per kilomole to get an answer in kilojoules per kilogram. I could have also taken the two kilograms and converted that into a number of moles or kilomoles but it would just be a different variation of calculations converting the two kilograms would actually be slightly faster here but I'm already writing so I'm kind of stuck. It'd be way too much work to go back and erase things. That's just impossible. So 6242 kilojoules per kilomole divided by 32 kilograms per kilomole. Kilomoles cancel. I get an answer of 195. So my specific internal energy at day one is 195.063 kilojoules per kilogram. So I'll bring down this equation up here and write U2 could be calculated by taking 72 kilojoules divided by my mass just bringing the mass over and then adding that to my U1. So 72 kilojoules divided by 2 kilograms plus 195.063 kilojoules per kilogram. So my U2 is going to be 231.063. So now jumping back to this table and I can use my U2. I could look up the internal energy and then figure out which temperature corresponds to that but in order to do that I'm going to need to convert this back into kilojoules per kilomole. So I'm going to use the molar mass again and say that my molar specific U2 could be calculated by taking 231.063 kilojoules per kilogram and then multiplying that by 32 kilograms per kilomole as per the molar mass. Kilograms cancel and I get 7,394. So now I'll jump back to my tables and use that 7,394 number to figure out a temperature. So starting from the first page of table A23 I scroll down this internal energy column and find 7,394. That's going to be somewhere between 7,303 and 7,518. So I know that my temperature at state 2 is going to be between 350 and 360 Kelvin. So in order to figure out the actual number I'm going to have to interpolate. So if I jump back over here and I say that well U bar at 300 and keep forgetting to switch the screen. So U bar at 350 Kelvin and U bar at 360 Kelvin were 7,303 and 7,518 respectively. 7,303, 7,303, 7,303 and 7,518. I think 7,518. Yes indeed. Awesome. 7,518. I can interpolate here by saying that the proportion of temperature between 350 and 360 will be the same as the proportion of internal energy between 7,303 and 7,518. So my internal energy here, 7,394 minus 7,303 divided by 7,518 and minus 7,303 is equal to the temperature that I'm looking for T2 minus my temperature of my first known state here 350 divided by 360 minus 350. And if you're not familiar with what I'm doing, I'm just using linear interpolation. So if I were to draw this on a graph, if I drew internal energy as a function of temperature here, then I'm taking the fact that I know two known points. I know 350 Kelvin and 360 Kelvin and I know the internal energy of both of those, that's 7,303 and 7,518 respectively. Then I'm taking the fact that if I were to connect these with a line, linear interpolation and use my number of 7,394, which is probably going to be right about here, then I know that Y point on the line, therefore I can figure out the X point corresponding to it, 7,394. So what I'm doing is I'm saying the proportion of this distance to this distance T2 minus 350 over 360 minus 350 is the same as the proportion of this distance divided by this distance. 7,394 minus 7,303 divided by 7,518 over 7,303. That's what I'm doing. Now I can solve for T2. So T2 is going to be 7,394 minus 7,303 divided by 7,518 minus 7,303 multiplied by 360 minus 350 and then all of that plus 350. So my temperature at state 2 is 354.233 kelvin. And what I did in that calculation by the way is I just took this 360 minus 350 quantity, brought it up to the numerator of the other side of the equation and then added 350 kelvin to both sides. So I can now write in an answer to part A. I know my temperature at the end of the process 354.2. What if I had done option 2? What if instead of going through the tables this longer process with more calculations, what if I had assumed a constant specific heat? Well, first I would have to list that. So I'm going to add an assumptions for option 2, constant specific heat capacity. Down here I would be using the fact that I know for ideal gases, Cp is defined as being the change in enthalpy with respect to temperature and the Cv is defined as being the change in internal energy with respect to temperature. So remember for an ideal gas, unlike a regular gas or liquid, this does not require a constant pressure or a constant volume condition. For an ideal gas, Cp is always del h del t regardless of what's happening and Cv is always del u del t regardless of what's happening. So here, even though it happens to be a constant volume, I'm going to be using this proportion. I'm going to be saying that because Cv is equal to del u del t, then del u is equal to the integral of Cv del t. And then because I assumed Cv was constant, I'm bringing that out. The integral of del u from u1 to u2 would be delta u is equal to Cv, since that came out of the integral, times the integral of del t, which would likewise be delta t. So I can plug that into this equation over here and say that 72 kilojoules is equal to mass, which is known, times Cv of oxygen times the quantity t2 minus t1. So I could find Cv, which would mean that I could calculate or determine t2 by taking 72 kilojoules divided by the mass times Cv of O2 plus t1. The question becomes, what is the Cv of O2? Well, I can figure that out by using the tables. So I jump over to the tables and then look through the tables. One of the tables is going to give me ideal gas specific heats. So for oxygen here, this is able, table A20 by the way, oxygen, I have Cp and Cv for a variety of temperatures. So the best thing to do here would be to use the Cv at a temperature halfway between t1 and t2. That would be ideal. By using the Cv halfway between the two, I get a better number, but since I don't know t2 in my analysis yet, I can't. So I could kind of eyeball. I know that it's not going to be increasing in temperature much, so I could use say 750 or 725, but that would kind of be metagaming since I already know that number from my previous analysis. So if I were doing this for the first time, the better thing to do would be just to use 300 Kelvin. So if I just looked up a value at Cv at 300 Kelvin and said I'm inducing some inaccuracy by assuming Cv is constant anyway, I might as well assume a constant value from 300 as opposed to the better number of 325. So Cv of oxygen at 300 is 0.658. That's in units of kilojoules per kilogram Kelvin. So Cv at 300 Kelvin was 6258. That doesn't seem right. What was that? I can't remember numbers. Or anything else for that matter. 6.658, man. 0.658 kilojoules per kilogram Kelvin. So I can take 72 kilojoules divided by 2 kilograms and 0.658 kilojoules per kilogram Kelvin and then add 300 to that number. That'll give me my temperature at state 2. So 72 divided by 2, rather, 72 divided by the quantity 2 times 0.658 gives me 54.7112. That number plus 300 gives me 754.7. So using option 2, I got a temperature of 354.711. So doing option 2 used way fewer lookups, way fewer calculations and it got me an answer that was within half a Kelvin. So is that far enough to be significant? That really depends on the application. If you're just using this to get an approximate number for an approximate purpose, that's probably just fine. Even in relatively precise engineering, a half a Kelvin wouldn't be super significant but if you were taking the time to do this in an actual project it would be better to use tabular data. You'd probably be solving this using software anyway so you wouldn't even have had to deal with the approximations over here in option 2. But for this exam, I accepted option 2 from the students. That's a perfectly understandable assumption to make. It's perfectly reasonable. So an answer of 354.2 or 354.7 were both acceptable. But regardless, using either option we now have a T2 and we can use that T2 with my ideal gas law to come up with a P2. So P2 is just going to be P1 times T2 over T1. So I had 1 bar times T2 which was 354.233 Kelvin divided by T1 which is 300 Kelvin which gives me an answer of 1.18078 bars. And I believe that the problem asked me for an answer in kilopascals. So in order to get this to kilopascals I'm going to have to remember that 1 bar is 100 kilopascals. And if I didn't have that conversion memorized I would flip back to my textbook and go all the way to the very front of the book where I have a variety of conversions here. So under pressure here I see that 1 bar is 10 to the 5th Newtons per square meter. I also know that a Newton per square meter is a Pascal. So 1 bar would be a 1 followed by 5 zeros which would be 100,000. I also know that there's 1,000 Pascals in a kilopascal so that would be 100,000 Pascals or just 100 kilopascals. So that gives me a pressure at state 2 of 118.078 kilopascals. And I've solved problem one. So if I had used my other temperature too if I had used 354.7 that would have given me a slightly higher pressure at state 2 but not an extremely significant difference.