22. The First Law of Thermodynamics

Heat Equations for Special Processes & Molar Specific Heats

# Heat Equations for Isobaric & Isovolumetric Processes

Patrick Ford

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Hey guys. So up until now, whenever we've seen Aisa barrick Orissa volumetric processes, the only thing we could calculate was the work done. And in some rare cases if you were given the heat transfer, you could also calculate the change in internal energy. But a lot of problems won't give you that heat transfer. For example, the problem we're gonna work out down below here. All we know about this problem is that we have some kind of a process on the PV diagram. We have the moles and we want to calculate the change in internal energy, but we don't have that heat transfer. So in these kinds of situations you're gonna have to calculate it and that's what I want to show you how to do in this video. I want to give you the two heat equations that you need to know for Aisa barrick and ice, a volumetric processes. And we're gonna see that they're very similar to an equation that we've already seen when we started kalorama tree. So let's go ahead. We're gonna keep on filling out our table here with some equations and we'll do an example. So basically what I'm referring to is an equation that we were, what we talked about when we talk about calorie mitri, which is the Q equals M cat equation. So remember that this worked really well for solids and liquids. We did lots of problems where you have water that's warming up from 0 to 20 or ice that's melting or something like that. But it actually doesn't work very well for gasses, so we're gonna need a different equation and it's usually because we're not given the mass of the gas. Remember that this little see in this M cat equation referred to the specific heat per kilogram, but we're usually not given the mass of our gasses in kilograms. So instead for gasses we're gonna use this big C. Here, which is the specific heat per mole. So this is also known as the moller specific heat. This big C. Over here. So basically what happens is that our two heat equations are not going to be M little C delta T. They're going to be any big C. Delta T. So N C. Big delta T. Now these two processes are different. And so these C values these big seas are also going to be different. We give them special names. So we do the one for barbaric. This is going to be C. P. This is the molar specific heat at constant pressure. Right? Aisa baric means constant pressure. So we use cp so volumetric means constant volume. And so we're going to use C. V. So these are just always the values that we're gonna use for these two equations. Alright, so that's all there is to it instead of delta T. We use any big C. Delta T. Now, what are the values for these big seas? Well, if you remember from kalorama tree? This little, see the specific heat per kilogram was always different depending on whether you had water or steam or aluminum or something like that. Right, It was always different where these big seeds are actually much, much simpler. They only actually depend on what type of gas you have. So I've come up with a little table here that shows you all of the different values that you need to know. It really just comes down to whether you're dealing with a mono atomic or a di atomic gas. And here are the values for them. And so you just basically look up on this table here which value you need. And then just plug and chug. So it's three halves are and five halves are form on atomic and it's gonna be five halves and seven halves are for diatonic. Right? So you just look on the table and then just plug it into the equation that you need. But that's basically all there is to it. So let's just jump into our example and see how this works. Alright, so we have three moles of a mono atomic gas and we have this process over here. And what we're asked to do is find the change in the internal energy. So we want to calculate delta E. We start off with our first law equation. So remember that's just the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. So if you want to calculate the change in internal energy, we just have to figure out these other two variables. Alright, so let's get started here. Now, we actually know that this is an ice, a volumetric process because it's a straight horizontal line. So this is S. A. V. And because of that, we already know what one of these terms is going to be. Remember what's special about ice. A volumetric. Is that the work done by the gas in these in these processes is zero. And so because of that, this whole term here will go away, so there's no change in volume. So there is no work done by the gas. So really a change in internal energy comes from the heat transfer. So this change in internal energy of the gas equals the heat transfer to the gas. Now we don't we're not given that heat transfer, but now we have an equation to go calculate it, that's the whole thing here. So basically we're just gonna plug in our new equation. Now this is our delta E. Internal of the gas is going to be our new heat equation. Right, so we have n C V, delta T. So this is gonna be n C V times delta T. So do we have everything we need while I have the number of moles here and then this C V here is just gonna come from this table, it's gonna be one of these values. And then I need the change in the temperature. The only thing I know about this problem here, I don't even know any of the values of pressure and volume but I do know that this process goes between two ice a therms 300 Kelvin and 350 Kelvin. So basically this is my initial temperature and this is my final temperature because the process goes up like this. So I actually do have what these values are for delta T. So I'm just gonna go ahead and plug and chug. So my change in internal energy of the system is just going to be I have three moles and then what CV. Well what kind of gas am I dealing with? Is a mono atomic or die atomic. Right, this is just my values from the table. I'm dealing with a mono atomic gas. And so that just means I'm gonna read off this table CV. This is gonna be three halves are and that's the equation that's the value I'm gonna use so three halves times are which is 8.314. And now I can just use the change in the temperature. Now I'm going from 352 started 303 150. So this is final minus initial. You also could just could have plugged this in as delta T equals 50 because that's the change in the temperature. And basically what you end up with is 1000 871 jules. So that is the heat transfer. That's N C V delta T. But because this is an ice, a volumetric process, that is also the change in the internal energy of the gas. Alright, so that's all there is to it. Now. Now we can calculate the heat transfer. Alright, So I just want to have one last point to make here. Which is that if you remember from our discussion on ice, a thermal processes, the farther you are away from the origin, the higher the temperature. So basically as you go farther and farther away from the origin, the temperature will increase in this direction. So generally what happens is that if you have a process like this that goes away from the origin, then your heat transfer and DELTA T. Are going to be positive. Which is exactly what we have in our problem here. If you have the opposite, if you're going towards the origin then it's going to be negative. So that's it for this one. Guys, let me know if you have any questions

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