22. The First Law of Thermodynamics

# Heat Equations for Special Processes & Molar Specific Heats

1

concept

## Heat Equations for Isobaric & Isovolumetric Processes

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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

2

Problem

How much heat energy is needed to increase the temperature of 5 mol of an ideal diatomic gas by from 273K to 300K if the a) pressure is held constant; b) the volume is held constant?

A

a) 3928 J

b) 2806 J

b) 2806 J

B

a) 3928 J

b) 1683 J

b) 1683 J

C

a) 135 J

b) 135 J

b) 135 J

D

a) 3.97 × 10

b) 4.36 × 10

^{4}Jb) 4.36 × 10

^{4}J3

example

## Calculating Temperature Changes

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Alright folks, welcome back. So in this problem we have a mono atomic gas that's initially at some temperature to 93 Kelvin. And what happens here is that there's two processes that are going to happen to this gas. The first one is we're gonna add some amount of heat at constant pressure until it reaches point b. And then we're gonna remove the same amount of heat 2000 jewels at constant volume until we reach point c. So we're taking two different processes. And basically what we wanna do is we want to calculate the temperature when it reaches that final point over here. Now, the first thing we want to do actually is just draw the process is out on the PV diagram. In fact, that's what we're told to do. So let's go ahead and do that. So from A to B, we're going to add 2000 heat jewels of heat at constant pressure. So that's going to be an ice A barrick process. It's just going to look like a flat horizontal line. So it's going to look like this from A to B. We're going this way, and then from B to C, it's an ice. A volumetric process because we're removing heat at constant volume. So it's just gonna look like a flat or so it's gonna look like a straight vertical line. Except we're gonna go downwards like this because we're removing heat. Alright, so here's what's going on. I know the temperature at point a. I'm gonna call this T. A is equal to 293 Kelvin. But then I have to heat transfers and with heat transfers are going to come some temperature changes and I want to figure out ultimately, well, what's the temperature here? When you finally hit points see now what happens is remember according to our new equations that we've seen for Aisa barrick and ice, a volumetric processes. When you add or remove heats, there's going to be some temperature changes for both of them. So there's delta tease. So what's going on here? Is that when you're adding heat from A to B, there's gonna be a temperature change delta T. From A to B. Then when you remove heat from B to C, you're gonna have another temperature change, which is delta T from B to C. I'm just gonna call those delta T. Is, right, So here's what's going on. If I can sort of write an equation for this, my final temperature, this TC over here is gonna be my starting temperature. The 2 93 plus the two changes, it's going to be plus delta T. From A to B. Plus delta T. From B to C. So, really what happens is if I'm starting off at 2 93 I just have to add the two temperature changes from A to B and B. Two C. And then basically that's gonna be my final answer. Right? So when you add those two things, that's gonna be whatever the temperature is at sea. Okay, so that's kind of what's going on here, The rest is just figuring out the right equation and then just calculating a bunch of stuff. So let's start off with this delta T. From A to B. So we've got A to B. And remember this is an ice, so this is an ice, a barrack process. So which heat equation we're going to use, we're going to use Q equals N. C. P times delta T. So it's gonna be Q equals N. C. P times delta T. From A to B. In order to calculate this delta T. From A to B. You just have to move this stuff over to the other side and then just start plugging in some numbers here. So what's the Q. What's the heat transfer? Well we're going to add 2000 joules of heat, so this is going to be plus 2000 divided by and the number of moles which is just three. And then cp remember cp just depends on what type of gas it is and what kind of process we're dealing with a mono atomic gas here that's undergoing a constant pressure process. So we're going to use this value here. So we're gonna use three times five halves times 8.314. And that's going to give you your change in temperature When you work this out. What you should get here is you should get 32.1 Kelvin. Alright, so that's the first number over here. So I'm gonna write 32.1 now, we just have to do the same exact thing from B to C. So from B to C here, this is gonna be an ice a volumetric process. So now we're just gonna use this equation this Q equals and C V delta T from B to C. And we're gonna do the basically the same exact procedure, we just have to divide this N C. V over to the other side and start plugging in some numbers. So what's the queue here? The queue is we're removing 2000 joules of heat. So be very careful here because you're not adding 2000 joules of heat, you are subtracting it. So this queue here needs to be negative, this is minus 2000 divided by the moles is still three. And now we're going to use CV. So we're actually just gonna use three halves are instead of five halves. So this is gonna be three halves times 8.314. That's going to give me my delta T from B to C. And this is gonna be negative 53.5 kelvin. So what I want to point out something really quickly here, what you'll see is that for these two different processes we added or subtracted the same amount of heat but the temperature changes were different, it was 32.1 For the ice a barrick and negative 53. 3.5 for the ice of volumetric. And that's because remember this molder specific heat is a lower number for ice. A volumetric. So that means that for the same amount of heat transfer, there's actually more temperature change. All right, so basically I just have to plug in this, this is gonna be negative 53.5. And when you go ahead and work everything out to 93 plus 3 32.1 plus this number over here, what you're gonna end up with is to 71.6 kelvin, and that is your final answer. So basically, if I can sort of draw these sort of ice affirms here, you're gonna have these sort of icy Therms, it's kind of gonna look something like this, where this one here is the 2 93. This one here is going to be the um it's gonna be 2 plus 32.1. Um So that's just going to equal 325.1. And then you're gonna have um this is gonna be just to 71.6 over here. Right? So you end up sort of at a lower ice a therm than when you originally started. Alright, so that's it for this one guys, let me know if you have any questions

Additional resources for Heat Equations for Special Processes & Molar Specific Heats

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