Heat Equations for Special Processes & Molar Specific Heats - Online Tutor, Practice Problems & Exam Prep

Heat Equations for Isobaric & Isovolumetric Processes

Hey, guys. So up until now, whenever we've seen isobaric or isovolumetric 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 going to work out down below here, all we know about this problem is that we have some kind of a process on a 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 going to 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 2 heat equations that you need to know for isobaric and isovolumetric processes, and we're going to see that they're very similar to an equation that we've already seen when we studied calorimetry. So let's go ahead. We're going to 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 talked about when we talked about calorimetry, which is the \( q = mc\Delta T \) 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 gases, so we're going to need a different equation, and it's usually because we're not given the mass of the gas. Remember that this little \( c \) in this \( mc\Delta T \) equation refers to the specific heat per kilogram, but we're usually not given the mass of our gases in kilograms. So instead for gases, we're going to use this big \( c \) here, which is the specific heat per mole. So this is also known as the molar specific heat, this big \( c \) over here. So basically, what happens is that our 2 heat equations are not going to be \( mc\Delta T \). They're going to be \( nC\Delta T \), so \( n\cdot C\cdot \Delta T \).

Now, these two processes are different, and so these \( c \) values, these big \( c \)'s, are also going to be different. We give them special names. So, we do the one for isobaric. This is going to be \( C_p \). This is the molar specific heat at constant pressure. Right? Isobaric means constant pressure, so we use \( C_p \). Isovolumetric means constant volume, and so we're going to use \( C_v \). So these are just always the values that we're going to use for these two equations.

Alright, so that's all there is to it. Instead of \( mc\Delta T \), we use \( nC\Delta T \). Now, what are the values for these big \( c \)'s? Well, if you remember from calorimetry, this little \( c \), 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 \( c \)'s are actually 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 monoatomic or a diatomic 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 \( \frac{3}{2} R \) and \( \frac{5}{2} R \) for monoatomic and then it's going to be \( \frac{5}{2} R \) and \( \frac{7}{2} R \) for diatomic.

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 Jb) 2806 J
• B. a) 3928 Jb) 1683 J
• C. a) 135 Jb) 135 J
• D. a) 3.97 × 10⁴ Jb) 4.36 × 10⁴ J

Calculating Temperature Changes

Alright, folks. Welcome back. So in this problem, we have a monoatomic gas that's initially at some temperature, 293 Kelvin. And what happens here is that there are two processes that are going to happen to this gas. The first one is we're going to add some amount of heat at constant pressure until it reaches point b, and then we're going to remove the same amount of heat, 2,000 joules, at constant volume until we reach point c. So we're taking two different processes and basically what we want to 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 processes 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 2,000 joules of heat at constant pressure. So that's going to be an isobaric 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 isovolumetric process because we're removing heat at a constant volume. So it's just going to look like a flat or so it's going to look like a straight vertical line, except we're going to 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 going to call this T_A = 293 K. But then I have two 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 point C? Now, what happens is, remember, according to our new equations that we've seen for isobaric and isovolumetric processes, when you add or remove heats, there's going to be some temperature changes for both of them. So there's delta T's. So what's going on here is that when you're adding heat from A to B, there's going to be a temperature change delta T from A to B. Then when you remove heat from B to C, you're going to have another temperature change, which is delta T from B to C. I'm just going to call those delta T's. Right? So here's what's going on if I can sort of write an equation for this. My final temperature, this T_C over here, is going to be my starting temperature, the 293, 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 293, I just have to add the 2 temperature changes from A to B and B to C, and then, basically, that's going to be my final answer. Right? So when you add those two things, that's going to be whatever the temperature is at C. 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 isobaric process. So which heat equation are we going to use? We're going to use q = nC_PΔT. So this is going to be q = n C_PΔT 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 2,000 joules of heat. So this is going to be plus 2,000 divided by n the number of moles, which is just 3, and then C_P. Remember, C_P just depends on what type of gas it is and what kind of process. We're dealing with a monoatomic gas here that's undergoing a constant pressure process. So we're going to use this value here. So we're going to use 3·52·8.314, 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 K. Alright. So that's the first number over here. So I'm going to 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 going to be an isovolumetric process. So now we're just going to use this equation, this q = nC_VΔT from B to C. And we're going to do basically the same exact procedure. We just have to divide this ncv over to the other side and start plugging in some numbers. So what's the q here? The q is we're removing 2,000 joules of heat. Be very careful here because you're not adding 2,000 joules of heat. You are subtracting it. So this q here needs to be negative. This is minus 2,000 divided by the moles is still 3, and now we're going to use CV. So we're actually just going to use 32R instead of 52. So this is going to be 32·8.314. That's going to give me my delta T from B to C, and this is going to be negative 53.5 K. So I want to point out something real