Hey, guys. So by now we've seen how to calculate the change in entropy by using this equation here, Δs=qt. Remember, this only works when you have isothermal processes where the temperature remains constant. Unfortunately, you may run across some problems in which the temperature will not be constant. So, we're actually going to need some other entropy equations for special thermodynamic processes and that's what I want to show you in this video. The bad news is that unfortunately some of your textbooks may or may not show the derivations for these equations. The good news is the derivations don't really matter and I've looked through every possible situation you might run across and I've basically summarized it in this table here. I'm just going to show you a bunch of equations for some special processes, and we'll do some examples together. Let's check it out.

So the first process remember is just an isothermal. We actually have seen this equation. This is just Δs=qt. Nothing new there. The second type of process we saw is where a substance changes phase. Like, for instance, if you have water that's going to ice or vice versa. We saw a couple of these and basically you're just still using q/t. Remember because the temperature remains constant for a phase change, you're just basically substituting this equation for ml. That's a calorimetry equation.

Now one situation that we haven't seen is the next one where a substance changes temperature, instead of phase. For example, if you have water that goes from 0 to 100 degrees but doesn't actually change into steam or something like that. In this situation, in this process, you can't use Δs=qt because the temperature is not going to remain constant. So we're going to need a new equation for this. Now again, some of your textbooks may or may not show this. I'm just gonna give it to you. This Δs=mc×lntfinaltinitial. This may look a little bit familiar to you. Remember, the equation for heat transfer and changing temperatures from calorimetry was the q=mcta equation. Basically, all that's happening here is instead of a Δt, you just have an lntfinaltinitial. So that's one way you might remember that.

Let's actually just go ahead and look at our first example here. We have to calculate the change in entropy when we have some amount of water that warms from 20 to 80 degrees Celsius. We have a substance that is now changing temperature, so we're going to use our new entropy change equation here. Alright. So we have Δs=mc×lntfinaltinitial. We're going from 20 degrees to 80 degrees, that's my tinitial and tfinal, and I have the amount of mass here, which is 0.25. All I have to do is convert these temperatures to Kelvin. So this is 20 plus 273, which is 293, and this is going to be 80 plus 273, which is going to be 353. So we have everything we need here. We have the mass, we have the temperatures, and we also have c, which is the specific heat that's 4186. So just gonna start plugging in some numbers here. Δs=0.25×4186, and now we have, we have ln353293. What you end up getting here is a 195 joules per Kelvin. Notice how we got a positive number, and that makes sense. We have to add some heat to the water to warm it from 20 to 80, so therefore you've spread out a little bit more energy and the change in entropy is positive. Alright. So that's pretty much all there is to it. Let's go ahead and take a look at some other processes here.

The next one you might see is an adiabatic process. Sometimes the problem will tell you that it's adiabatic, and this one's actually really straightforward. Remember that in adiabatic processes there is no heat transfer, which means q=0. Now what that means is that in your qt equation, there is no q. So if there's no heat transfer, that means that there is no change in entropy. Δs=0. This is one of the rarer cases where you're going to have no change in entropy for the system or the universe or something like that.

The next one is called a free expansion. Now this is really unfortunate because some textbooks may refer to this as an adiabatic free expansion, which kind of sounds like an adiabatic. But these are actually different processes because in these processes, the q is not equal to 0. So what happens is, in a free expansion, the difference is that you have a gas that suddenly expands to a larger volume. The example I always like to think of is kind of like a balloon. It's like a balloon like this, and all of a sudden the gas the balloon pops and all the air just suddenly rushes out to a larger volume. There is a change in entropy because there's more randomness now in the system. And the equation for this is going to be nr×lnvfinalvinitial. You can kind of remember this because in an adiabatic free expansion you have a sudden rapid change in volume, and this equation here is going to involve the new volume that it rushes out to divided by the initial volume that it was originally in.

Alright. So that's one way you can kind of remember that. Now the last one is called, is basically when you have the gas that's changing at, at any, constant volume or pressure. Basically, this is an isobaric or isovolumetric process. I'm just going to give you these equations. They're ncv×lntfinaltinitial, so that's one of them. The other one is ncp×lntfinaltinitial. Alright. So basically, the only difference here is that you have a cv or cp. Remember, those are just values that we can read off of this table every year. It just depends on what type of gas you're working with.

Now we're going to run we're going to go ahead and work out the second example here, but I have one final point to make. This is a lot of equations, so I kind of come up with one way to help you remember them. If you look at these last sort of 4 out of the 5 equations, you'll notice there's a pattern. So the first letter is either an n or an m. It's either moles or mass. It's basically how much stuff you have times a constant like c, that's the specific heat, or r which is remember the universal gas constant. So it's some constant here, c or r or cv or cp and then the last thing is always an ln. And in this case, you have an ln of some final minus or over initial. So tfinaltinitial or vfinalvinitial. So that might be some way you might remember those equations. Alright. So let's take a look at our second example now. We have 3 moles of a monoatomic gas and it's cooled from 350 to 300 at constant volume. So what's the change in entropy? Well, we're going to start off with Δs, that's change in entropy. Which type of process is this? Let's just go down our list. Is it changing temperature? Well yes. It actually is. So do we use this equation here? Well, what happens is this is when you have a substance that's changing temperature, but our our values here are m and c. We have mass and the specific heat. In this problem all we're told is that we have 3 moles of a gas. So that's just n not m. So we're probably not going to use this equation. Now this is not an adiabatic process because it's not adiabatically, it's not told we're not told that it's adiabatic. We're told that it's cooled at constant volume, so it's not this one either. It's not a free expansion. This thing isn't just suddenly expanding out to a larger volume. In fact, we're told that it's actually a constant volume, so it's definitely not a free expansion. So it's actually just gonna be this last one here. It's either one of an isobaric or an isovolumetric process. We actually know which one it is because it says isovolumetric constant volume. So we're actually just gonna use this equation over here. Alright. So we're gonna use mcv×lntfinaltinitial. That's the equation you wanna use. So we know the number of moles. This is just gonna be 2. The cv here depends on what type of gas it is. We're told in this problem that it's a monoatomic gas, which means the cv value we're gonna use is 3 halves r. So this is gonna be 3 halves times r which is 8.314. Now we're gonna have to multiply the ln and what's the initial and final temperature? Well, it's cooled from 350 to 300. So what happens is this is actually my final is my 300 here. My 350 is my initial. Don't get those confused. So you're going to do 300 over 350 and when you work this out, what you're going to get here is negative 5.8 joules per Kelvin. So in this case, the entropy decreased and that makes sense because in order to cool this gas from 350 to 300, you had to extract or remove some heat and there's a decrease in entropy. Alright. So that's it for this one, guys. Let me know if you have any questions.