Entropy Equations for Special Processes - Online Tutor, Practice Problems & Exam Prep

Entropy change can be calculated using different equations depending on the thermodynamic process. For isothermal processes, use Δs = q/T. For phase changes, apply Δs = ml/T. When temperature changes, use Δs = mc⁡ln(T_final / T_initial). In adiabatic processes, Δs = 0. Free expansion involves Δs = nR⁡ln(V_final / V_initial).

Entropy Equations for Special Thermodynamic Processes

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.

3 moles of an ideal gas are in the left side of an hourglass-shaped container, separated by a thin barrier. The right side is completely empty, but the volume of the left and right sides are equal. The barrier is suddenly removed, and the gas freely expands into the vacuum. What is the change in entropy?

• A. 17.3 J/K
• B. 0 J/K
• C. –17.3 J/K
• D. 49.9 J/K

Entropy & Calorimetry

Hey, everybody. So in this problem here, we have a mixing of 2 different temperatures of water. We have 195 kilograms at 30 Celsius and 5 kilograms of boiling. In the first part, we're going to calculate the final temperature of the mixture of water. So remember, this is going to be like a classic calorimetry problem. In the second part, we're going to calculate the total change in entropy. So let's go ahead and get started here. Before I get started with an equation, I just want to go ahead and draw a diagram. We've got this hot tub like this, and I filled it with some massive water. This is going to be 195 and this temperature here, this T, is going to be 30 Celsius. I'm going to convert this to Kelvin real quick, which is going to be 303. Then what I'm going to do is I'm going to add some hot water to it. This hot water here is going to be 5 kilograms at an initial temperature of boiling. Remember, the boiling point for water is just at 100. So in other words, it's going to be 100, which is going to be 373 Kelvin.

So what I'm going to do is here I'm going to call the hot water MH and TH. The colder water is going to be MC and TC. Alright? You mix these two waters together, and then ultimately what you end up with here is you just end up with a total amount of water of 200 kilograms. Right? The 5 plus the 195. And it's at a new equilibrium temperature. We've got these two waters that mix at different temperatures. There's going to be an exchange of heat from the hot to the cold, and they're going to end up at some equilibrium temperature. And that's actually the first thing that I want to calculate here. In part a, I want to calculate the equilibrium temperature.

Now we've done this in an earlier video on calorimetry, but we've basically come up with an equation for the equilibrium temperature. It's this really long one that it's like MCTs, MCTs, and MCs, and MCs. You basically just go ahead and plug and chug this. This is kind of a tedious part of this problem, but basically, this is just going to be the MC for water and TC. We're going to use C for water for everything. Plus, this is going to be MH, C for water, TH, and then you divide this by MC_w+MH_w. So if you plug this in, what you're going to get is 195 * 4186 * the initial temperature. So this is going to be the 303. Alright. This equation you can use Celsius or Kelvin. I'm just going to use Kelvin. So this is going to be plus 5 * 4186, times 373. And then you're going to divide by, 195 * 4186 plus 5 * 4186. Alright. So it's kind of tedious, but when you plug all that stuff in, just make sure you enter it carefully in your calculator with parentheses and all that stuff. What you're going to get here is 304.8 Kelvin.

Now if you look at this number, this should make perfect sense to you. Remember, this colder water here is at 303. There's much less of the hotter water which is at 373. So when you mix them together, the final temperature should be pretty close to the initial temperature of the colder water. Right? So we got 304.8, and that makes total sense. Alright. So now let's move on to the second part here. What is the change in entropy?

So in this problem here, what we have is we have multiple objects that are exchanging heats and therefore changing entropies. Right? The hotter water gives off some heat to the colder water. So in other words, there's a ΔS total here. That's the second step, which is right our ΔS total equation. And it's going to be made up of two things. The water doesn't exchange heat with the hot tub or the air or anything like that. The only thing that's sort of happening in this problem is that we just have the two waters that are exchanging. So in other words, the ΔS total is going to be ΔSh + ΔSc.

So now what we have to do here is to solve for this variable here. We're going to have to figure out which equation we use for ΔS. So remember that we've actually come up with a couple of different equations for different processes. So the first one we should check is if you can even use Q over T. Is it an isothermal process? And the thing is it's not. Remember, it's not an isothermal process because both of these waters here start out at their initial temperatures, and then the final one is going to be different. So this is not a constant temperature process. So, is it a phase change? Well, we don't have water changing to ice or gas or anything like that. It's not a phase change. What about the temperature change? And actually, this is the equation we're going to use. So it's going to be MCLn of T_finalover T_initial. That's that, and that's actually going to be the case for both of these terms here. So the first one is going to be MH* CW * Ln of T_final, that's our equilibrium temperature, over T_initial, which is just the temperature of the hot water, plus the MC * CW * Ln of T_final over the temperature of the cold water. So you just add those MCLn terms together, you plug everything in, and when you add them together that should be your ΔS total.

So when you plug everything in, what you should get is you should get 5 * 4186 * the Ln of T_final, which is the 304.8 divided by the 373. Plus or sorry. Yeah. Plus this is going to be 195 * 4186 * Ln of 304.8 divided by 303. And then when you work this out, what you should get here is that this number here just becomes negative 4,222. And this number here works out to 4,835. So when you add them together your ΔS total should be 613 joules per Kelvin.

Alright. So this is basically just confirming the second law of thermodynamics. We had these two processes, and basically, one of them resulted in a decrease in entropy, but there was another part of the problem that increased more entropy. So the overall change here in entropy of the universe is positive, and that's exactly what we should expect. Alright, folks. So that's it for this one.

Entropy & Ideal Gas Processes

Hey, everybody. So in our example here, we have an ideal monoatomic gas that's going through this process that's shown in our PV diagram. In other words, we're taking a 2 step process here from a to b and then from b to c. And ultimately, we want to calculate the change in entropy for this 2 step process. So, what we want to calculate here is Δs total. Alright? Now when we've used Δs before in the past, we usually have 2 objects that are exchanging heats. But here, we actually have 1 gas that's just going through 2 processes. So we're just gonna use the Δsab plus the Δsbc. That's the total change in entropy. It's just the change from both of the processes. So we have this list of equations here that represents all of the different equations for different processes and to basically calculate this Δs total, we just have to figure out which one of these equations apply to both of these processes. That's the whole thing here. Okay? So let's get started. We're gonna look at the first process, the one from a to b. So what type of process is it? Can we use the isothermal equation? Is it a phase change? Well, if you look at this process here, it's not gonna be isothermal. Remember, anything on a PV diagram is just this curve over here, and this represents constant pressure. But a to b actually goes steeper than that. It doesn't remain on the same line. It's not isothermal. It's also not a phase change. We don't have a gas that's changing to a liquid or something like that. We could use this temperature change here, but usually what happens is we're not given we don't have mass and the specific heat in this problem. We're only just given moles. So it's probably not gonna be this equation. However, this process is adiabatic, because we're told that this curve is an adiabatic. So because this process here is adiabatic, then we have a special value for Δs. It equals 0 for adiabatic processes. Remember these are sort of the special cases or the special processes where there is no heat transfer, q=0. So in other words, what happens is this whole entire term just goes away and it becomes 0. So in other words, your Δs total just becomes the change in entropy of the second process. So we're just gonna jump down to the second process. What type is it? Well, if you look through your diagram, you should be pretty familiar with this because this is just a flat horizontal line. So in other words, this is an isobaric process. So which equation do we use for Δs? Again, it's not isothermal. It's not a phase change. It's not a temperature change. It's not adiabatic, and it's also not a free expansion. This gas isn't suddenly exploding or, you know, into a new volume or something like that. It's basically just going along until it just hits another isotherm. So, really, it has to be this equation over here, the equation we use for ideal gases. So this is gonna be n*C_P *ln(TC/TB). That's final over initial. Right? So this isotherm here is TC. This one over here is TB. Okay? So that's basically the equation that we're gonna use. We have the number of moles and the C_P here we can figure out from this table, because remember this is a monoatomic gas. So really, the only thing we need to figure out with this problem is what T_C and T_B are. What are the two temperatures that we're going between in this process here? So let's figure that out. So let's look at T_C first. Okay? So T_C. Well, the only thing we know about T_C is the pressure. We know the pressure is this value over here, but remember that's not enough to actually solve for this temperature. So what we're gonna have to use is we're gonna have to use the fact that T_C is actually the same as T_A. Remember both of these, these points are along the same isotherm. So T_C is actually equal to T_A, and this is really important because this T_A here we have a bunch of information about. We know what the pressure is and the volume, so we can use PV=nRT to find it. So in other words, PAVA=nRTA. So T_A is gonna equal P_AV_A divided by nR. Alright? And we actually have all that stuff. So the pressure at point a is 1.5 times 10⁵. Volume is 0.022. And then we divide by the number of moles, which is 0.8 times the R, which is 8.314. When you work that out, what you're gonna get is 496 Kelvin. So that's 496. That's what we're gonna plug in over here. So now we just do the same thing to figure out T_B. So how do we figure out T_B? Well, this point over here Or sorry. This point over here, we also have the pressure and the volume. So we're basically just gonna do the exact same thing. So we're gonna use PBVB=nRTB. So what happens here is that T_B is equal to, we've got P_BV_B over nR. So in other words, we've got 0.5 times 10⁵ and we've got the final volume which is 0.042. And then divided by, 0.8 and then 8.314. When you work this out, what you're gonna get for T_B is you are gonna get 316. Now this kind of makes some sense here. Right? So this is 316. This is 496. As you go away from the origin, the temperature should be increasing, so that makes total sense. So that's the number that we're gonna plug in for over here. Okay? So now we just have to go ahead and figure out or plug in our last equation. So this is gonna be n which is 0.8. The C_P that we're gonna use is 5/2 times 8.314, and now we have times the ln of T_C is 496 divided by T_B, which is 316, and you work this out what you're gonna get is that a Δs total. The total change in entropy for this process here is gonna be 7.5 joules per Kelvin. Alright. So that's it for this one guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is the entropy change equation for an isothermal process?

For an isothermal process, where the temperature remains constant, the change in entropy (ΔS) can be calculated using the equation:

$\mathrm{\Delta S}=\frac{q}{T}$

Here, q is the heat added to the system, and T is the absolute temperature in Kelvin. This equation is straightforward because the temperature does not change during the process.

How do you calculate the change in entropy for a phase change?

During a phase change, the temperature remains constant, so the change in entropy (ΔS) can be calculated using:

$\mathrm{\Delta S}=\frac{\mathrm{ml}}{T}$

Here, m is the mass of the substance, l is the latent heat of the phase change, and T is the absolute temperature in Kelvin. This equation is derived from the fact that the heat added or removed during a phase change is used to change the phase, not the temperature.

What is the entropy change equation for a substance changing temperature?

When a substance changes temperature without a phase change, the change in entropy (ΔS) is given by:

$\mathrm{\Delta S}=\mathrm{mc}ln(\frac{T}{}$_finalT_initial)

Here, m is the mass of the substance, c is the specific heat capacity, T_final is the final temperature, and T_initial is the initial temperature. This equation accounts for the fact that the temperature is not constant during the process.

How is the change in entropy calculated for an adiabatic process?

In an adiabatic process, there is no heat transfer (q = 0). Therefore, the change in entropy (ΔS) is zero:

$\mathrm{\Delta S}=0$

This means that the entropy of the system does not change during an adiabatic process, as there is no exchange of heat with the surroundings.

What is the entropy change equation for a free expansion process?

For a free expansion process, where a gas expands to a larger volume without doing work, the change in entropy (ΔS) is given by:

$\mathrm{\Delta S}=\mathrm{nR}ln(\frac{V}{}$_finalV_initial)

Here, n is the number of moles of the gas, R is the universal gas constant, V_final is the final volume, and V_initial is the initial volume. This equation reflects the increase in randomness and disorder as the gas expands.