 next heading condition for spontaneity, condition for spontaneity. The first one is energy, delta H. Okay condition for spontaneity the first one is energy, right on. Exothermic process, in exothermic process, in exothermic process the energy releases, energy releases and hence it is spontaneous. Okay, next line, same point, next I do this only. There are few processes which is endothermic in nature but is spontaneous. There are few processes which is endothermic in nature but is spontaneous. Exothermic is to clear that we are going towards less energy so stable and process takes place. There are few endothermic reactions which are process which are endothermic but is spontaneous. For example, you write down this reaction, conversion of this liquid into gas. Okay, conversion of liquid into gas, decomposition of CaCO3, right. It also requires energy and this dissociates into CaO sorry it is solid. This is also solid and this is gas, CaO2. But burning of fuel we just need an initiation. You want to see, you want to see, wait but I have burning of fuel. No, burning of fuel is just, which one, this one. No, in this also it is not required. Once, see actually when any reaction takes place, right, so in this room also we have oxygen and carbon but the reaction is not taking place, carbon plus CaO2, CaO2 form carbon. Because for any reaction we require some minimum amount of energy. That energy once you provide, it takes place. It goes on its own and that minimum energy is activation energy. Okay, this is for example like I said carbon and oxygen is present but this is what I am not reacting because the minimum amount of energy that is required is not here. When you provide that energy reaction takes place. No, that is not permission. Formation in that way we will discuss. Think I said this, you provide heat into it, it converts back into this. So point is this process, this process is a spontaneous, it is a spontaneous process and the factor is not because of energy, because we are providing energy into it. The factor is randomness. We define this process, why this process is spontaneous is because of randomness. So in the product side we have gaseous molecules which is more random, right. So what we say the entropy of this delta S is greater than 0. Entropy we will see what is entropy. For H2 also we have the same thing. Yes, more entropy, right. So process either goes towards the lesser energy state or more randomness, more entropi state. So why do you like delta S? Is it the enthalpy? Yeah, energy means the enthalpy, the energy which is associated with that process. But we said that enthalpy is energy only when x1 measures the constant. No, no, enthalpy is also the energy content of the, energy content of the system. The system, the energy associated with the system is the enthalpy of that system. So it is not that, see it is, see we write here as delta H, right. Like there are, delta H. It is the energy content of the system. Yeah, that energy which is associated with the system. We write a delta H enthalpy of the system but you can also consider as u. That is not internal energy. Okay, that is what we write here. So this process, this two process takes place because the entropy is increasing this side. Energy it is consuming, endothermic. But since the entropy is increasing, that is why this process is taking place, right. So the driving force for this process is entropy, not energy. Process takes place when we are going towards lesser energy state or more randomness, towards more randomness. Okay, then only it is possible, right. So now to understand this randomness, we have a term given and that we call it as entropy. So next you write down entropy. Entropy is represented by s. Just you need to know, more randomness, more entropy. So if I write down, if I compare the entropy of a gaseous system and entropy of a liquid system, which one will have more entropy? Gas, more randomness. Entropy of a liquid system and entropy of a solid system. Obviously liquid will have more entropy. Entropy of gas and again entropy of solid and entropy of gaseous. Right, so lesser intermolecular force, more randomness, more entropy. Okay, definition of this you write down. It is also a state function like internal energy and all. Okay, state function depends upon initial and final state, not the path followed. Okay, so write down. Entropy, it is a thermodynamic state quantity. It is a thermodynamic state quantity which gives us the idea about, which gives us the idea about the randomness or disorderness, randomness or disorderness, disorderness of the system. It is an extensive property, extensive property and depends upon initial and final state. Initial and final state. Right, so delta s will write entropy at final state minus entropy at the initial state. Unit of entropy is what? Unit, if you have any doubt you can compare with this. This formula you will memorize all the useful things. Delta g is to delta s, mass state, delta s. This energy, energy, this should also be energy, right? The whole t, del s term should also be energy. Delta s is what? Energy per this temperature, so it is joule per Kelvin. Okay, next you write down, next line. Mathematically it is defined as... Delta s is the heat content of the system per temperature. Delta h. No, delta h is randomness. How does it change? I will give you some more energy in a moment. So you write in more random state. That is random property. That's alright. See, it is the... See the physical significance of the time. It is the energy for unit temperature. So when you provide the energy to the system that will also change the temperature. Energy increases and increases. So the issue of that is nothing but that problem. It's very typical actually, probably if you try to define. Just one word if it's randomness or disorderness. That's why I have given this comparison. If the molecule has more tendency to move, it is in more random state. That's how we do it. Okay, so write down, mathematically it is defined as... It is defined as the terms involved in heat exchange. Mathematically it is defined as the integral of all the terms involved in heat exchange. That is Q in bracket you write down. Divided by the absolute temperature in reversible isothermal process. Okay, so delta S is defined by... Mathematically it is defined by dS is equals to dQ reversible. Divided by T at integral of this. Okay, this is the mathematical definition of this. So if you find out this, it becomes delta S, the change in entropy. Divided by 1 by T dQ, reversible isothermal process. Okay, now if energy absorbed by the system. It means Q reversible is positive or negative? Positive. Delta S has to be positive. By the system. It means Q is positive greater than 0. It means delta S is also positive. This means what? Entropy. Entropy increases. Delta has positive means entropy increases. If heat release or energy release by the system. Release by the system. So Q is what? Less than 0. Delta S is also less than 0. And hence entropy decreases. See, all that energy is involved in reversible process. It is nothing but Q reversible by T. For this expression if you solve this, you can write this also. You can write this. All kind of energy will add first and then divide by 1 by 2. Okay, fine, let it just be right for you. Okay, I know next. What do you mean by this? It is the heat exchange. It is possible. Then we can write delta S total change in entropy, total entropy is equals to the change in entropy of the system plus the change in entropy of surrounding. Okay. Total delta S is equals to delta S of the system plus delta S of the surrounding. If system receives energy, its entropy increases. Delta S will be positive. Delta S greater than 0. And for delta S of the system greater than 0 and delta S of surrounding less than 0. System receives energy, its entropy increases and surrounding entropy decreases. Delta S less than 0 for surrounding greater than 0 for system. Okay, make sure I write down. Delta S total always 0. Delta S total always 0. No, delta S total always greater than 0. It always increases. At equilibrium it is 0. At equilibrium it is 0. Entropy changes in isothermal reversible process. Entropy changes in isothermal reversible process. Write down into this. Suppose the system absorbs Q amount of heat. Suppose the system, write down here. If the system absorbs Q amount of heat, amount of heat, temperature T. So delta S of the system is what? It goes to plus Q by T. Isn't entropy increases? Right? And delta S of surrounding is what? Minus Q by T. Total entropy change is what? S of system plus delta S of surrounding. And that will be the total S delta S will be 0 in this system. So for reversible isothermal process, the total change in entropy is equals to 0. Reversible isothermal process. Reversible isothermal process delta S total is equals to 0. Similarly, in the process is reversible adiabatic. Since we have adiabatic process, so Q reversible is what? Zero. Adiabatic process we have? Q reversible is 0. So again delta S is what? Delta S of the system. Not that important. Next, write down entropy change in irreversible process. Suppose the system here is at temperature T1 and the surroundings, again we are assuming T1 is greater than T2. Two different temperature we have, so T1 is greater than T2. So system has at higher temperature, so this will lose energy, heat and the surrounding gains. So delta S of the system is what? Minus Q by T1. Can you write this? It loses Q amount of heat and surroundings gains Q amount of heat. It loses energy, so minus Q by T1. Why T1? Because it is a T1 temperature. Delta S of surroundings is what? Plus Q by T2. So what is the total entropy change? Delta S total will add this to it and that will be Q by T1. And that will be Q by T1 plus Q by T2. So this becomes Q. So what is the formula? See delta S formula is what? Mathematical definition is what entropy change. Q reversible by T. It means in reversible process whatever the change in heat we have, energy that divided by the absolute temperature. So all these are for reversible process. But this we are assuming for irreversible process. Similar kind of formula we have. Since it loses energy Q amount, so minus Q by T1. Heat releases by the system, minus negative. When this accepts the same amount of heat, it becomes Q by T2. So delta S total is this plus this. This two will add and we will get this. Now T1 minus T2 is what? Negative or positive? Positive because T1 greater than T2. So what we can say is this term is greater than zero. So delta S total is what? Delta S total is greater than. Even if you assume T2 greater than T1, then also you will get the same expression. Delta S total is greater than zero. If it is T2, then this loses heat. Right, so you will get the same expression here. So it does not matter that whether this temperature is high or this temperature is high. Right, what we can conclude here that in irreversible process, the delta S total always increases. Greater than zero, positive. So right after this, for irreversible process, entropy always increases. Entropy always increases. That's because H always increases. So that gives free energy can be less than 0. Can be less than 0. Can be less than 0. So in an isolated system, energy will increase. Entropy of the universe is actually increasing. But isolated system does not interact at all. There is no exchange of heat. So isolated system, if you see delta S, if you see it will be zero. In an isolated system. In an isolated system. There is no exchange of heat. But in an open system it depends. Open system it depends whether it is taking heat or releasing heat. Okay, isolated system is what? The boundary is completely isolated. It is not interacting with the surroundings. There is no heat exchange. So delta S should be zero in that. But you don't know whether it is interacting or not. When you say system is isolated, it means it is not interacting. Its boundary is isolated. So within the system, what if you had some... That's another case. If you are assuming that in the system there is some mixing of gases. Because of that mixing, that probably will increase. So condition if you put the conclusion will be different. Okay, if you have a system. Right, system, completely isolated system. Then this for this delta S will be zero. There is no increase in entropy. Provided that what? That we are considering within the system that there is no reaction. There is no mixing. Right, but when we have gaseous particles. And if it is mixing, then delta S will be greater than zero. Understood this? Okay, what I find? Understood. Okay, doubt. Okay, understood this? So when the system, within the system with gaseous are mixing. Then delta S greater than zero. If it is not for isolated system, then the change in entropy is always zero. Right, now there are like to calculate this entropy. We have certain formulas. Okay, that formula you have to directly apply. Just give me the formula. So write down the first thing. Calculation of entropy, write down. Calculation of entropy. First case, when temperature and volume are variable. D and V are variable. The formula of delta S is equals to. 2.303 N C V log of T2 by T1 N C V. N C V log of T2 by T1 plus 2.303 N R log of derivation you want derivation? Not right. Technically. We can derive it also. But that is not right. Okay. Let us see. N R log of V2 by V1. Okay, this is for temperature and volume variable. We will have the application directly. The question will be given. You can directly use the formula. Now the second case, if temperature and pressure are variable. This part will be same. Delta S is equals to 2.303 N C P. N C P log of T2 by T1 plus 2.303 N R log of T1 by P1. These two basic formula we have. All other formula you can derive from this. Right? Okay. These are two basic formula. Third one now you see from this two we can write down all the formulas. If the process is isothermal. For isothermal process what we can say? T1 is equals to T2. This term would not be there. This term would not be there. Right? Delta S is equals to this or this is equals to this. So formula will be delta S is equals to 2.303. N R log of T2 by P1. And similarly you see isonal process we know pressure volume in universal proportion. That also in terms of this you can write 2.303 N R T1 by T. For isomeric process. This is equal to right? Yes equal to this and equals to this. Both formula you can use depending upon what value is given. I am also very this term would not be there. Right? So this is equals to this. Delta S is equals to 2.303 N C P log of T2 by T1. In terms of volume what we can write? 2.303 N C P log of V2 by V1. Fourth one is isochoric delta S is equals to 2.303 N C V log of T2 by T1. For isobaric pressure constant this term will be 0. How will be only the second part of the first? We are not writing it down with the help of this formula we wrote this. Now the pressure isobaric so what we can write? T2 by T1 is equals to V2 by V1. So instead of T2 by T1 I just written V2 by V1 because of process isobaric. That is the same thing I did here also. So why can't we just write the first formula on V? So what is the difference in this? No this is... So basically that is what we are talking about. So that means this case 1 and that fourth one are the same thing. So the second point is in this case they will give you the temperature and volume both are variable. In that case either they will give you change in temperature or change in volume. They won't give both. If they give you both then you have to use this. Both then you will use this. Cp. But you can always find that out. Yes Cp by Cp. Are they both equal? No they both are equal. This we have plus here. This equals to this equals to this. That has a case 1. That has a case 4. No it won't be. When both are changing it is given in that question we will use this formula. When the temperature is changing Cp is given we will use that formula. Temperature is changing Cp is given we will use this formula. Like that. If mixing of gases are there in case of mixing of gases this equals to minus 2.303 are X1 and 2 log of X2 and so on. Right. N number of gas we have so all those will add. Well X is the mole fraction. X is the mole fraction and N is the number of moles. See actually like the derivation we do not have. But if you see this term 2.303 Nr log something right. If you see in walking also we get the same kind of expression. Right. So basically we can understand this part is the energy that is exchanged. Right. So if you have mixture of gases so this change will be there for all the gases. Total change will have because of that only the change in entropy will be there. Because it is nothing but randomness. How fast or how slow the gases molecules and particles are moving. Right. So walk done is nothing but changing energy. So energy change means walk randomness will change. That is why the every energy change we are adding up. That is why we get this. Maybe because of that but derivation if you want I can give you. I have not done this derivation but we can think about that also. That is what it is a part of derivation. Why we are getting negative over there also in reversible isothermal expansion work done. We have got some negative top over there. Right. So this is the part of derivation that is why we are getting negative. If you want derivation I can give it to you. Next class I can give it to you. This negative thing we are getting because of derivation. But again derivation is not required you will get this question directly. We will see next class I will tell you the derivation. Okay.