 Hello and welcome to the next segment of our discussion about thermodynamics. So, over here so far we have discussed the first law and second law of the thermodynamics. We understand the connection between the energy, work and heat. We have understand two important parameters enthalpy and entropy. We looked into the effect of entropy how it can lead us to the spontaneity of a reaction and how entropy can also define the third law of the thermodynamics. So, in this segment we will talk about the free energies. So, let us start our discussion about that particular factor in thermodynamics. So, when we talk about the entropy we say during this process, this process is spontaneous when the entropy actually increases. Now this system whose entropy is increasing it is present around the surrounding. Now when we increase the entropy of a system at the same time the surrounding entropy is also decreasing to balance it out because we are talking about a reversible process. An ideal reversible process this will always be true. So, even if we look that in a very minute scale this will remain same. Say dS is the entropy change in minute amount on the system plus dS the energy the entropy change on the surrounding will be equal to 0 if it is a reversible system. But what happens if we include the irreversibility what happens if the system is not reversible? That was the question posed by the scientist Clausius whom we already discussed for the second law of thermodynamics statement. So, what Clausius said that if it is a irreversible system and if this is happening then the overall entropy will increase that means dS of the system plus dS of the surrounding will increase in the irreversible system. Now if we compare and bound these things together we can say entropy change in minute amount for a system plus entropy change in minute amount on the surrounding is always greater than 0. Where the equal term defines the irreversibility whereas the greater than term talks about the irreversibility. And this particular equation is known as the famous the Clausius inequality. So, now we want to find out what is actually happening in terms of the heat exchange with this Clausius inequality. So, we already know the Clausius inequality says dS plus dS surrounding is greater than equals 0. So, we can just rearrange that and write dS is greater than equal minus dS surrounding. Now over there with this equation we know some heat exchange is happening. The heat is actually going to the system from the surrounding. So, that means dS surrounding I can write it as minus dQ by T. Again this minus sign signifies that heat is going from surrounding to the system. And as we just know once the heat is actually coming to the system it is positive. But now we are talking with respect to the surrounding. Now the surrounding we are talking about and surrounding is losing heat to the system. So, with respect to that we can say surrounding is actually losing heat. So, that is why this negative term is coming over here. Now if we all put this together equation this and equation this one we can say dS the minute change in the entropy of the system is greater than equal dQT because the two negative signs actually cancel each other out. And this is another way we can write the Clausius inequality. So, it says that the change in the entropy is always greater than or equal to the heat exchange during the experiment at that particular temperature. And this is actually set the field for understanding the change in the free energy. Now over here this heat transfer can happen in two different scenarios. First scenario will be the heat transfer at a constant volume and the second one will happen heat transfer at constant pressure. So, these are the two scenarios we can play with and find out what is the effect of entropy change with respect to the heat and temperature. Let us take a look into them one by one. First take a look into that heat transfer at constant volume. So, which means delta V is equal to 0. So, there is no other work is going to happen because everything is happening with respect to the change of the energy and heat. And if we look into the first law of thermodynamics it says E equal to Q minus W and over here W is nothing but P into delta V and our P delta V is equal to 0. So, this W becomes 0. So, we can over here write energy is equal to Q and we are writing QV to say that it is happening at a constant volume. Now if you look into a this minute amount of that a smaller value of that we can say D E is equal to D Q V. Now we already know D S is greater than equal to D Q by T and we are doing this at constant volume where this D Q becomes D E. So, from there we can rewrite further and write D S is equal to greater than equal to D E by T. So, just exchanging the D Q with D E over here and all those things this equation if I just rearrange it I can write T D S is greater than equal to D E or I am bringing the D E to the left hand side D E minus T delta S and putting a minus 1 multiplication altogether less than equal to 0. So, that is the function I am getting at this moment. So, change in energy minus temperature into the change in entropy will always be less than and equal to 0 equal to 0 when it is reversible process less than equal to 0 when it is irreversible process. So, that is what we got from the heat transfer at a constant volume we will come into that in a while. Let us go to the next part where we are going to talk about heat transfer at a constant pressure. So, over here we can say from the first of thermodynamics E is equal to Q minus W because over here it is constant pressure that means I can still change the volume. So, that means it becomes delta E sorry it should be delta E equal to Q minus P delta V or Q will be equal to delta E plus P delta V and we all remember delta E plus P delta V is nothing, but change in enthalpy. So, over here the overall heat change at a constant pressure is nothing, but the enthalpy change. So, we can write D Q P, P defines that it is happening at a constant pressure is nothing, but D H again going to the Clausius inequality D S is greater than equal to D Q by T. So, from there we can write D S is greater than equal D H by T again if I rearrange that we can write T D S is greater than equal D H or D H minus T D S is less than equal to 0. So, that is the function we found. So, it is the change in enthalpy minus temperature into change into the entropy is less than equal to 0 equal to when it is reversible less than when it is irreversible. Now, if we draw all those two factors together D E minus T del is less than equal to 0 it is happening at a constant volume and D H minus T D S is less than equal to 0 when it is happening at constant pressure. So, these things are actually giving me an idea when an reaction will be thermodynamically more spontaneous especially when it is irreversible less than equal to 0. However, we need to define these two terms separately and that has been done by using these two function. First function is known as A is defined as E minus T S which is known as Helmholtz free energy and over there it has been done at a constant temperature because from this equation we can write D A at constant temperature and constant volume is equal to D E minus T D S which is less than equal to 0. So, D A T V is less than equal to 0. So, it is giving me an idea when and reaction will be spontaneous. Similarly, when you doing that at a constant pressure this function is given by this function G is defined at H minus T S and G is known as Gibbs free energy and what does it define? It defined change in D G at a constant temperature and pressure then it becomes D H minus T D S which is less than equal to 0 or D G T P less than equal to 0. So, it defines when a reaction will be spontaneous at a constant temperature pressure if Gibbs free energy goes to lower energy. So, that is the mathematical functions that we have derived, but what is the physical significance of both this Helmholtz energy and Gibbs free energy that we want to discuss it further. First take a look into Helmholtz free energy what does it actually means? So, as we know we have defined E equal A equal to E minus T S. Now, if we want to find out the differences in change in the Helmholtz free energy from initial to final for a process we can write is delta A equal to delta E minus T delta S it is happening at a constant temperature. So, that is why T remains constant over there once we write it we know delta E is equal to Q minus W. So, delta E minus Q we can write is nothing, but minus W. So, that we keep in our mind. Now, for a reversible process what happens over there delta A equal to delta E minus T delta S for a reversible process this is nothing, but delta E minus Q why because as you remember D S is greater than equal to D Q by T. So, for reversible process is D S equal to D Q by T or del S is equal to del Q by T. So, that is the del Q is nothing, but the heat exchange. So, we write it as only Q and that is becoming this. So, T delta is nothing, but Q. So, that we have interior over there and over here we just find out delta E minus Q is nothing, but minus W that the work can be done by the system. So, we can write it is nothing, but minus W. So, what we can write that delta A is equal to minus W or minus delta A is equal to W which is nothing, but the maximum work possible for a system during a process is actually given by this term delta A. So, that is the maximum work possible and that is why A is known as maximum work function and the German term for work is Arbeid and that is why the term A actually is denoted over here for this Helmholtz free energy. So, Helmholtz free energy is nothing, but giving you an idea how much work can be possible in the maximum if you are using this function and as you know it is greater than or equal to. So, that is the maximum work possible. So, you can have less than work, but the maximum will be when it is a reversible system and that is what is defined by this Helmholtz free energy nothing, but the maximum work possible for a particular system. What happens for the Gibbs free energy? For a Gibbs free energy we can do similar calculations and find out delta G is equal to delta H minus T delta S that is again from this equation G equal to H minus T S and if you are doing that at a constant temperature delta G equal to delta H minus T delta S. Now, as we know at a constant volume sorry at a constant pressure we know delta H is equal to delta E plus P delta V. So, if we include that in this equation we goes to delta G is equal to delta E plus P delta V minus T delta S a little bit of reorganization bringing E and T S together delta E minus T delta S plus P delta V. So, now over here is delta E minus T delta S is nothing, but the Helmholtz free energy function delta A plus P delta V and we already know that delta A is nothing, but minus W the maximum work possible. So, with respect to that we can write minus delta G is equal to W minus P delta V where W is the maximum work possible whereas P delta V is nothing, but work done by the system due to the expansion because at a constant pressure we can still change the volume. So, it is expansion process and there I am losing some work done by the system. This is the overall work possible and this is the work I am losing by this expansion work. So, what is remaining is the maximum useful work possible is the maximum useful work doable other than the expansion work. So, Helmholtz function gives me the overall work possible it does not define what is the mechanical work what is the actual work whereas Gibbs free energy gives me the difference between the actual work minus the mechanical or the expansion work. So, actual useful work that can be used during a process from a system is given by delta G. So, that is why delta G or Gibbs free energy becomes a very unique quantity, a unique parameter because that defines us how we can find what is the overall maximum work we can use without doing the mechanical or expansion work for a system. So, that is the importance of the Gibbs free energy. Now, with respect to that we come into the system the concept of spontaneity. So, now we have discussed over here that the delta G is equal to delta H minus T delta S and it can be less than equals 0 and for a irreversible process it is less than 0. So, over there the entropy is actually increasing and according to second law of thermodynamics that is where my spontaneous reaction will be. So, that is why we define delta G should be less than 0 if I want to have a spontaneous reaction. Now, how the delta H and delta S factors into here? So, now there are 3 conditions first say delta H is negative and delta S is positive and delta G is a summation of delta H minus T delta S. Now, if is delta H is already negative delta S is positive and it is multiplied that with minus T. So, the overall delta G value will be negative. So, the reaction will be spontaneous. So, if enthalpy is actually less than 0 for exothermic reaction and entropy is increasing it is going to be a spontaneous reaction. What happened? Delta H is positive delta S is also positive, but in such a way that T delta S is actually larger than delta H. If that is the case now imagine delta H is positive delta S is also positive, but delta S into minus T is a huge negative term and magnitude wise T delta S is greater than delta H. So, the overall term becomes negative. So, delta G still remains negative and the reaction will be spontaneous. That is actually what is happening during the melting of ice that we have discussed earlier that, enthalpically it is a endothermic reaction delta H is positive, but delta S is so positive that at the temperature of melting it is giving a value of T delta S such that it is overcoming the delta H and it becomes a spontaneous process. How else we can actually achieve spontaneity? Say delta H is negative, delta S is also negative. So, in the case delta G delta H minus T delta S, so delta H is negative, delta S is also negative. So, T delta S is going to give me a positive value minus T delta S, but in such a way that the magnitude of T delta S is actually less than the magnitude of delta H. In that case what will happen? The delta H negative will overcome the positive term given by T delta S and delta G will still remain negative and the reaction will be spontaneous. So, over here we can find that it is a combination of both entropy change and enthalpy change come into the picture and then it actually defines whether the reaction will be spontaneous or not. It does not matter whether it is a endothermic or exothermic reaction and if we can play with the entropy term such that the overall delta G is negative then the reaction will be spontaneous. Similarly, a delta S positive always go for a spontaneous reaction, but a delta S negative that means a change in the disorder such that disorder is decreasing that always does not mean the reaction will not be spontaneous. If the enthalpy factor can cover up the difference of T delta S we can still get a delta G negative value and it can still support us with a system which can be spontaneous. So, with respect to that we will try to conclude this particular section where we will discuss in this particular section we have discussed the Clausius inequality and we talk about two different important free energy function one is the Helmholtz free energy function one is the Gibbs free energy function. Helmholtz free energy function is given as energy minus T S factor Gibbs free energy is given as H minus T S factor. Helmholtz free energy function has been developed from a constant volume calculation and the Gibbs free energy is actually derived from a constant pressure calculation. The Gibbs free energy gives us the actual work that can be found from a system removing the expansion work whereas the Helmholtz function give us the idea of the overall work possible to get from a particular system during a process. So, that is why delta G and delta A value always give us an idea which side the reaction is going. We generally follow the Gibbs free energy because it gives us mostly the work that can be done by a system without the expansion work or the mechanical work. So, that is why Gibbs free energy is widely used and that becomes very famous to us that the delta G negative value means a spontaneous reaction. We have also discussed how delta G is connected with delta H and delta S and what are the different combination are possible to get a delta G negative with respect to the delta H and delta S. Thank you.