 Now, let us take a closer look at entropy generation due to internal irreversibilities. First thing that we talk about is mixing. Mixing is a highly irreversible process. So, we can take for example, let us say we have a container like this which is fully insulated. So, let us say initially there is a partition and we have gas on side A and gas at a different state on side B. So, we remove the partition, allow the gases to mix. Notice that since the vessel is insulated, there is no entropy transfer. Integral 1 to 2 delta Q over T is 0 because it is insulated. However, because of the mixing, there is going to be internal irreversibility and entropy generation. So, if I take this to be my system, entropy change in the system is positive because sigma int is positive as a result of mixing. Now, in some cases, you may not have the insulation on the vessel, there may be some heat loss to the surroundings in which case delta Q over T is negative, but sigma int is positive. So, you may ask the question whether delta S can become 0. If you recall, this is the expression for the change in entropy of the system. So, in the case of mixing, sigma int is positive and if I allow heat loss to the surroundings, then delta Q is negative. So, this is negative, this is positive. Is it possible that we could end up with a situation where delta S is equal to 0? Theoretically, yes, but in reality probably not because sigma int is very high in positive in the case of mixing. So, in most situations, even if you have loss of heat to the surroundings, entropy of the system will still increase. And it is possible that generation of entropy due to the internal irreversibility may be in some cases offset by the loss of heat to the surroundings. Now, work does not appear in this expression at all. So, entropy change is due to entropy transfer plus entropy generation due to internal irreversibility. Since work does not appear here in this expression, there is generally a misconception that work does not contribute to change in entropy of the system. That is incorrect because work can still contribute to entropy change through the sigma int term. For example, stirring work. So, let us say that we have the same sort of situation. We looked at this example before. So, let us say that we have a vessel which is again insulated. Let us say that we have a stirrer. So, we stirred the contents of the system. So, let us say this is my system. Since the vessel is insulated, there is no transfer of entropy delta Q equal to 0. However, because of the stirring, the thermodynamic state of the system changes and stirring being a highly irreversible process, sigma int will be large and positive. And here, there is only work transfer, nothing else. So, you can see how work transfer can also cause change in entropy of the system because mixing is highly irreversible. Sigma int is large and positive. Large and positive. So, entropy change of the system is also large and positive. So, work can still influence entropy change of the system. And the same goes for electrical work. You may recall that we had an example like this. So, if I take this to be my system, then I am transferring electrical work into the system. Remember, there are two ways of defining a system for this. One where we treat the interaction of the system with the surroundings as transfer of electrical work and one where we treat the interaction as a transfer of heat. So, if we write it like this, then this is transfer of electrical work and ohmic heating of course, is highly irreversible. So, again, if you treat it as electrical work, the first term is 0 and sigma int is, when the interaction is treated as work, then the first term is 0, second term, sigma int is large and positive. So, the overall change in entropy of the system is also positive as a result of that. On the other hand, if we treat this as a heat interaction, then delta Q is positive. So, that means this can be the increase in entropy of the system can be considered as result of entropy transfer. Sigma int becomes equal to 0 in that case. In both cases, entropy change can be calculated accurately. It is only a matter of how we account for it. In one case, we account for it through ohmic heating which is highly irreversible. In the other case, we say that heat is transferred. So, entropy is transferred to the system and as a result, its entropy increases. So, it is the manner in which we account for it, either the first term or the second term. Now, displacement work in the absence of friction and any other internal irreversibility, if there is no internal irreversibility, friction is also absent, then sigma int is 0. Then, if the system expands, its entropy will decrease and if it is compressed, then its entropy will increase. Now, if the process is carried out adiabatically, then the first term in this expression is also 0. Already, we have said that it is internally reversible, sigma int equal to 0. That means the second term is 0. If you carry out the process adiabatically, then the first term also becomes equal to 0. So, there is no entropy change as a result of displacement work, if carried out in an adiabatic and reversible manner. If it is carried out in an adiabatic and irreversible, internally irreversible manner, then sigma int will be non-zero whereas the first term, entropy transfer term will be 0. So, you can understand the importance of this equation and the value that this equation brings, although we may not be able to exactly evaluate sigma int. We have not really given an expression for evaluating sigma int that is not required because this expression by itself allows us to make a lot of influences on the entropy change of the system, which are very, very useful. Entropy change itself can be calculated in many other ways. What we are mostly interested in, generally, is entropy change of the system as a result of its interaction with the surroundings and internally reversible is if there are any. So, this is a very, very powerful expression and for each problem, you can actually try to see which one of the terms on the right hand side is 0 and from that infer whether entropy is going to increase, decrease or remain the same. So, when you do the calculation, it serves as a consistency change. Now, in an adiabatic process, since sigma int is greater than so, if you see delta s equal to integral 1 to 2 delta q over t plus sigma int. So, in an adiabatic process, delta q is 0. So, that means this term becomes equal to 0. Since sigma int is greater than or equal to 0, delta s can be positive or remain the same. So, for an adiabatic process, the entropy can increase or remain the same. It cannot decrease. That is very, very important. For an adiabatic process, for just for an adiabatic process, the entropy can increase or remain the same. For general process, as we wrote down before, the entropy of a system can increase, remain the same or decrease for any general process. So, this has to be borne in mind. What we are saying is for an adiabatic process, the entropy can increase or remain the same because the first term goes to 0 and the second term is positive or equal to 0. Now, if you have an adiabatic and reversible process, then the second term is also equal to 0. In fact, what happens here is more than saying this, we can actually say the following. ds is equal to, if you write it on a per mass basis, ds equal to delta q over t plus delta sigma int, delta q is 0 because it is adiabatic, delta sigma int is 0 because it is reversible. So, we get s2, the initial and final entropy remains the same, but we also see that the entropy remains the same throughout, ds is equal to 0. That means entropy remains the same, it is not only the same between the initial and the final state, it is also the same throughout the process. That is very important, that is an additional concern. There may be many processes for which the initial and final entropy are the same, but the entropy changes during the process. For an adiabatic reversible process, the entropy is the same in the initial and final state, it is also the same, it remains the same during the process also. It is a very special process and it is called an isentropic process. So, we will have more occasions to make use of the isentropic process as a sort of an ideal process. We can think of it as an ideal process under certain conditions for certain types of process. So, what we will do in the next lecture is figure out how to actually calculate the entropy change value, calculate a value for the entropy change. So, this equation tells us qualitatively what the entropy change for the system is, whether it is increasing, remaining the same or decreasing. What we will do in the next lecture is actually try to derive expressions that we can use to assign a numerical value for delta S. This expression only tells us positive, negative or equal to 0. What we will do is try to derive expressions for calculating the numerical value for delta S.