 So, we have seen first law of thermodynamics for a cyclic process as relatively straightforward for a non-cyclic process, we wrote this as delta E equal to q minus w. And basically delta E is nothing but as we saw delta U plus delta P E of the system plus delta K E of the system. We know how to calculate delta P E of the system is basically m times G times change in elevation with respect to a datum. And delta K E also we can easily calculate it is half m v square or half m v2 square minus v1 square. So, that is also easy. Now, what we want to do is how do we calculate this? How do we calculate change in internal energy of a system which is associated with the molecular motion inside the system? We need a framework for calculating change in internal energy of the system and the framework is what we will discuss next. So, that is provided by this notion of pure substances. So, we have to say certain things about the material or the matter inside the system so that we can calculate the internal energy and change in internal energy of the system. So, that is the framework that we are going to look at next. So, the definition of a pure substance goes like this. A pure substance is a thermodynamic system that is a uniform in composition. What we mean by composition is that the relative proportion of the individual elements that make up the substance. So, in other words this has so many atoms of carbon, so many atoms of hydrogen, so many atoms of oxygen and so on. So, that is the relative proportion and that should be uniform everywhere in the system. So, that is uniform so that it is uniform in composition and B uniform in chemical aggregation. What we mean by this is so many atoms of oxygen or combined with so many atoms of hydrogen in a particular way or the atoms must be combined in the same manner. So, the number of atoms must be the same that is A, B they must be combined in the same way because depending upon temperature and other things the same set of elements may be combined in different ways to get different compounds. So, what we insist on in B is that it should be uniform in chemical aggregation. So, basically in a given system when I draw samples from different spatial locations. So, I may have let us say a vessel like this filled with something and this is my system. So, if I draw a sample from this corner or this corner or this location or some other location the samples that I draw must all have the same composition they must all have the same chemical aggregation. So, if that is satisfied then we say that the substance that is inside the system is a pure substance. You may recall that we said something very similar when we talked about macroscopic approach. We said that for macroscopic approach to be valid when I measure pressure at this point in the system this point this point or this point anywhere in the system pressure, temperature, any other property it should all be the same irrespective of where I measure. So, now in the definition of pure substance what we are saying is the composition and chemical aggregation must be the same irrespective of the spatial location in the system. So, once that is satisfied we can then define whatever is inside this as a pure substance. Let us now take a closer look at this definition through a series of examples it is best understood through a series of examples. So, here we start with liquid water. So, basically let us say we have a vessel like this this contains. So, this is our system. So, we look at each the system being comprised of each one of this in turn. Let us say that the system consists entirely of liquid water which means that the proportion of the hydrogen and oxygen atoms or 2 to 1 sorry is 2 to 1. So, we have for every 2 hydrogen atoms we have one oxygen atom that is A. So, A is satisfied. Now 2 hydrogen atoms are combined with the single oxygen atom in the same manner anywhere in the system no matter where I draw the sample from they are combined in the same manner. So, B is also satisfied chemical aggregation is also satisfied. So, it is a pure substance and we make the observation that it is a single component in the liquid phase it is water single component in the liquid phase. Now, let us say that you know the system or the vessel consists entirely of water but in the way performed. Then also the proportion of the elements is the same for every 2 hydrogen atoms we have one oxygen atom. So, A is satisfied and the hydrogen and oxygen atoms are combined in the same manner 2 hydrogen atoms are combined with one oxygen atom. So, aggregation is also same. So, B is also satisfied and this is a pure substance single component in the gaseous phase. Now, let us say that the vessel is filled with nitrogen you can easily see that the composition is the same no matter where I draw the sample from I always get 2 atoms of nitrogen the proportion is the same. And no matter where I draw the sample from 2 atoms of nitrogen or joined in the same way. So, both A and B are satisfied this is simply a gas later on we will say that this is actually an ideal gas that idealization will come later on for now we recognize the factor this is a pure substance and it happens to be a gas. Now, let us say that you know the vessel is filled with air instead of a single component gas let us say that it is filled with air which means oxygen and nitrogen in this proportion for every kilo mole of oxygen we have 3.76 kilo moles of nitrogen. So, irrespective of where I draw the sample from I will always have 1 kilo mole of O2 and 3.76 kilo moles of N2. So, the proportion is always the same. So, that means A is satisfied and oxygen atoms 2 oxygen atoms combine to form O2 molecule 2 nitrogen atoms combine to form an N2 molecule and they are separate. So, these are two components. So, the chemical aggregation remains the same oxygen atoms are combined in the same way nitrogen atoms are combined in the same way everywhere in the system. So, aggregation part is also satisfied. So, the only difference between this and this is that instead of being a single component gas we now have a mixture of gases, but it is a pure substance it satisfies both A and B and the requirement A and B. Now, let us take an interesting example let us say that the vessel is filled with let us say liquid water here and water vapor here. So, here we have. So, let us say we have vapor here and we have liquid here. This is a pure substance that is what we are going to try to answer. Now, if I draw a sample from this part, then I get water in the liquid phase. Now, if I check requirement A that is proportion of different elements for every two atoms of hydrogen I have one atom of oxygen regardless of whether I take it from here or take it from here I have the same proportion of elements for every two atoms of hydrogen I have one atom of oxygen and now let us check for requirement B. So, if I draw a sample from the liquid side then two atoms of hydrogen are combined with one atom of oxygen. If I draw a sample from the vapor side I notice that again two atoms of hydrogen are combined with one atom of oxygen. So, aggregation is also the same irrespective of whether I take it from liquid or vapor. So, both the properties are satisfied. Only differences they are in different phases. One is in the liquid phase, other one is the vapor phase. So, it is a two-phase mixture. So, we saw single component in the liquid phase, single component in the vapor phase, single component gas, mixture of gases. Now, we have two-phase mixture. Both are, it is a pure substance because both are two phases of the same substance namely water which is why the requirement A both requirement A and requirement B are satisfied. And this is the pure substance that we will encounter very, very frequently. You remember we talked about working substances we said gases and two-phase mixtures of water and water vapor or refrigerant liquid and refrigerant vapor. So, this is something that we will encounter quite extensively and it is a pure substance. So, two-phase mixture is also pure substance. Now, let us look at something different. So, let us say now our system is filled with something like this. So, we have, so now it is filled with 2 kilo moles of hydrogen, 1 kilo mole of O2 and water vapor. So, let us check for requirement A. So, if I draw a sample from any location, I get the same proportion of hydrogen, oxygen and water vapor. So, A is satisfied and once again hydrogen atoms are combined in the same manner respective where I take it from oxygen atoms are combined in the same manner and two hydrogen atoms are combined with oxygen atom in the same manner. There is no change there also. So, the aggregation is the same. So, this is simply a mixture of gases containing three different components hydrogen, oxygen and water vapor. Earlier we had a mixture of two gases, here we have a mixture of three gases. So, B is also satisfied. So, this is also a pure substance. Now, let us say that we have this mixture in our, so this is liquid water and this is nothing but H2 comma O2 in this proportion. So, now if I draw a sample from the liquid side, I see that for every two atoms of hydrogen, I have one atom of oxygen. If I draw a sample from here, I have the, I see that I have for every two atoms of hydrogen, I have one atom of oxygen. So, property A is actually satisfied for this case. However, if I check for requirement B, then I notice that if I look at the sample that I have drawn from the liquid side, then two atoms of hydrogen are combined with one atom of oxygen. Whereas, if I draw it from the gas side, I notice that two atoms of hydrogen are not combined with oxygen. Hydrogen appears to be separate, is a separate gas, oxygen is a separate gas. They are not aggregated in the same way. So, which means that property requirement B is not satisfied in this case. Now, if I change the 2 kilo mole of hydrogen, so this is not a pure substance, if I change the 2 kilo mole of hydrogen to 1 kilo mole of hydrogen, then even A is not satisfied because if I draw a sample from the liquid side, two atoms of hydrogen, for every two atoms of hydrogen, I have one atom of oxygen here that is not the case. So, even the requirement A is not satisfied in this case. So, this is also not a pure substance. So, in order for us to calculate the change in internal energy of the system, the system must be comprised of a pure substance. That is the framework that we require in order to calculate the change in internal energy of the system. That is what we had said earlier. So, if you recall, we can calculate change in PE, change in KE very easily. Now, to calculate change in delta U, we now require that the system be comprised of a pure substance. This is a very, very important requirement. Once we have this, we can then evaluate, we can then go on to evaluate changes in property of each one of this type of pure substance. How do we calculate change in internal energy of a single component gas? How do we calculate, say, change in internal energy of a mixture of gases? How do we calculate change in internal energy of a two-phase mixture? So, these are things that we will take up in the next lecture. So, we will see how to calculate properties of ideal gases and ideal gas mixtures first in the next lecture.