 Welcome back. We are now going to look at properties of a simple compressible system. We all know that a simple compressible system means a system containing either a solid or a liquid or a gas. One possibility is that our system contains only a solid or only a liquid or only a gas. In that case, we will call it a single phase situation. On the other hand, it is possible that we may have a combination of the three distinct phases, either solid and liquid or liquid and gas or even solid and gas, sometimes all three together. At least two of these three are present, in which case, we call it a multi-phase situation. In this lecture, we look at this situation, the single phase situation. Whereas this situation will be considered later when we consider properties of steam. So, let us restrict ourselves to a single phase situation. Now, remember that for any simple compressible system, we need any two properties to determine the state. The choice of the pair is with us. We can use temperature and pressure or temperature and volume or pressure and volume. Actually, any two properties, thermodynamic does not put any restriction on that. Put some minor restrictions, which we will see later. Then, once you decide, for example, if you decide on temperature and pressure, then all other properties, thermal energy, entropy, specific volume, enthalpy, any other property you define, will be considered to be a function of temperature and pressure. If you consider T and V independent, then pressure will be a function of temperature and volume. Thermal energy would be considered a function of temperature and volume. So, would be the enthalpy, so would be the enthalpy and so on. Which pair to choose is left to us or because of convenience. Let us consider the thermal energy U. And let us consider T and V as the independent variables. Why T and V? Because it turns out that this is a convenient pair when thermal energy is the property considered. Now, since U is a function of two variables T and V, we can expand the differential of U in this form. Partial of U with respect to T at constant V into dt plus partial of U with respect to V at constant T into dV. And now we notice that the first partial derivative here, this is defined at CV. So, we can write this as CV dt. We do not have much to do with the second derivative. But we should notice that the second derivative will in general be nonzero, but will be zero for an ideal gas. Let us consider another property of interest, the enthalpy H. Now, here let us consider H to be a function of T and P. Why T and P? Because this is a convenient pair for H. Again, the way we expanded the differential of U, let us expand the differential of H. So, partial of H with respect to T at constant P dt plus partial of H with respect to P at constant T dP. And then we know that the first derivative we have already defined at CP. And hence we can rewrite this expression as dH equals CP dt plus the second term. The second term has no specific name, but we notice that this term will be nonzero in general, but it will be zero for an ideal gas. Now, the ideal gas is a spatial case of a simple compressible system. And we know that because of Joule's law, the thermal energy of an ideal gas is a function only of temperature. And so is the enthalpy. And hence for an ideal gas, we should remember that the variation of U with respect to V at constant temperature will be zero. And also the variation of enthalpy with respect to pressure at constant temperature. Actually, if you consider this to be an ideal gas, then the variation of U with respect to even pressure at constant temperature would be zero because U is a function only of temperature. Similarly, the variation of enthalpy with even specific volume would be zero because enthalpy is a function only of temperature. Another spatial case, the other extreme, see low density gases form a spatial case known as the ideal gas. Similarly, liquids and solids are almost incompressible. If we consider them to be really fully incompressible, we end up with an idealization known as the incompressible liquid or incompressible solid. Now for such a system, the number of two way work modes is zero because you can do nothing with them. You cannot compress or expand them and they are simple systems so there is no other work mode. Consequently, there is only one property which will decide the state and that property obviously will be temperature because remember any system which is of interest to thermodynamics as a consequence of zero law will have a property associated with it called temperature. And hence, any property other than temperature will have to be a function only of temperature. So specific volume would be a function of temperature, specific thermal energy U will be a function of temperature, specific entropy would be a function of temperature. We have not studied entropy as yet, just consider it as some other property. However, once you remember that enthalpy is a special case and an expression. That is because of the definition of enthalpy itself. Enthalpy is defined as U plus PV. We will notice that because U is a function only of temperature, V is a function only of temperature. These two components do not change with temperature but pressure is directly present here. And because of this enthalpy turns out to be a function of both temperature and pressure even for an incompressible liquid or an incompressible solid. Later on when we consider relationship between thermodynamic properties in detail, we will be able to show another relation and that is for such systems Cp and Cv lose their distinction. They have the same value and hence we may simply call them specific heat. Thank you.