 Welcome back. Well, we have derived an expression for S2 minus S1 for an ideal gas with constant specificity. Let us see how that expression looks in the state space. Let us sketch first a PV diagram for an ideal gas and to get our bearings right, let me sketch two isotherms. This is one isotherm, let us say this is another isotherm, T1, this is T2 greater than T1. How will the isentropic line look like? Isentropic lines are steeper than isothermal lines. So, for example, one isentropic line could go like this, another isentropic line could go like this, making some sort of a, you know, a crooked twisted parallelogram between two isentropic lines and two isotherms. So, let us say this is S1, this is S2. If I say S2 is greater than S1, you should check that out, why it is so. Then let us sketch this the other way on a TS diagram. Now here, how will constant volume lines and constant pressure lines look like? Let me first sketch constant pressure lines. The constant pressure lines are almost like exponential curves, e to the power something x that will be reminded of that function. So, these are two constant pressure lines P1, P2 greater than P1. The constant volume lines are sharper than these, they are steeper than these. For example, one constant volume line could be like this, another constant volume like could be like this. Let us say this is VA line, this is VB line. One thing to remember that for any system, if you sketch the PV diagram, take one isotherm and one isentropic line, they will intersect each other just once because the value of S and T defines a unique state. While sketching, do not make that mistake of letting an isentropic line and isothermal line intersect each other twice, just once. Similarly, on the TS diagram, a constant pressure line and a constant volume line will also intersect each other just once. I will leave it to you as an exercise to determine whether this S2 greater than S1 is right and which one of these is higher, VB or VA. Thank you.