 All right, friends, so today I am going to take a different kind of problem, okay? So you're going to see this kind of question many a times, okay? So there are one set of problems in thermodynamics wherein you are asked to find the specific heat, right? So as we know that for the gas specific heat depends on what kind of process you are following, isn't it? So specific heat at constant volume and specific heat at constant pressure, you will get two different values, all right? And similarly there can be many different kind of processes, right? So practically I can have any kind of process, all right? I can say that I am following a process in which temperature cube divided by pressure square is a constant, all right? So that is neither constant volume process nor constant pressure process, all right? So similarly today we are going to take a generic process for which pressure into volume raised to power n is a constant, fine? And I am going to find out the specific heat for this process for which PV raised to power n is constant, all right? And don't confuse this with the adiabatic process. For adiabatic process PV raised to power gamma is constant. Gamma is a fixed value for a gas, gamma is ratio of Cp and Cv, right? But here I have taken that it could be anything, need not be the ratio of specific heats, all right? So let's try to derive the specific heat for this particular process, all right? So if specific heat is C, then if you supply dq amount of heat, it should follow this equation, right? dq should be equal to nc dt, all right? And one more thing we are assuming while solving this problem is that we are dealing with ideal gas, okay? So basically I am trying to find out the specific heat for an ideal gas which follows this process for which the process equation is PV raised to power n is constant, all right? So continuing from this dq should be equal to nc dt, okay? And according to the first law of thermodynamics, dq should also be equal to change in internal energy plus this small amount of work done, which is PDV, okay? So dq is nc dt, right? And du doesn't matter what process we are following, du will be always ncv dt, right? Plus PDV, okay? Now using this particular relation, you can say that specific heat should be equal to cv plus p divided by n dv by dt, okay? So I'll leave it here and then try to see what else I can do, fine? So first of all, I have an ideal gas equation which is PV is equal to nRT, fine? So if I differentiate with respect to volume, okay? I'll get p plus vdp by dv is equal to nR dt by dv, okay? So if I rearrange the terms, I will get pdv plus vdp is equal to nR dt, fine? So this let me call as equation 2, fine? Now I have a process equation also, so I can utilize that as well. So PV ratio power n is constant. Now for this equation also, I will try to get a relation like that, okay? So basically what I'm trying to achieve is that I use these two equations. So when I differentiate this, let me first differentiate, it will be clearer than. So v ratio power n dp by dv plus nv n minus 1 into p will be equal to 0, isn't it? So if I rearrange the terms quickly, if I divide by v ratio power n minus 1, I'll get v over here and vdp plus n times pdv is equal to 0, all right? Now you can see that using equation 2 and equation 3, if you use these two equation, you can get the value of pdv, the value of pdv in terms of dt. What that will help us is that we'll substitute that value instead of pdv over here and then I'll cancel out the dt from left-hand side and right-hand side, okay? So then I'll not get an expression like this, fine? So let's quickly do that. So if I subtract third equation from the second equation, okay? I'm going to get this. 1 minus n pdv is equal to nR dt, right? Because vdp will get cancelled away when I subtract, okay? So pdv will be then equal to nR divided by 1 minus n dt, fine? Now this value of pdv, I'm going to substitute over here, okay? So when I substitute there, this is what I'll get. Let me use some other color over here. So I'm going to get ncdt is equal to ncvdt plus nR divided by 1 minus n dt, okay? So I can cancel away dt and number of moles as well, all right? So I'm going to get c is equal to cv plus R divided by 1 minus n, okay? And from here, cv I know is R divided by gamma minus 1, right? The gamma is the ratio of the specific hits, constant pressure and constant volume, all right? The c will be equal to this plus R divided by 1 minus n, okay? So this is how you derive the expression for the specific heat for this particular process, okay?