Thermodynamics part 6

Added: 2026-05-16

[00:00:00]now coming to the next one that is enthalpy as i study about enthalpy enthalpy can be also defined in this way also that the thermodynamic quantity u plus pv is also called the heat content or the enthalpy of the system the thermodynamic quantity u plus pv is also called the enthalpy or the heat content of the system i said you already that enthalpy is nothing but known as a heat calculated under constant pressure we can say that under constant the pressure remains constant all right so uh in a particular cylinder and is it closed the pressure is constant isn't it clear so that is the enthalpy of the heat content of the system which is defined as nothing but h is equal to u plus pv we can say now you know let us say ah that is h1 because as i know that enthalpy is a state function it's not a path function it's a state function so hence i can calculate the initial enthalpy and the final enthalpy so initial enthalpy h1 h1 is equal to u1 plus pv1 now you'll see you'll ask me say why not p1 because pressure remains constant remember enthalpy p is constant so p is constant here it will not be p one p two if i'm writing p one p two it's variable how it is constant okay try to understand that try to understand everything here so next is this is your initial enthalpy h1 is equal to u1 plus pv1 final enthalpy h2 is equal to u2 plus pv2 final enthalpy now change in enthalpy delta h is equal to h2 minus h1 that is u2 plus pv2 okay minus that is u1 plus p v one we can say that okay so we can write it in this way u two plus p v two i can open the bracket and i can write in this way your u two minus u one you can see if i open the bracket minus plus p goes common v two minus v one u two minus e one is w and p v two minus v one is del v so hence i got it del h is equal to w plus beta understood now from where i got the formula i tell totally we'll derive it all right so del x we can write it as q p and this i can address q b i already said q p is equal to q b plus l v clear so now this is we got it now one more thing i want to tell you here that if the work is done in a cyclic process got it so in a cyclic process when the system returns to its original state after the number of changes then delta u or delta h is equal to zero all right in case of cyclic process cyclic means when the when the system returns to its original state after number of steps we can say that then delta u or delta h is equal to zero solution fine now that you must remember it in case of cyclic process delta u is also 0 and delta is also 0 and remember it work done is also 0 got it work done is also 0 now del u is equal to q plus w so if w is equal to 0 q is also 0 all right then work done is also will be 0 in case of cyclic process fine now we'll find a relationship between that is uh the reaction a particular reaction and take it between heat of reaction at constant pressure and that at constant volume what i said i will be calculating a relationship between ah heat of reaction at constant pressure and heat of reaction at constant volume that means in simple words i should say that i am i am going to calculate a relation relationship i am going to find out the relationship between q p and q v q p and qb all right i'm going to find out a relationship so to say qp i told you just not to you it is delta h and qv is nothing but delta u i said it already to you and we also know just now we came to know the formula uh we derived the formula rather delta h is equal to del u plus p del v we can say that clear it can be written as now that is when we are going to that is find out a relationship that is the condition should be ideal conditions the condition should be ideal condition because what i said you a reaction i am going to find a reaction and it is possible when the gas is an ideal gas the condition is an ideal condition where the initial volume is v one and the final volume is v two so when i say the condition is ideal condition then in that case i can write p v one pv is equal to n nrt so n one rt similarly pv2 is equal to n2 rp all right clear so now here that is uh volume and number of moles is definitely it is varying differently but because pressure remains constant in case of an ideal gas how it is idealized pressure means constant because what i said i'm going to calculate the relationship qp and qv all right qp and qb qp is heat calculate a constant pressure all right then you can say it calculate constant volume if it is constant volume that's v1 and v2 it is remaining constant then the work done will be zero then how will you calculate it so hence i took the condition as an ideal gas i took it as an ideal gas so where i am considering the pressure is constant our universal gas constant of course it is constant and temperature also is constant so to say and considering it clear i'm repeatedly saying you now here it is p v 2 minus p v 1 is equal to nothing but n 2 r t minus n 1 r t so p v 2 minus p v 1 is p v 2 minus v 1 here r t goes common it is n 2 minus n1 now for a particular reaction this n2 stands for that is nothing but that is here this is initial and this is the final that means this is standing for the reactant and this is standing for the product so here i can write np minus nr where p is a product and r is the reactant i've said you already n1 is the number of moles for the reactant and n2 is the number of moles of the product so here instead of n2 i have written np and instead of nr n1 i have written nr number of moles of product minus reactant so np minus nr delta n change in number of moles so it is change in number of moles of gas i have written it for gas why it is you will come to know g stands for here gas all right and this is your p del v we can say that p del v is equal to delta n g r t now we know the equation already we derived it just now del h is equal to del u plus p del v now when i put this value here so we get it del h is equal to del u plus delta n g r t del h is equal to so instead of del h i can write as q p is equal to q b plus delta n g r t and also i can write it down okay so many relations if i give it to you that is equal w plus p lb now i'm putting the value del h is equal to del u plus instead of p del v what do you got the value this is your p l v you got it isn't it the c is your p del p got it you put the value of it here delta n g r t where delta n g is nothing but change in number of moles of the gas when we calculate the change in number of moles now we calculate only for gaseous not for solid or liquid why because solids and liquids are not compressible compared to that of the gas they are less compressible solid is not at all compressible so design liquids to a certain extent but gases are highly compressible we can say that so hence though whenever there will be change in number of moles the change in number of moles will take place in case of gas only all right for example how to calculate the change in number of moles for example you see i am writing an equation h2 gas plus cl2 gas gives 2xcl gas so here delta n g all right is zero height is zero here number of moles of product is two and number of moles of the reactant is also two this is one this is one so one plus one it is two to minus two is zero done now i am taking another equation c solid plus o2 gas gives co2 gas let me calculate delta ng here so delta ng will be nothing but here 1 minus 1 it is also 0 you say sir this carbon it's a solid it's a solid so what i mean to say to you is that that if in a particular reaction liquid in solid is given in the equation that has to be neglected we will not consider the number of moles of liquid and solid will not translate we will consider only for the gaseous for that reason i have written it here very clearly i have written written g g stands for gas so i am going to consider the change in number of moles of gas only you can see very clearly this is one mole this is one mole so one minus one is zero got it i am not considering this the carbon solid it's just solid i'm not considering it to calculate the change in number of moles i've already said the reason so in this way what we find is that here that is delta energy stands for the change in number of moles of gases so this is the formula we got it this is a relationship we got it that is the relationship between heat of reaction at constant pressure and at constant volume got it at a constant volume we got this particular relationship so to say now when a reaction is carried out in a closed vessel remember it when a reaction is carried out in a closed vessel the volume remains constant remember it del v is also zero definitely when it is carried in a closed vessel volume is changing initial final remains same now so del v is also zero in that particular case so to say and the value of r it depends if it is a unit it is eight point three one four joules per kelvin per mole the value of r is in case of s i units so let's see but in other units it depends whether it is zero point zero eight two one or zero point zero eight three all right so zero point zero eighty one will be what liter atm per kelvin per mole and zero point zero eight three will be liter bar per kelvin per mole already we have studied all these things all right so this is uh about the relationship so to say i said thank you