 of heat capacity. Calculation of heat capacity. Write down, it is a heat required heat capacity definition. It is a heat required to change the temperature by heat capacity. Write down, it is a heat required to change the temperature by. Temperature by 1 degree Celsius. Total heat capacity is C. The unit of C is heat capacity. Unit is, we represent this by C. And unit of C is joule per Kelvin. Because change in temperature we have. This is total heat capacity. So this is actually total heat capacity. See, now this total heat capacity we have two times. One is molar. Another one is specific heat capacity. What is molar heat capacity? It is the heat capacity per unit. For unit mole. For one mole. Write down, it is the. Write down, it is the. Molar heat capacity is 2 times Cp and Cv at constant pressure and constant volume. Specific heat capacity is represented by s. ms delta t. mc delta t. mc delta t also you write. Specific heat capacity also you write it. Because we are using this for C. So here we are writing it as ms delta t. Point is unit you must take care of. If per gram if it is there then it is specific. For mole then it is. Write down molar heat capacity. It is the amount of heat required. Yeah, it is the amount of heat required to change the temperature by 1 degree Celsius for one mole of a system. So molar heat capacity we can write this as cm. Specific heat capacity is s. So molar heat capacity the unit of this is joule per mole kelvin for one mole it is. And this is for one gram. So joule per gram per kelvin. OK. Specific heat capacity is for one unit gram. For my gram should it be. There should be kg. We define it as per gram. If you convert this into kg then the value will only change. Then like it should be s. Actually we define it as per unit mass. Per unit mass. Gram cannot be properly. Per unit mass is specific. Per unit mole is molar. That would be better. Per unit mass. So unit of mass you can take anything. Gram, kilogram anything. OK. Now write down in isothermal process. First one I don't know in this only. Isothermal process. What is isothermal process? Delta t will be 0. Right. So there is no change in temperature then. Then molar heat capacity will be what? Q is equals to we can write c delta t by definition. What is c here? Total heat capacity. And this is given by q delta t means q is what? q is the energy or heat required. And this is change in temperature. So heat capacity is what? It is a heat required energy required to change in temperature by what dimensions? So this is the formula of this. q is equals to c delta t. Where c is the total heat capacity. If it is molar heat capacity we can write this as n c m delta t, n number of moles, m s delta t. m is the mass for molar heat capacity and specific heat capacity. So s is the specific heat capacity, c m is the molar heat capacity. So now when you have isothermal process. Isothermal process delta t is equals to 0. And when delta t is 0 c is what? delta t is 0. So q by 0 is it? So isothermal we have c. See first of all we are taking this condition of isothermal process. What happens? You are providing heat equal amount of work here to temperature will be constant. That is what it is. So how the temperature won't change. The energy that you are providing it. So point is this is the condition we are taking. Because what is happening? Delta t will be 0 c will be infinity. In adiabatic process what happens? Adiabatic q is 0. Delta q is 0 and when delta q is 0 c is 0. Sir of that s is specific heat capacity, c m is molar heat capacity. So m is mass. m is this m. This is mass. Because per unit mass we have this s is per unit mass. Isochoric process right down next. Isochoric process. Isochoric is volume. So in Isochoric process the heat capacity we write it as c v. So right down in Isochoric process the molar heat capacity is c v. Now c m becomes c v. See there are many terms into this. If you don't understand this properly it will always haunt you in the center. Sir what did you say? I what I said c m is the molar heat capacity. Now I am considering Isochoric process. So in Isochoric process volume is constant. So molar heat capacity in Isochoric process we write it as c v. We will see that. First of all you write down. In Isochoric process the molar heat capacity is c v. Write it down. And by the basic definition of this c v what we can write? Bq by dt right. This is the energy required to change the temperature by 1 degree. And this is at what? Since we have v here. So this is the constant volume. Now from first law of thermodynamics when constant volume is there then work done will be what? 0. Work done will be 0 first law of thermodynamics. Work done will be 0. So dq is equals to what we have? dq is equals to what we can write? dq is equals to d u plus d w. First law of thermodynamics. So when volume is constant this becomes 0. So dq is equals to d u. So d u I can substitute here. And then also it is fine by the system on the system. I a dq is equals to d v i. So dq I can write here as d u. So c v is equals to what we can write d u by dt. So this we can write it as what? d u is equals to c v dt. This is for one mole. Correct? Because molar capacity. So for n number of moles what we can write? d u is equals to n c v dt. Generally if you see in the book they are derived like this only. But here it is not very sure that it is applied for only ideal gas or with rigid boundary. So the previous expression that we did there we have taken the condition of ideal gas and rigid boundary. So this is like this also we can derive this particular thing. So d w is 0 because volume is constant. And work done is what? p delta v. So volume is constant so work done is 0. Now write on next the last one we will take a break after this isomeric process. Isomeric process is what? constant pressure. So like we have constant volume c becomes c m becomes c v. So your constant pressure c m becomes c p. Yeah what happened? Isomeric process write down the molar heat capacity is c p. And definition of enthalpy we haven't done but we will do it. So what we can write c p is equals to again the basic definition is what? d q by dt at constant pressure. What is the definition of enthalpy? It is a change in energy at constant pressure. Enthalpy of any system it is a change in energy d q at constant pressure is nothing but d h. Enthalpy definition we will see. See all these things you need to Now we are understanding the topic and terms. So we have to remember this. What is enthalpy? It is a change in energy at constant pressure. That is the definition. Mathematical definition of d h is what? d h is equals to d u plus p delta v. This is mathematical definition. This equation is mathematically defined like this. But theoretically it is the change in energy at constant pressure. So we will see this. So at constant pressure d q becomes what? d h. So c p is equals to d h by dt at constant pressure. So d h is equals to what we can write? c p dt. For n number of moles d h is equals to n c p dt. We will do enthalpy and then we will see this expression. That is the same for all of them. So in this we will see how this enthalpy is related with pressure and volume temperature. We will see that. All of them are equal to the number of moles in terms of the capacity in terms of change in energy. Yeah, because we have only constant pressure so we will write c p and not c p. But the expression is equal to d h and d q. d h and d q we will do that after this we will do that. But if you want you can write this as the same. d q is equals to u plus d delta v. So constant pressure we will write this as the same. Just a second. Float is what d q is equals to d u plus d w. So constant pressure. So d q is equals to d h. So constant pressure. So d u is equals to d w. And d w is equals to d p into v. Work done to p into v. Pressure to constant. So we can take this out. So d h is equals to d u plus p d v. Okay. Unless you are d h is equals to d u plus p d v. But we will do this. We can take this out. That is the definition of enthalpy. Enthalpy is the change in energy at constant pressure. Definition. So taking a look at enthalpy first. Okay. Right on the question. Find the work done when 18 ml of water is getting vaporized. 18 ml of water is getting vaporized. At 373 kelvin in open vessel. In open vessel. Assume ideal gas will be available. Number of moles is not given. There is one mole. It is not given in STP. So this is 18 ml. So 18 ml. So it is one mole. Correct. See one mole you have to find out volume. Because what you need to find out work done right. Work done is what feed head does. Pressure is given. Open vessel. So pressure across there. That is constant. So just you need to find out final volume. 18 ml is given. Find out V and then find out P delta V. Oh and we can find final volume. Yes. Dv goes to M. So what is R? R when volume is written. So 0.08. See pressure is going at most. Volume is written. R when you should be 0.08 to 1. Which is 1 by 12. 1 by 12. You can assume that. Approximation you get. So this is 30. 30. 30. 30. 30 point. Yeah. It is almost. Approximately 30, 31 you will get. So P delta V will be minus V into delta V is 30 minus. 18. 18 ml we have. So this is also 30 ml. No 30 ml you are getting. It should be liturized. 30 liters. 30 liters. So 18 ml and 30 liters you have to subtract 18 ml from that. So just you neglect that. So 30 you are getting. What is the final volume you are getting? Around 31. Around 31 right? So walk done is what P delta V? Correct. So walk done is equals to see first of all you need to find out PV is equals to NRTV alumni. Initial volume I is given which is 18 ml. Open vessel we have so pressure is nothing but 1 atmospheric equals to the atmospheric pressure. Right? So this number of moles is 1. R is 0.0821. Temperature is 373 Kelvin. It is equals to pressure into volume or pressure is 1. So 1 into V. At most can you give the glittery tip. So what is final volume here? Approximately. 31. 31 you are getting. Approximately. Right so walk done is equals to. Minus P delta V. Right P is given that is 1. Delta V is 31 minus 18 into 10 to the power minus 3. You can neglect those depending upon the option what is given. So approximately the answer will be. 0 to 100. The unit of this is what? 80 liters. This is 80 liters. Right? So depending on the option you can change it to 2 cells. Minus 31 into 101.