 Okay, that's how you see this formula is true P into DTSP where it is. No, no, no, you see. One second, I'll go back. One more, right? This one. This one. No, it's not this internal precedent. This is the is formula. You see. This is the is formula, which we use to differentiate when a given variable depends upon more than one variable. I suppose y is equals to fx if you have then what will do D, y, y, dx directly will write to FDAS X. But why an X only one in 21 independent variable and one dependent variable. If you have more than one independent variable, then we use Euler's formula to differentiate. In this, what we do, do you will differentiate with respect to one independent variable keeping the other variable constant. This expression means differentiation of you with respect to T keeping the constant. Same thing. Differentiation of you with respect to be keeping T constant into TV. This is the formula we have. We have another variable. Then we'll write doh u by doh p keeping the another variable constant into D like this will add. Okay. So you don't, you know, you don't have to put some put and don't have to understand this actually it is not required because not it is not in this levels. Right in engineering college once you study engineering mathematics know, there you will study all these things I lose formula, and then no. It's simple. Domains partial differentiation. That is it. Partial derivative. We are differentiating you with respect to T keeping the constant. It means this partial derivative to doh u by doh t. Nothing else. Yes. So domains partial derivative. So we had discussed q, w and you based on this will see the first law of thermodynamics first law of thermodynamics based on what? In short, we read this float first law of thermodynamics. It is based upon conservation of energy conservation of energy. DQ is the heat supplied or heat, you know, released. Simple you need to balance the energy. And that is it. So, okay. So you have already done this first law of thermodynamics question that we have simply I'll quickly explain this and then we'll move on. See you have initial internal energy of any system that is you I we have state one to state to the system is going. The final state we have initial internal energy is you I and final internal energy is you. You I and you. So change in internal energy is equals to you F minus UI. This is a change in internal energy. Right. And this can be done in two different way either you provide heat to the system, or you do some work on the system or system will do work. Okay. The change in internal energy is possible in two different way. First is heat flow into the system into or out of the system that's there will be heat exchange. So internal energy will change the second way is what either work done on or by the system on or by the system. Okay. So suppose what happens first case I'm taking here heat absorbed heat absorbed and work done work done on the system heat absorbed and work done on the system. Okay. It means what you F is the final internal energy we have is equals to UI initial internal energy heat absorbed so plus Q it is given to the system and work done on the system that is also positive so plus W. So you have minus UI is Delta U so Delta U is equals to Q plus W. So if you see this expression it means work done by the sorry on the system we have. If it is on the system, Delta U is equals to keep the stuff. If it is by the system heat absorbed and work done by the system, then it is minus W over here. Second case is heat absorbed and work done. Yeah, one second. One done by the system. So everything is same work done by the system is negative I'll go back one second. So it will be equals to Delta U equals to Q plus instead of W we have minus W which further we can write Q is equals to Delta U plus W. So this is the expression clear. This is the first law of thermodynamics based on the energy conservation. So look at some question will see no question number seven only all of you are getting be it is eight into point three zero into 101. Okay, so all of you are getting be so I'll go with me. This question you see I'll write down here one question. You have a piston cylinder system piston cylinder system expands against against a constant pressure. Constant pressure of one atmospheric external pressure is this from volume four liter to 14 liter. In this process, it absorbs 800 Joule of energy. Find out Delta U change in internal energy. So answer you got minus 213.2 close. Yeah. So first of all, we'll find out the work done right in this expansion process work done will find out and work done equals to W is minus P delta V. So this will be minus one delta V is 14 minus four. That is minus 10 unit is ATM leader. This is the energy is given in Joule, we have to convert this in Joule and that would be minus 10 into 101.3252. Okay. Now this delta you because this expansion whenever you have cases of expansion it is work done by the system work done by the system expansion means work done by the system. Work done by the system the relation we have delta you is equals to delta Q. Plus W. Work done by the system is negative so negative we are taking here so positive sign here we have so Q is 800 plus 1013.252. When you solve this you'll get approximately 213.7 Joule. This is the answer of internal energy, not this one. We haven't done this question number 14 you try. We'll do question number 13 also after some time. Try 14 first. Then 14. 15 can you do 15 try once. Question number 15 what is the answer 14 what you got all of you. Answer is A. It is pressure volume work done this area you need to find out 13 also will do after some time. Question number 15 you are not getting. But I'll do 15th one. See in 15th one what happens is simple questions not that difficult. Okay. Two identical containers A and B with frictionless piston contain the same ideal gas at the same temperature and the same volume. Right same ideal gas same temperature same volume. No words D is not correct. Okay. The mass of the gas in a is MA and that of B is MB gas in the cylinder now allowed to expand isothermally to the same final volume to be a change in pressure and be given then mass relation we need to find out. So first of all, can we write the pressure of a in for gas a. No, B is not correct. A is equals to we can write any by any RT by B. Yeah, so D is not there B is not there. Then C and a two options were left sees an RT by B. Okay. Two identical containers and we with frictionless piston contains this same temperature same volume to be an RT by B. And if you represent in terms of mass, because mass relation we need to find out MA RT by M V capital M is the molar mass. Right so same ideal gas we have so same molecular mass also we have. Right. Similarly, PB is equals to what we can write MB RT by MB. This is not what an order you are not reading the question also properly you see same gas, same ideal gas. Correct. Now expansion till the volume becomes to be this to be will substitute here. New pressure would be what it is P dash a is equals to M a R to M V and P dash B is equals to MB RT by two M V. Now the last time the change in pressure in a and B are found to be delta P. So a in a we have delta P and delta P is equals to what we have this minus this. So half minus this it would be minus MA RT by two MB. Can you write this just to subtract this to subtract this to you'll get this. Similarly for be it would be 1.5 times delta P, and that would be the same minus MB RT by two MB. Can we take the ratio of the two. Can we take the ratio of the two this entire thing will get cancelled so we'll have two by three is equals to MA by MB, which is nothing but two MB is equals to three M. Option C is correct for this one. Now the next is, we'll discuss enthalpy, and then again we'll do some question enthalpy is represented by H capital H enthalpy it is mathematically defined as H is equals to you plus PV internal energy plus pressure volume work done you plus PV correct. It is what it is the heat content, it is the heat content of the system of the system at constant pressure, heat content of the system at constant pressure. Okay, constant pressure is the condition we have. How do we get this expression you see D u is equals to we know it is DQ plus DW first law of thermodynamics FLOT. So DQ plus DW is PV which further we can write DQ plus PDV plus VDP. Here, since we have constant pressure, so DP would be zero. So D u is equals to DQ plus PDV, the relation we have when we have constant. Okay, so since we apply the condition of constant pressure, so heat content of the system. Once again, okay guys we'll take a break now there's some drilling going on okay. Okay, we'll take a break now we'll resume at 625 okay, we'll resume, by then we'll be finished. Yeah, take a break we'll resume at 625, 620 we'll resume guys, 620 we'll resume okay, take a break yeah.