 No, this is the question. Yes, what is the answer? Yes, see the question is 2.8 gram of N2 gas at 300 Kelvin and 20 atmospheric was allowed to expand isothermally against a constant external pressure. That is one atmospheric. Okay. So obviously when expansion is there. So volume will increase, correct? Could you tell me what is the initial volume? P1, V1 is equals to N1 RT1 we can write initial condition is this right initial volume is you can substitute all the values here. Number of moles would be 2.8 divided by 28. R is 0.0821. Temperature is. Temperature is 300 Kelvin. And this entire thing is divided by and this entire thing is divided by the pressure which is 20 atmospheric initially it is given. When you solve this you'll get initial volume and initial volume is 0.123 liter. Approximately. Now when the expansion is there. Against this pressure. So again we can write P2 V2 is equals to N2 RT2. And RT is same. It is not changing. So volume final volume would be 2.8 divided by 28 into 0.0821 into 300. And this whole thing is divided by divided by 20. Sorry, one atmospheric. Okay. Pressure is one atmospheric. You'll get volume and the volume is 2.463 liter. So this is the expansion to change in volume. So what is the work done in this process? Could you tell me? Work done is minus P external into delta V minus P external. We have that is one delta V is 2.463 minus. 0.123. This is a change in volume we have. So when you solve this again you will get minus 2340. Atm. Which again you can convert it into Joule. And that would be minus 236.95 Joule approximately. Okay. We need to find out delta you since the question it is given that it is an isothermal process. So delta you for isothermal process. We know it is zero for isothermal process. And then we can find out delta Q it is expansion. So work done by the system. Okay. So Q is equals to or delta you is equals to Q plus W we can write. We can substitute the value. So Q is equals to plus 236.95. Joule understood. So this is how we find out nothing. So expansion was there all conditions were given. We when we do will find out first once you have even we do you have pressure you can find out work done and then delta you isothermal process zero and hence Q we can find out using first law of thermodynamics. Okay. Now, when you have PV graph given. Okay. Then the work then would be what AD under curve that you know. Okay. Look at this question here. This is suppose volume B. I'll give you some value. This is two, four, six, and this is eight. This is volume given in leader one, three, four pressure in ATM. Right. The process is A to B, C, D and A. You need to find out work done in this process. Total work done minus 18 you are getting right so we'll find out the will find out the area under curve right so work done in PV graph is nothing but area under. The curve only in PV graph it is only in PV graph area under curve. So what is the area here length into breath right rectangle so length is six and breath is for right. So three, so area is nothing but the work done so area is we have three into six so we have 18. Okay. The unit is ATM leader here, which you can convert in Julie if you want easily. Okay, ATM leader, but so the next is here you see work done total area though you can find out simply, but total work done would be what if you see whether it is negative or positive and what we'll do. We'll have work done in A to B, plus work done in B to C, plus work done in C to D, plus work done in D to A. This is the entire thing. So B to C and D to A you see it is a constant volume process so work done would be zero here in these two process constant volume A to B and C to D you see the expansion is taking place. Against the constant pressure for so minus four into change in volume is six. Plus, this is zero. This would be C to D, we have compression against one. So minus one into two minus eight that is minus six. So minus 24 plus six we are getting minus 18 ATM leader. So when you solve like this step by is you will get the sign. Okay, whether it is negative or positive, but area under the curve if you find out area under the curve if you find out, then we'll have only the magnitude, not the sign. Okay, minus four into delta we know so eight minus this eight minus two six six into four is 24. Work done formula is minus P external into delta V in the formula itself we have minus P minus P external into delta V. Correct. So here you'll get the sign and the sign means the work done by the system because it's negative. So work done by the system. Okay, work done by the system. So what happens here, we have 18 ATM leader. Okay, whenever you have cyclic, sorry, clockwise process, right, you see it's clockwise. So when the cyclic process is clockwise, then it is always work done by the system. So what you do area you find out you'll get the magnitude. And then since the process is psych is clockwise, so you have to put a negative sign over here. Okay, and that is what the answer is. So what is the, you know, final conclusion here. If, if clockwise process is there, if the process is clockwise, then it is work done by the system by the system. If process is anticlockwise work done on the system, these two things you have to keep in mind. Okay, whenever you have graph, PV graph, find out area, if it is clockwise, put a negative sign, if it is anticlockwise, put a positive sign over there, and that is it. Understood. Yeah. Okay. Okay, can we move on next. So this kind of questions you will see clockwise anticlockwise process, you must remember. Now the next is next time you write down enthalpy represented by H. Mathematically, H is defined as the sum of the sum of internal energy and pressure volume work done. Internal energy and pressure volume work done. Mathematically it is defined like this. It is also defined as the heat content enthalpy is the heat content of system at constant at constant pressure. So if you look at this expression here of FLOT, first law of thermodynamics, du is equals to dq plus dw. So this is equals to dq, dw we can write minus dpv. So this is equals to dq plus dq plus minus pdv plus vdp, pdv plus vdp. Okay, when the pressure is constant because our condition is what constant pressure at constant pressure what happens this term would be zero. So what we can write here you see du is equals to dq plus sorry minus pdv at constant pressure at constant pressure. Okay, so what is dq here, dq is equals to du plus pdv. And that is what we have here if you see this expression. du is equals to du plus pdv is nothing but dh. If you see here dh is equals to du plus pdv at constant pressure. So here if I write down one more steps if I write down here at constant pressure at constant pressure, dh is equals to du plus p. So this is dh and this is nothing but dq over here. So from the first these two expression what we can write dq the heat at constant pressure is nothing but dh. This is what the mathematical definition of enthalpy we have it is the heat content of the system at constant pressure. Okay, done. Okay, now you'll see like we did for internal energy enthalpy also. Yes, h is the enthalpy and it is also the function of last part okay the last part. Second last step second last it is nothing we just see dq is what du plus pdv and du plus pdv from this expression is nothing but dh. And dq we can equate this to that's what we did here. That's why the enthalpy is nothing but the heat content of the system at constant pressure because all these things we have applied at constant pressure the condition we have taken at constant pressure. Correct. So this is the thing we have. Now you'll see enthalpy is the function of function of pressure and temperature. It depends on pressure as well as the temperature mainly. Okay, it depends upon the number of moles also and volume also to some extent, but for closed system, the number of moles and volumes are constant so mainly these two factors we have which affects the enthalpy. I list here on what we can write dh is equals to again doh by doh p differentiation with respect to one variable keeping other constant constant into dp plus doh by doh t keeping p constant dt. Okay, this is the formula we have correct. This is the formula doh by doh p at constant temperature. This is cp and for any number of moles you see dh is equals to ncp and cp dt plus doh by doh p at constant temperature dp. Okay, this is also true and valid for all the processes, valid for all kind of process. This is in general formula we have. You must have seen the formula of dh is equals to ncp dt, isn't it? Have you seen that formula? dh is equals to ncp dt. Right, but this formula is the actual formula of dh for all process. This is applicable for all process and dh is equals to ncp dt. We'll apply one condition and we'll get that relation. Did you understand till here? Yes, done till here. Okay. Now, condition you've seen this is the expression we get. If you apply the condition for ideal gas here, for ideal gas, we have h is equals to u plus pv. This is the expression we have of enthalpy. u is a function of temperature, internal energy and pv for ideal gas is nothing but nrt for ideal gas. So this pv also we can say it is a function of temperature. Okay, so overall we can say h is a function of temperature. Right, h is a function of temperature. It means the differentiation of enthalpy with respect to pressure at constant temperature equals to zero. Yes, and this if you substitute in that general formula, we'll get dh is equals to ncp dt, ncp dt. Right, why? Because this doh by doh p at constant temperature, we have substituted here in this expression. This entire term is zero and dh is equals to ncp dt. Right, this dh is equals to ncp dt. It is valid for ideal gas only, for ideal gas. For ideal gas, all processes. It's not like only constant pressure, all processes because we did not apply any condition of constant pressure over here. Okay, whether pressure is constant or not, we can apply this. If the pressure is constant, suppose it is given in the question, the pressure is constant, then obviously this expression would be zero. This dp would be zero and when this dp is zero, the entire expression is zero and again dh is equals to ncp dt. So it is applicable for constant pressure as well, but at delta h or dh is equals to ncp dt is applicable for all processes. So this is one thing. Now in chemical reaction, how do we use this expression? You see this. In continuation, you write down what chemical reaction what happens for chemical reaction. Any chemical reaction, it takes place at constant pressure and temperature. Okay, so generally what happens, generally chemical reaction takes place at constant pressure and temperature. Constant pressure and temperature. We have constant pressure. We know at constant pressure, q is nothing but the enthalpy that is delta h. And if the constant volume is there, at constant volume, q is nothing but the change in internal energy delta u. W, if you see, work done is equals to minus pdv integral. Two, from initial volume to final volume, vi to vf, initial volume to final volume. So pressure is pdv is what we can write. We can write pdv is nr dt or nrt we can write. So if you can write the nf final number of moles only for gases. This is only for gases. R into t minus ni, initial number of moles only for gases, r into t. This we can do. Vf minus vi is this. So this would be equals to minus one negative sign is already there. So this would be equals to minus nf minus ni for gases, rt, which is equals to minus delta ngrt. pdv is minus delta ngrt. Because we are applying pdv is equals to nrt here, in order to find out the value of pdv, pdv is equals to nrt is applicable only for gases. The chemical reaction generally takes place at constant temperature and pressure. This is what the definition of enthalpy or at constant temperature heat content of the system is nothing but the enthalpy change. It's not like I got it. I have done it already. I've just written it over here. Similarly, this also we have done delta u. So pdv is equals to delta ngrt. So the expression of delta h would be for chemical reactions, gases or species delta h is equals to delta u plus delta ngrt. This is the formula we have. We can use this formula for chemical reaction, chemical reaction. Delta ngrt will calculate only for gases. Once again, Sudeesh. Delta ngrt is number of gaseous product minus the number of gaseous reactant, number of gaseous product and gaseous reactant. Look at this example here. Suppose we have a reaction that is 302 gas gives 203 gases. So delta ngrt for this reaction would be 2 minus 3, product minus reactant, that is minus 1. So delta ngrt is what? Delta ngrt is less than 0 for this reaction. And when delta ngrt is less than 0, work done is less than 0, work done is negative. Work done is negative means it is work done by the system. Is it clear? Because you see, work done is what? This expression you see, work done is minus pdv, which is nothing but minus delta ngrt for gaseous reaction. Delta ngrt is negative. Once again, one mistake I made. Delta ngrt is negative. No. So negative negative becomes positive. So work done is positive. Work done is positive means work done on the system we have. One correction. Right? Delta ngrt is negative. So negative negative positive. So work done on the system. Next you see, if you have this reaction, N2 gas plus 3H2 gas gives 2NH3 gas. So for this, you see delta ngrt out of the reaction like this, CaCO3 we have solid, gives CaO solid plus CO2 gas. For these reactions you see delta ngrt is greater than 0. Right? Delta ngrt is greater than 0 means work done is negative. Work done is negative means work done by the system. Okay? When you have negative, it means work done by the system and when system is doing work, it is a case of expansion. When work done on the system means we are doing work on the system. So it is a case of compression. Done? One note you write down. In case of change in state, in case of change in state, the change in volume delta B is almost 0. There's no change in volume when there's a state change there. We have one exception in this. That is the melting of ice. In this case, what happens when H2O solid, that is ice, is converting into H2O liquid. Right? Ice, what happens in case of ice, in case of ice, we have open cage-like structure. Open cage-like structure. We have some void vacant space over there in ice. That's why density is less and it can float on water. Okay? It's an open cage-like structure. We have some void space present over there. It's not like, you know, solid we have from inside. Right? And we have, when this ice melts, we have hydrogen bonding present in liquid, right? So due to hydrogen bonding, what happens? Due to hydrogen bonding in water, volume decreases. This is the information sort of things. You should keep that in mind. Okay? Usually when there is a change in state, we don't observe much change in volume. Delta B is 0 almost. Right? But in case of ice and liquid, we have open cage structure due to hydrogen bonding in water, volume decreases. And it is a case of compression. It is the case of compression. And compression means work done on the system. Right? So it is work done on the system. Hence it is positive. Okay? This one information you must remember. See, when there is bonding, right? When there is bonding, there will be a compression, right? Molecules are attracting, no, HOH. These molecules are attracting because of this bonding, right? So usually the kind of expansion that we observe, we don't have in this case. We don't, you know, observe the expansion. Actually, we have almost similar volume. Change in volume is not that great. Okay? But because of this attraction, hydrogen bonding, we observe some compression in this case. Ice to water. And that's why we say it is compression and work done on the system. It is not getting compressed. Okay? Relatively, we are talking about, means the volume that we are expecting, we don't get the exact same volume, but it is slightly lesser than that. So ice is converting into liquid, but the volume is not that great according to our expectation. That's why we say in this process, there is a sort of a compression we have. It means work is done on the system when you convert ice to liquid. Okay? So this is for enthalpy and the relation. We have important relation delta H, delta U and delta NGRT, very important relation we have that you must keep that in mind. Now, so we'll take a break now. And after the break, we'll see some more relations here, which is important for the other concept that we are going to see. So we'll take a break now. We'll resume at 6.30. Take a break guys.