 Right on 10 decimetre cube of O2, 10 decimetre cube of O2 at 101.325 kPa and 298 Kelvin is heated to 348 Kelvin, is heated to 348 Kelvin, 10 decimetre cube of oxygen O2 at 101.325 kPa and 298 Kelvin is heated to 348 Kelvin. Calculate the heat absorbed, calculate the heat absorbed delta h and delta u of the process at first question constant volume, constant volume and constant pressure, constant volume and constant pressure. CP is given for this. CP is 25.72, 0.013, it is temperature minus 3.86 into 10 to the power minus 6 d square. So this is CP, you need to find out heat at constant pressure and constant volume. We will see one thing and then we will solve this for just a small thing, you see the delta h is helping, it is defined as the heat content, heat contained, constant pressure, constant pressure, heat contained, constant pressure, constant volume and constant pressure. So for constant pressure, CP value is given which we need and for constant volume what we need? CV which is not given in the question. So first of all we need to find out CV to solve this question. Now how to find out CV, you see heat content of the system at constant pressure and then we can write it as dh is equals to du plus vdv. This we are writing down from the first law of thermodynamics which is d cube is equals to du plus dw, work done. Constant pressure if you take this dw is nothing but we can write du plus dp into v because work done is pressure into volume. First of all we can write it as du plus vdv plus vdv and this is nothing but d cube. Now if you take the condition of constant pressure because this is what we have at constant pressure. So this d cube becomes what? dh and dp is equals to what? 0. So when I substitute this to here dh is equals to du plus vdv. This is the mathematical definition of enthalpy. Now in this you see if I divide this by dt this is by dt and this is by dt. dh by dt is what? dh by dt is cp and du by dt is cv and this one is what? r. This becomes cp minus cv is equals to r. This is the relation of cp and cv. Right cp minus cv is equals to r. Correct? cp is given so we can find out cv here from this equation. Correct? This is one thing. Okay. Understood? Yeah so much. We get cv also instead of expression of temperature. Right? Now to solve this question first of all we need to find out the number of moles which is pv by rd. Right? Pressure volume r and temperature is given. Right? So this number of moles if you substitute all the values what is the value given? So 101.325 and volume is given 10 liter divided by r is 8.314 will take and temperature is 290 in Asia. When you solve this what is the n u again? 4 by 10. So 0.4 u again right? Approximately 0.4 u again. Okay? Now what we need to find out? At constant pressure we need to find out delta h. So delta h is equals to what? dh is equals to what? Ncp dt. Right? And cp is a function of temperature. That cp will substitute here and integrate this. Right? So dh is equals to n u let it be and cp is nothing but 25.72. This is 15, make it 50 and then use. Inside that I put this square. And when you integrate this what is the limit we have here? 298 to 340. So this will be 25.72 0.013 5 to 8 to the square minus t1 square. No, no, no, that's not about it. Okay? And this becomes minus t2 times 3.86 and then 1 minus 4. No, I think you can do it. Okay. So this is like that. Sir, last time you said we said we have to do it. Sir, this is high 70 for you. That's a 10. 600 across, actually, that's it. Okay. Method you understood. Yes. You won't get this kind of expression in the exam. Okay, that complex calculation here. Okay, but you should know the method how to do it. Okay? So whenever the cp is given in terms of temperature, so we'll write it down like this, integrate, and we have to put the value of temperature, we'll get the answer. Right? The answer this h you will get approximately around 604 or something. When you solve this, don't do it, let it be. Sir, do you eyeball it and say 604? No, sir. No, I don't know. Delta less than enthalpy is this. Method is this. Right? Now, if you find out this delta u, then what we have to do? Delta h is equal to delta u plus v delta v. Right? Delta v we need to find out. Okay, we'll see one more thing with this. Okay, I see here. Delta h is equal to delta u plus v delta v we have. Yes. And p delta v, we can also write it as delta ng rt. Delta ng rt, can we write this? That's what states of matter, right? NG. Yes. No, no, wait, I don't know if you can say that. But it's what you mentioned. Suppose we have for a, for a gaseous reaction. I remember that. That's it. For a gaseous reaction, what we'll write? Tv1 is equal to N1 rt. N1 is the number of moles of gaseous particles. Tv2 is what? N2 rt, number of moles of gaseous particles. Okay? So any reaction if it is supposed to be given, like suppose I write down one reaction here. 3. We didn't do it in the same way. I'm going to take an example later on. So when you subtract this, let's say we have p delta v. And here we'll get rt, N2 minus N1. Who will do it? Okay, this is for reaction. Like in this case, we are going to use this. So p delta v is nothing but minus of this. Minus of this is also the workup. So we multiply minus sign both sides. So what we can write? Tw is equal to, or delta w is equal to minus delta ng into rt, where this delta ng is equal to N2 minus N1. Can you write this? Delta ng, N1, N2 is the number of moles of only gaseous particles. Right? It depends on the reaction. No, we'll come to this question. But this is the general thing I'm giving you. Right? It's a concept that we have. This we use, this we use the expression of organ we use this when reaction is given. Okay, so you need to find out delta v into that. Right? So the expression, this delta h, this expression becomes what? Delta h is equal to delta u plus minus delta v. Why p delta v is the only one? The delta ng into rt. This is another expression of delta h and delta u that we use when chemical reaction is given. We only take gaseous particles. Okay? Right? Now, two more things you can understand here. See, work done is work done delta w is equal to minus delta ng into rt. Now, if delta ng is less than 0, then work done is what? Positive. Greater than 0? Positive? Positive means work done on the system. I'll tell you what happens. Work done on the system. Now, for example, you see delta ng less than 0. If I write this one, 3 or 2, gas and it converts into 2 or 3. This reaction. So for this reaction, delta ng is what? 0. N2, that is 2 minus this. Which is less than 0. 2 minus 1. Understood this? The delta ng thing, it depends upon what reaction you have. Okay? That will ignore. We'll see that. Understood this? When chemical reactions is given, then delta w we use this expression to solve the question. Like in this case, delta ng is this. Now, suppose if I write down here, if delta ng is greater than 0, it means what? Delta ng greater than 0. It means work done is less than 0, work done negative. It means what? Work done by the system. So just to find out delta ng, you can say whether work done by the system or on the system we have. For example, if I say this one. Expansion. CSU3. Solid. When it is associated, it gives you what? CSU3. Solid plus CO2 gas. Can you tell me delta ng for this one? This is one. This is solid to ignore this, to ignore this. So 1 minus 0. Oh yes sir, this is like expansion because there is a gas being involved. Yes sir, you can say that. The point is with this delta ng, you can understand where the work done by the system or by the system. Number of work done by the system. It is not a reaction, no. Delta t is given. It is not a reaction. No sir. Nr delta t. Nr delta t. Got it? Because the reaction is not given in the question. Then w is equals to what? Nr delta t. If reaction is given, then only we use this. Okay delta ng. Okay. So these two conditions we write down. 1 and 2. By the gas we have already done. In this case, expansion. And in this case, contraction. Number of moles for this n value in the product line decreases, contraction decreases. Right on next question. Calculate the work done. Right on calculate the work done. When one mole of zinc dissolves in hydrochloric acid. Calculate the work done when one mole of zinc dissolves in hydrochloric acid at 273 Kelvin in an open vessel. 273. Zinc dissolves in hydrochloric acid. Atmosphere. Atmosphere in juice what we have produced. Always write down the answer in juice. Atmosphere in juice. Atmosphere in juice. Atmosphere in juice. Pressure will be the atmospheric pressure. Pressure will be the atmospheric pressure. Just need to find out delta v. You see here the reaction is Zn with ZCl. It gives you ZnCl2 plus H2 gas. Right. This is the only gas we have here. And if you see the condition temperature is 273 Kelvin. Pressure is 1 atmosphere. It means is the condition of? Sir, one second. Sorry. Go ahead. This is the only gas we have here. Condition of STP right? So, one more of H2 gas we are getting. So, this gives you what is the volume here? 22.4 litre of STP. Right. So, work done is what? Minus p delta v. Volume is given. Right. Pressure is 1 atmosphere. Okay. You can find out easily. So, you will do with a litre ATM. And then you have to convert this into? June. Right. What is the volume initially we have? Sir, all you do there. All you do there. Delta NGR. Wow. Delta NGR. Yes. That also you can do. Correct. Yes. Because it is not directly significant. Condition is that only. Delta NGRT also you can do. Right. This way also you can do. Or you can also do Delta NGRT. Okay. So, this is the only gas we have here. Right. This way also you can do. Or you can also do Delta NGRT. What is Delta NGRT value for this? One. One. Because this is the only gas. One minus zero. So, one into RT value you can substitute. 8.314 into temperature. No, sir. No, sir. We want back to RT. Yeah, we want back to RT. RT you substitute 8.314. Directly you will get in June. Yes. Right. So, in this case you do not have to convert this into a litre ATM. Pressure volume gives litre ATM. This gives you directly in June. Okay. So, that also is very important. When you are using this expression. Right. So, this comes in June. Yeah. Right. This also you have to take in June. Sir, but this is not in June, sir. No, sir. Why? Zero 8.314 if you take in June. Delta NGRT is what? See, the point is if you take it in litre ATM. Then also you can do. If you have to find out this in June. Directly you put 8.314. Zero 8.314. All these things you take in.