 Next is the increase of compression. Again we have two cases here, when same volume, final volume and same final pressure. So two graph I am drawing here. This is P, this is D, final pressure is same, this is PF, case one, when final pressure is same and the other one. Can you see this? Who is singing what? What do you say? How do you do just that? No, they are both are smart. Controversial, you know what I am saying? Other tell me, which graph is this? This is the slope thing. Top one is adiabatic for both. So this is isothermal, slope also you can compare. And this is adiabatic. This is adiabatic. And this is isothermal. Do you see this? It was now. The lower graph is now isothermal. This question is here. Which graph represent which gas? Before, before, again before. What is the reason for the intercept? When gamma is higher than the slope will be less. No, gamma is higher means more smoke. Minus also we have this. Minus gamma P by P. So good link. So then the last one is what? Last one is what? Helium. See first of all when the graph intercepts it means all these are adiabatic for adiabatic process. It is not isothermal. It is adiabatic. Now isothermal and adiabatic if you compare this. This is expansion. So upper one will be isothermal. This is what we did. Isothermal is what? PV constant. And here is what PV? Gamma constant. Here you see PV to the power one we have. And here the gamma is what? So when the power of V increases the graph shifts downwards. That also we can conclude. So for monatomic gas the gamma value is what? 1.66 for diatomic 1.40 and for helium it is 1.33. So when gamma value increases the graph shifts downwards. It means the lower graph is for? P4 is 1.33. That is reversed. So this is P4. And this is N6. Right? It means this graph is for maximum gamma and that is for helium. And this is code 2. This is P4. Question right now guys? 2 moles of an ideal monatomic gas. 2 moles of an ideal monatomic gas. CV value in bracket write down monatomic gas after that write down CV in bracket. 12.55 joule per mole per Kelvin. 12.55 joule per mole per Kelvin by close. Expands irreversibly and adiabatically. Irreversibly and adiabatically from an initial pressure of 1.013 Pascal. 1.013 Pascal against a constant external pressure of 0.1013 Pascal. Until the temperature drop from its initial value of 325 Kelvin to a final value of 275 Kelvin. So how much work is done? And what is the final volume? Irreversible adiabatically. 20,000. Final volume 17. What is the unit? Everything. Because 17 is the immediate length. 17,000. 723 milliliters. 720 milliliters. Oh yeah, that's right. You are getting? Yeah. See, first of all, work done we need to find out. So any adiabatic? What is the process we have? Adiabatic irreversible. So work done is delta u. Adiabatic process. And this is equal to what? NCV. NCV, delta t. And it's given 2. CV is also given 12.55. And t2 minus t1 is what? 3. 275, right? So this is the... That's just 1255. 1255. 1255. 1255. 1255. 1255. 1255. 1255. So work done is what? So it's negative 1255. So because given a, we should have pictures. What's it? It's plus, it's minus. Yeah, it should come negative. Anyway, it's work done by... It should be negative. So this is work done. Now the initial volume, we can find out what we can write. Work done minus 1255 is first equals to minus p external into delta v v2 minus v1. This v2 we need to find out. So this will be 1255 divided by p external plus v1 is equals to v2. And v1 is like... This v1 is not given. So v1 is equals to what we can write? v1 is equals to nr t1. So v1 we can find out from this number of forces to r value is... Okay, major. 275. 275. 275. 275. 275. No sir, t that is 225. 225. 225 divided by... Pressure is what? From this you'll get value of volume, substitute here p external you already know, substitute you'll get p2. See while calculating work done you don't have to put negative sign anywhere. Right? Right on the form right is minus p external delta v. But when you solve like this whatever answer you'll get here, sign. According to that you can say whether it is worked on by the system or on the system. You don't have to put negative sign or positive sign, okay? Next question you write down. So v2 expression will be 125 divided by 1.013 plus v1 is 2 into 8.314 into 325. But the p external is 0.015. 0.015. What is the xn of 0.015? Is it 0.015? 0.015. 0.015. 0.015. 0.015. 0.015. 0.015. 1 is right. That's right. Right. So when you solve this you'll get final volume. Sir this is a liter. Here it is whatever. This is the method. I don't know what answer you'll get here. From here the answer will be liter. This is personal. Whatever answer you'll get the unit will be liter. Okay? Okay? Next question write down. Calculate the work then when 1 mole of zinc calculate the work then when 1 mole of zinc dissolves in 1 mole of zinc dissolves in 1 mole of zinc In the other case In an open beaker and close beaker in an open beaker in an open beaker In a B.M. & N. Close beaker and push me Xn E press 0 Then the fielder Right on this question, an ideal gas, an ideal gas, an ideal gas, an ideal gas, an ideal gas, an ideal gas, expands against the constant external pressure of the amb device, an ideal gas expands against the constant external pressure of the two atmospheric from 20 liter to 40 liter. to atmospheric from 20 litre to 40 litre, constant external pressure at 2 atmospheric from 20 litre to 40 litre and absorb 10 kilojoule of energy from surrounding and absorb 10 kilojoule of energy from surrounding, 10 kilojoule of energy from surrounding, what is the change in internal energy? Do it fast. The external pressure is given and volume is given, find out what that means. And 10 kilojoule of energy is the value of Q, use first law then what I am saying. In this case it will be minus bigger, work is expanded. First of all, what you did? 10,000 minus 40. Work is atm litre. That you need to convert into 10 kilojoules. See the work that you find P, that the unit is atm litre that you have to convert into joules. See the formula is this delta u is equals to Q plus w that is what you need to do. When you use the delta v formula here, the unit will be atm litre, but this Q is given in kilojoules. So this you have to convert into joules or kilojoules. So the Q is 10 into 10,000 joules plus external pressure to delta v is what? 40 minus 20 minus p delta v. What is this 80 or 80? This is 201. That is 20 to 40. What is this? Isn't it past 20 or past 20? It's 20 to 40. That you can convert into joules. It's too much. 6,000 approximately. Because you said kilojoules I had to convert into joules. I thought it was joules. So yeah, it's correct. So it is 10,000. Minus what we get? What is this value? 40 into this? Like somewhere more than 4,000 you get here. Approximately 4,000. Is it 590? 590, right? So around 6,000 joules you will get here. You can actually use 590 something it is given. Exactly. 590, 480 it is given. So always keep that in mind. In the end you have to convert into joules.