 So we had discussed the Van der Waal equation and Van der Waal equation is p plus a n square by b square into v minus nb is equals to nRT. n is the number of moles, a is the pressure correction factor, stands for intermolecular interaction, b is the volume correction factor, stands for the size of the gaseous molecule, more size, more will be the value of b, okay. Now we'll see next the calculation of compressibility factor in different condition, in different condition. So for all these things we are assuming one mole, so n value is one here, okay, so first condition is at low pressure and moderate temperature, moderate temperature. So we have this Van der Waal equation that is p plus a by b square because n value is 1 into v minus b is equals to RT, okay. Now we have low pressure, it means the volume is high because volume and number of moles sorry volume and pressure is inversely proportional. So at low pressure what condition we can apply, vm is high and when the vm is high then we can take the approximation that is vm minus b is almost equals to vm where we are assuming, b is negligible in terms in comparison to vm and we can take this approximation. Now we substitute this here and we get p plus a by vm square into vm is equals to RT, we need to find out z, okay. So we'll multiply vm here, so it is pvm plus a by vm is equals to RT, we'll divide RT both side, so it becomes pvm by RT is equals to 1 minus a by vm RT. This is nothing but the compressibility factor, so z is equals to 1 minus a by vm RT. So under this condition low pressure and moderate temperature, okay. The value of z we are getting it is less than 1 and we have seen this condition when z is less than 1 it is easier to compress, once again, oh it should be vm this one, wait let me see what exactly qualifies as moderate temperature low pressure. No actually it is not actually defined moderate temperature and all, when you get the questions they'll mention the same term that moderate temperature, low temperature or high temperature because for different different gases we have different different value, okay so we cannot define this, it's not a general term, okay so yes, yeah so this is the first condition we have, okay. Now the second condition you see at high pressure and moderate temperature, high pressure and moderate temperature, so when the pressure is high we can take the assumption here a plus p plus a by vm b square is almost equals to p, we can neglect this term here because the pressure is high here, right and then we substitute this in Vendor-Ball equation, so p into vm minus b is equals to RT, pvm minus pb is equals to RT, divide RT this side so it is pvm by RT is equals to 1 plus pb by RT, okay so the value of z here 1 plus pb by and that is what the value of z we require, so under this condition, so under this condition the value of z is greater than 1 and when it is greater than 1 it is difficult to compress but easier to expand, repulsive force dominates the attractive force here that also we can write easier to expand, the last condition you see for the calculation of this the value of z compressibility factor that is c at low pressure and high temperature, here we are taking very low pressure actually at very low pressure high temperature, so for the low pressure we have the same condition so because of this we have vm is high and hence we can write vm minus b is almost equals to vm, okay and because of high temperature when temperature is high then we can write ke is high, kinetic energy is high, with high kinetic energy the velocity of the gaseous particle is high and what we can write then interaction is less, interaction is less or we can say negligible and hence we can equate a to this almost equals to 0 because of less interaction or no interaction, so because of this all these things what we can write p plus a by vm square is almost equals to p, this two condition if you apply into Wendervoll equation you will get p into vm is equals to RT and hence p vm by RT which is nothing but g, z here is equals to the Wendervoll compressibility factor is equals to 1 and this is the condition of ideal gas behavior you have, remember in the beginning I said that any gas at low pressure and high temperature behaves as ideal gas and this is because the value of compressibility factor here it is coming out to be as 1 and hence the, then guys no we don't consider all those inaccuracy because the exact thing we won't calculate okay those inaccuracy also we are you know it's not that great so we are neglecting that okay you won't get any question where you have to take the actual value or exact value of z so that is not required at all without approximation you won't be able to calculate it otherwise the calculation will be a bit tough very tough in fact okay you required all the value of a b and everything so that calculation will be very tough that's why we don't do it do that okay look at this question here try these questions all four done okay you see the first question is the internal pressure loss of 1 mole of Wendervoll gas that is real gas over an ideal gas so we know that this p ideal p ideal we had discussed is equals to p plus a n square by b square this is the pressure of the real gas we have here and this is the pressure of the ideal gas right so difference is what internal pressure loss of 1 mole of Wendervoll gas and real and this so we need to have the difference here is p ideal minus p that is p real is equals to a times n square by v square n values given one more a by v square hence option c is the correct answer we have here the Wendervoll equations for CH4 at low pressure at low pressure means what just now we did low pressure means volume is high vm right so vm minus v we can equate to v so what we can write nothing you have to memorize just p plus a by v square into v is equals to RT so now from this if you solve pv by RT is z is equals to this so z is equals to 1 minus a by v RT this is the expression we have that's now we did so pv plus pv plus a by v is equals to RT option b is correct which of the following can be most pretty liquefied we haven't done this yet okay it comes under the liquefaction of gas liquefaction of gas will be easier when the interaction will be more right when a value will be more so more value of a easier will be the liquefaction we'll discuss this later okay so which one has the maximum a value that is for SO2 I guess yeah SO2 so SO2 can be liquefied easily okay out of NH3 and N2 which will have larger value of a NS3 has hydrogen bonding larger value of a so for this we have NH3 larger value of b what we can say has large molecular size so for this one it will be N2 size of N2 is more this kind of question is you know a bit difficult to understand like which one has more size or less but usually you see nitrogen and hydrogen we have here the other atom here we have nitrogen nitrogen okay but when you solve some questions you should have the understanding of it means a bit of memory like you know you have to memorize this kind of thing at N2 and other gas all these things you will understand once you solve some questions okay whenever you get this question take a note of it and revise before the test that's what you can do in this type of question okay understood right one more question you see just now this one you try you need to find out the pressure 5 p divided by 5.277 this value you need to find out it's a numerical type question yes obviously a value is given no so you have to use Vendor one see volume occupied by CO2 molecules negligible right it means what we can say b is negligible here so you can you can you know remove b in b minus beta and then you can multiply this and find out pressure from there you will get a quadratic in v in v and that quadratic you need to solve to get p one condition you need to apply over there you're getting 8 to 8.5 okay but the question is something else Sastro I think you're close question is p divided by 5.277 isn't it see what we'll do here we have the equation Vendor wall equation that is p plus a n square by v square into v minus nb is equals to nrt okay one mole is given so n is equals to one we have to substitute and volume is negligible right volume occupied by CO2 is negligible it means b is equals to zero it is given in the question so we can write v minus nb is almost equals to b right and we substitute it here so we'll get p plus a by v square into v is equals to rt okay now p is equals to what we have p is equals to we have rt by v minus a by v square you will get a quadratic in v multiply both sides by v square so you'll get v square into p minus r into t into v plus a is equals to zero so v is equals to minus b that is rt plus minus b square r square t square minus four ac so four a into p root over of it divided by two a right two times p this is the quadratic equation we'll get since for a pressure we have one value of v two value of v for one single pressure is not possible and that is the condition satisfies when this term equals to zero right when discriminant is equals to zero so for the single value of p what we can write r square t square minus four ap should be equals to zero so p is equals to r square t square divided by four a all value you need to substitute here value of p is zero point r is zero point zero eight two one zero point zero eight two one square into t is given 300 square divided by four a value is given 3.59 you can assume this as 3.6 question is three divided by 5.277 so this into 5.277 what is this value you are getting you can use calculator because you see for a single pressure we cannot have two different volumes now p and v for a given value of pressure v should be unique so that plus minus term should be zero so what is this value you are getting okay you are getting it as eight i think you eight is the answer that's what someone we got no shashod you got eight only right because the question is this only p divided by 5.277 that's the value you need to find out that's why i divided yeah approximately the answer would be eight you will get is it are you getting it at eight yeah roughly so it is eight will get right adhija i didn't get you coming at the question is p divided by 5.277 understood right yeah yeah okay one last question you see the question is calculate the diameter of of helium from its from its van der wal from its van der wal constant b the value of b is given it is 24 centimeter cube mole inverse try this okay so you see here b value is given and formula of b is what b is equals to b is equals to four into n a into b right b is the volume of the one molecule so four times n a n a you can take six into 10 to the power 23 approximate into four by three pi r cube so this would be four into four 16 16 into six is 96 so 96 into 10 to the power 23 into r cube pi is 3.14 divided by three so for approximation like this two you can cancel out i'm not doing it here but you can cancel out this two and we can see the formula of b in terms of radius directly you can use approximately is 10 to the power 23 into r cube this also you can use directly sometimes even in some books they give this formula like this one right then you substitute the value of b here and you will get r cube as b divided by 24 into b divided by 96 into 10 to the power 23 24 divided by 96 into 23 90s what i like 96 into 10 to the power 23 and then you can find out r from this diameter is 2 into r so we can find out that also okay we can do one more thing just to solve this we can multiply here by 8 and this side also you multiply by 8 and this becomes from here we can write 2 r to the power cube is equals to 24 into 8 divided by 96 10 to the power 23 so this is four times this is two times so two into 20 into 10 to the power minus 24 this approximately you can solve two r to the power cube is equals to this which you can solve so 20 cube root is it is somewhere around 2.5 to 2.6 okay and then you can get the answer understood guys exact answer is 267 picometer it is given so like this is some basic formula we have that we'll use and we'll find out the answer in all these kind of questions okay so next class we'll finish this chapter there are a few things left okay we'll finish this in the next class and then we'll start a new chapter yeah i'll share the dbp i'll share the dbp meanwhile you can solve the center module uh prana view cross check this uh just let me know okay method is this only by solving this you'll get the answer anything else tell me different different values you are anyways you understood the method how to do this okay you try this once you'll get the answer fine guys thank you so much see you in the next class okay and submit your assignment on time okay whatever you have done submit it on time okay thank you