 So, what we do is this we assume and rightly show that ambient air is a mixture of both water vapour and other vapours I would say which are dry. And then we say that mass of the air is going to be mass of the two components dry air and wet air therefore mass is equal to volume into density therefore volume of air into density of air this will be equal to the individual volumes of dry air and water vapour. So, now we apply the law so just take VA on that side just take VA below do you think I should dry this here or you will be able to understand generally what do you feel we can do it here also if you want. But I have taken lot of trouble to go step by step here so that you can understand how it is being derived. And once again I am saying just observing it on the screen will not help you much it is better that you also do it down do it as we as we proceed. So, density of air I just divide by VA on both sides and then what I will do is I will take this VA inside so it will become VDA by VA times rho DA and VW times VA times rho VW. Then we apply those two laws so from there you can get the ratio of dry air volume to the total volume is equal to the pressure of dry air and total air similarly the ratio of volume occupied by the water vapour upon total volume is equal to the ratio of pressure of pressure partial pressure of the water vapour divided by the total pressure of the whole system. These two equations come from application of the previous law so I will replace it there now so this VDA by VA there is VDA there is VA here so this will be replaced so I am just going to write this as PDA into rho DA plus PWV rho WV upon PS is this clear we just replacing applying Dalton and Amagat s law we are able to get a relationship which relates ratio of partial volumes to ratio of partial pressures. So that we put here and we get the expression for rho A and going one step further now the density of water vapour can be expressed as density of air into some relative density term called as Rd because density of water and density of air are fixed quantities at standard conditions so therefore you can say that there is something called as a relative density and therefore you can replace the expression rho of water vapour by rho of dry air times Rd water vapour where Rd is the relative density the purpose of this is to use the numerical information that we have from atmosphere in arriving at the effect of humidity. So what do you get if you do this please put in the expression and tell me so rho A is equal to PDA rho DA plus PWV rho VW rho VW will become RdV RdVW rho DA so there will be two rho DAs you can take it out so you will get rho A is equal to PDA plus PWV pressure of water vapour times RdVW and both terms have rho DA which comes outside so you have rho DA outside by PS agreed. So things will actually become crystal clear to you only when you read it when you go back home and when you derive it yourself so this expression rho A times this I am going to just copy and paste in the next slide so the first thing is the same expression. Now looking further PDA is the pressure or the partial pressure of the dry air that will be equal to the total pressure minus pressure of the water vapour correct that is how we define partial pressures so you can replace for that PDA there. So rho A will become PS minus PWV plus PWV RdVW rho A by PS and now there is one more term which is common that is PWV pressure of water vapour is also common term in the two so you take it out and simplify the expression. Now finally we have to apply the simple relationship for any gas so P1 is equal to rho N R T1 and P2 is equal to VdVR T2 but this R is the universal gas constant so with that for any gas rho 1 can be taken as P1 divided by P2 okay and if you divide these two together you will knock off R you will simply get the expression which says density at any condition is equal to density at any other condition with the pressure ratios and temperature ratios. So PS minus I take PWV common 1 minus Rd no it is correct we need one more bracket yeah yeah yeah that is fine we need another bracket here there you go so applying this now you this you should do yourself I will not like to show it to you so what you do is replace rho DA dry air density so rho 2 will be rho 0 C level density P1 will be P0 T1 will be T0 and sorry I am sorry T2 P2 and rho 2 will be at C level and P1 T1 rho 1 will be at dry air so replace that in the expression and tell me what you get is this point clear that using the universal gas law for two conditions C level condition and the condition at any operating altitude by taking a ratio so I want to replace this rho DA which is the density of the dry air by parameters about air which I know at C level so therefore rho DA will be equal to PS by P0 T0 by TA into rho 0 right now substitute this value so this expression on the top this expression I want you to substitute rho DA with PS T0 P0 TA rho 0 here you will notice there is a PS already in the denominator here it will cancel out and what will remain will be it will become this this term in the numerator divide by TA PS will knock off and what will remain here will be T0 P0 rho 0 there you go I have not knocked them yet I have just written down as it is so all I have done is I have just copied and pasted the red block is the definition or the formula for rho DA and now you cancel and when you cancel you will get you will get this expression if you cancel because TA will come on the denominator PS will cancel with the PS which is below I have still not cancelled it now PWW is the pressure of water vapor and that is called E in our terminology and PS will be cancelled so if you if you replace it you get now a slightly elegant formula for density of air so the density of air is the ambient pressure PS which is known to you because you are flying at some altitude minus a term because of the humidity this is basically the pressure of the partial pressure of the of the water vapor inside divide by temperature of the ambient air under as the conditions into T0 P0 rho 0 now interestingly these T0 P0 rho 0 are constants so you can replace now these numbers by the numerical value of T0 which is 273.16 plus 288.13 P0 which is 101 325 Newton meter square rho 0 is 1.256 if you multiply that you get a constant 0.003484 Kelvin by meter square second square Kelvin second square by meter square so you can put that in the expression secondly this RDW this is the relative density of water vapor with air and this number can be measured by experiments as 0.6220 density of water vapor as compared to the air relative density is 62.2 percent okay so 1 minus RD is 0.378 so therefore what you can do is you can replace it by another simple formula which only has known parameters after all when you are flying the airship at some altitude what do you know you know what is the ambient pressure ambient temperature and this E value can be calculated by the law based on ambient temperature any question RDW is basically the relative density so density of helium is X density of air is Y ratio is constant at any altitude it is just it is just a question of relative density one relative to the other so against air water vapor both of them have some density the relative density is constant that constant value is 0.622 so straight away you can now calculate no why should he be inside the bracket it is point yeah you are right it is 0.378 into E yeah you are right it is 0.378 into E so it has to be brought inside because let us see where we went wrong PWPWB should come inside here and therefore it will be there bracket is also missing one more bracket is also missing okay I will correct this thanks I will thank you for pointing out I will correct this so then it is going to be this number which is a constant times PS minus 0.378 into E divided by TA I think that I should correct immediately so that there is no confusion other things will take me some time good you are looking at it very carefully and critically that is good when I when I type this formula at that time I might have made mistakes in entering the data okay so now we are at the end of that particular place where we can take care of the humidity directly once again the E will go inside now let us see what happens if you ignore humidity we are doing all these calculations there must be some reason and if you there is no substantial difference then there is no point in doing all these things so you will notice that this the value of E as you saw in the graph is very low at low altitude low temperatures and it becomes very high at high temperatures so this error will be actually more prominent only at high temperatures at low temperatures see the value of E is around 1700 and you are multiplying by 0.38 and subtracting from P P is 1 0 1 3 2 5 so the amount that you minus is very small so you know it will not come out to be very different but let us just see so it is better to calculate that so now suppose I ignore the value of relative humidity which means I just do not do this reduction of 1 minus R D V W into E so I get rho A S P S by T A into T naught P naught by rho naught and this constant so the error in ignoring the humidity will be just the change in the pressure upon actual ambient pressure times 100 so you can just see it is going to be only 0.3484 times E upon P S.