 So, we will go to the next problem, third one, air contains 79 percent of nitrogen and 20 percent of oxygen. So, when I say air, it is itself a mixture, 79 percentage nitrogen, 21 percentage oxygen by volume. So, air itself basically, air contains CO2, argon, etcetera, but traces. So, major components in air is 79 percent nitrogen and 20 percent oxygen, ok. Now, when I, we have a rigid vessel, a rigid vessel contains 0.01 meter cube of air and 0.001 meter cube of, there was a lot of sound, you know, that is what I said, you know. So, in a rigid vessel, the problem states that a rigid vessel initially contains 0.01 meter cube of air and 0.001 meter cube of methane. Methane is CH4 and molecular weights are given here. So, air contains 79 percent of nitrogen and 21 percent of oxygen by volume. So, air itself is a mixture of several gases. It contains argon, CO2, etcetera. The major components are nitrogen and oxygen and the standard air, we have 79 percent of nitrogen and 21 percent of oxygen by volume and now the problem starts with a rigid vessel which contains 0.01 meter cube of air and that means, this air is again constituted by these two gases that you have to remember. So, the molecule of methane is 16 kilo gram per kilo mole. Initial pressure is 1 atmosphere that is 101, 325 pascals and initial temperature is 298 Kelvin or 25 degree centigrade. So, this is given. Now, calculate the masses of nitrogen, oxygen and methane present in this mixture. Now, this is the first part. Second is to this mixture 0.001 kg of hydrogen, molecular weight is 2 kilo gram per kilo mole is added. During this process, temperature remains constant. Temperature that is 298 Kelvin, it remains. What is the molecular weight of the mixture after the addition of hydrogen? What is the mixture pressure after the addition of hydrogen? So, when you add hydrogen, the pressure, the rigid vessel, so pressure will increase. So, what is this? So, this is what this problem is about. So, here partial volumes of nitrogen and oxygen in air can be calculated as 0.79, that is the mole fraction. When you say air, mole fraction of because by volume it contains 79 percent, mole fraction of nitrogen in air will be 0.79, so that into the volume, volume is 0.01 meter cube, 0.01 meter cube. So, partial volume of nitrogen VN2 will be equal to 0.0079 meter cube. Similarly, for oxygen in air, it is 0.21 into 0.01, this is for nitrogen, this is for oxygen. So, that means, VO2 will be equal to 0.0021 meter cube, partial volumes. Partial volume of methane is already given, so for methane, partial volume is given as 0.001 meter cube. So, this is how we got this, this is by Ammergat's law, so now I got the partial volumes. So, now I can calculate the mass of the each component, so mass of nitrogen can be calculated as pressure, 1 0 1 3 2 5 Pascals into the partial volume of nitrogen divided by the molecular weight of nitrogen, that is, universal gas constant 8314 divided by 28 into the temperature 298. So, that will be equal to 9.05 into 10 power minus 3 kg or 9.05 grams. Similarly, I can calculate M of oxygen equal to 1 0 1 3 2 5 into 0.0021 divided by 8314 by 32 into 298, which is equal to 2.75 into 10 power minus 3 kg equal to 2.75 grams. So, what is for methane, methane it is M CH4 equal to 1 0 1 3 2 5 into 0.001 divided by 8314 by 16 into 298, so we get totally 6.54 into 10 power minus 4 kgs or 0.654 grams. So, the masses of individual, that is what is asked, ok, calculate the masses of nitrogen, oxygen, methane, presently mixture, these are the masses. So, for that we calculated the partial volumes, because air itself is a mixture, ok, by Amagat's law, the partial, the, if the total volume occupied by air is 0.01, the volume occupied by nitrogen will be mole fraction into this volume, so mole fraction into that volume. Similarly, for oxygen, it is oxygen's mole fraction into the volume occupied by the air. So, from that, once the partial volumes are known, masses can be calculated, because P into partial volume divided by M into, that is, multivit into temperature, so using that we can calculate this, not multivit R, this is R. R equal to R U is R N2. So, pressure into volume divided by specific heat, specific gas constant into temperature, that is the, so ideal gas equation of state we use. So, I get, got the masses, ok. Now, to this mixture, 0.001 kg of hydrogen is added, temperature is still at 298 Kelvin, ok. Now, total mass of the mixture will calculate, mixture equal to 9.05 plus 2.75 plus 0.654 plus 1. So, this is actually, this is actually 1 gram, 1 gram hydrogen added, ok. So, that, that will be equal to 13.454 grams, ok. So, now, mass fractions, mass fractions of the components, ok. That will be Y N2 equal to 9, 9.05 divided by 13.454, which is equal to 0.673. Y of O2 will be equal to 2.75 divided by 13.454 equal to 0.204. Y of methane equal to 0.654 divided by 13.454 equal to 0.049. And finally, Y of H2 will be equal to 1 divided by 13.454, which is equal to 0.074. These are mass fractions. Now, molecular weight of the mixture, based upon the mass fractions, can be calculated divided by 1 by divided, 1 divided by sigma YI by MWI, which is equal to 1 divided by 0.673 divided by 28 plus 0.204 divided by 32 plus 0.65 sorry, 0.49 divided by 16 plus 0.074 by 2. So, that is equal to 14.19 kilo gram per kilo mole. So, now the volume, volume is 0.01 plus 0.001 partial, this is the volume occupied by air and this is the volume occupied by the methane and in this rigid volume only the hydrogen is added. So, the total volume is 0.011 meter cube. Now, what is, how is this pressure? P v equal to M mix R mix into T ok. Now, we can write from this P equal to M mix that is 13.454 into 10 power minus 3 kgs. So, grammed kgs M mix into R mix is nothing, but what this molecular weight is known. So, 8314 divided by 14.19 R mix into temperature is 298 divided by the total volume v 0.011. So, that is equal to 213551.5 Pascal's. So, that is the increase in pressure due to the hydrogen's injection, this is the problem. Next problem, yeah frictionless leak proof distance oriented device ok, device contains mixture of gases ok, gas mixture. Here the mole fraction of neon is 0.25, mole fraction of oxygen is 0.5, mole fraction of nitrogen is 0.25, diameter of the piston is dp is 25.25 centimeter, initial volume enclosed v 1 equal to 0.1 meter cube ok. Now, the temperature initial temperature is 10 degree centigrade 283 Kelvin, initial pressure P 1 equal to 200 kilo Pascal's. Heat is now added slowly and the piston moves by a distance of 50 centimeters. So, heat is added. The molecular masses, molecular weights, molecular weight is the molecular mass, molar mass of neon oxygen and nitrogen ok. So, Mw neon will be equal to 20, Mw that is kg per kilo mole and Mwo2 is 32 kg per kilo mole and Mw nitrogen equal to 28 kg per kilo mole. So, these are the given things. Similarly, CVs are also given, CV of the neon is 618 joule per kg Kelvin, CV of oxygen is given as 658 joule per kg Kelvin and that of nitrogen equal to 743 joule per kg Kelvin ok. So, calculate the heat and work interactions ok. So, for this problem first let us find the molecular weight as a mixture equal to sigma mole fraction of each component into its molecular weight which is given as 0.25 into 20 plus 0.5 into 32 plus 0.25 into 28 which is equal to 28 kilo gram per kilo mole that is the thing. Similarly, the specific gas constant of the mixture or mix equal to the universal gas constant divided by the molecular weight of the mixture which is equal to 297 joule per kg Kelvin then I can find the mass fractions y i equal to x i into molecular weight of i divided by molecular weight of the mixture. So, this is the formula. So, I can find y of neon as 0.179 y of oxygen equal to 0.571 and y of nitrogen equal to 0.25 ok. From this I can find CV mix. CV mix equal to sigma y i CV i because it is in joule per kg Kelvin. So, mass fractions are to be used. So, that will be equal to 0.179 into 618 618 plus 0.571 into 658 for oxygen plus 0.25 into 743 for nitrogen which will be 672.1 joule per kg Kelvin. Now, mass of the mixture is P V by RT which is equal to 200 kilo Biscuits is the pressure. So, 200 into 10 power 3 into volume is 0.1 given in the problem divided by the 297 is the R mixture which we have calculated into the temperature initial temperature is 283 Kelvin. So, this is R mix and this is T 1 ok this is V which is given ok. So, now, this will be equal to 20.238 kg ok. Now, as the piston moves up while heating pressure remains constant why you can see this there is no spring or any other opposing force and only atmospheric force plus some piston mass may be there. So, that has. So, the piston flows on the gas mixture which exerts a pressure of 200 kilo Biscuits. Now, when you heat it pressure rises, but the piston is free to move it is frictionless frictionless and leak proof and it can be free to move. So, it is going to move because there are no stops or anything. So, it will move equilibrating the pressure the pressure will remain constant as the piston moves through a height of 50 centimeters. So, pressure remains constant as piston moves through a height of 50 centimeters. So, now what is the work done? Work done will be integral p dv equal to p into delta v. Initial volume is given. So, I have to find the final volume. So, how to find the final volume? Final volume will be area into. So, I can say delta v it is simply calculated to be basically delta v will be equal to area of the piston into delta h or delta is the height moved by the piston. What is the area of the piston? It will be equal to pi by 4 dp square. The diameter of the piston is given 25.25 centimeters. So, you can use that water own bar. So, delta v. So, you can find W as 200 kilopascals. So, 210 power 3 into pi by 4 into 0.2525 square for the area into 0.5 for delta h. So, which gives 5,007 joules. Let us work done. Now, the final volume v 2 is known. So, the area of cross section into the difference in the height. So, that will be delta v. So, v 2 will be equal to v 1 plus delta v. So, that is known. So, then find t 2. t 2 equal to p 2 that is p 1 or p 2 are same into v 2 divided by m mix or mix. Now, from this v 2 we can calculate delta v you know. So, we can calculate v 2 and substitute this. So, you will get temperature as in the final state as 353.7 Kelvin. So, v 2 is v 1 plus delta v. Delta v is we already told that is A p into delta h. So, calculate that and substitute you get this. So, finally, what is delta u? m mixture c v mixture into t 2 minus t 1 which is equal to 11309.2 joules. So, now q equal to w plus delta u which is equal to 11 sorry 5007 plus 11309.2 which is equal to 16 316.2 joules or 16.316 kilo joules that is the heat transfer. So, you can see this in this the mixture properties again are used it is only a constant pressure process and we can calculate the values.