 So, now we will see the tutorial 4, which is on ideal gas mixtures. The first problem a gas mixture having volumetric composition of 50 percent CO2, 30 percent N2 and 20 percent O2 at 14 megapascals and 270 degrees centigrade in a rigid flask of volume 0.05 meter cube. It is added slowly to the flask until the pressure rises to 16.8 megapascals. Molecular weight of CO2, N2, O2 are 44, 22, 32 kilogram per kilo mole respectively. Similarly, for Cp, S in joule per kg Kelvin for these three components is given as 750, 746 and 696. Determine the heat interaction. Volumetric composition that is percentage volumes are given as say 50 percent CO2, 30 percent N2, 20 percent O2. So, now mole fractions same as volume fractions. So, mole fraction of CO2 will be 0.5, mole fraction of N2 will be equal to 0.3, mole fraction of O2 will be 0.2. Then we can get the molecular weight of the mixture that is sigma mole fraction of each component into molecular weight of that particular component which is equal to 0.5 into 44 plus 0.3 into 28 plus 0.2 into 32 which is equal to 36.8 kg per kilo mole. Now, specific gas constant of the mixture R mix will be equal to universal gas constant divided by the molecular weight of the mixture which is equal to 8314 joule per kilo mole Kelvin divided by 36.8 kilogram per kilo mole. So, which is equal to 225.92 joule per kilogram Kelvin, the specific gas constant R of the mixture. Now, first law for the system equal to Q minus W equal to delta U because delta E equal to delta U here in the absence of we have neglected the kinetic energy and potential energy changes. Now, rigid vessel that means W equal to integral P dv equal to 0 since dv equal to 0. So, we can say Q equal to delta U, delta U equal to mass of the mixture into Cv of the mixture into T 2 minus T 1. So, now, we have to fix the states, state 1 P 1 equal to 14 mega pascals, T 1 equal to 270 degree centigrade equal to 543 Kelvin, P 2 equal to 16.8 mega pascals and given data volume of the rigid vessel is 0.05 meter cube. So, we can find the, so this is the constant volume at every state, state 1 and state 2 the volume is only 0.05. So, now, we can do the calculation of the mixture mass, mixture mass we need in this. So, use state 1, M x will be equal to P 1 that is 14 into 10 power 6 into 0.05 P 1 V 1 divided by R, R of the mixture 225.92 into temperature initially is 543 that will give you the mass 5.706 kg. Now, state 2 temperature I need, so T 1 I have, this I have now state 2 temperature T 2 equal to pressure is known volume is same ok. So, 16.8 into 10 power 6 into 0.05 divided by M is now known 5.706 into R mixes 225.92. So, that will give you the T 2 as that is it. Now, we have to calculate the CV mix. So, let us continue in the next slide. Here for calculating CV mix, we have CV mix equal to sigma y i CV i because here this is in joule per kg Kelvin. So, I have to find the mass fractions. So, let us find the mass fractions and y C O 2, mass fraction of C O 2 is equal to mass mole fraction of C O 2 into molecular weight of C O 2 divided by molecular weight of the mixture which is equal to 0.5978. Similarly, y of N 2 will be equal to x of N 2, molecular weight of N 2 divided by molecular weight of mixture which is equal to 0.2282. Similarly, I can find y O 2 equal to 0.174. So, now CV mix equal to 0.5978 into the CV of C O 2. CV of C O 2 is given as 750 plus y of N 2 0.2282 into CV of N 2 is given as 747 plus 0.174 into CV of O 2 is 696. So, that will give CV of mixture as 739.92 joule per kg Kelvin. Now, Q equal to delta u equal to M mixer CV mixture into T2 minus T1 which is equal to 5.706 into 739.92 into T2 is 651.62 minus T1 is 543. So, that is equal to 458591.2 joules. This is the heat transfer. There is no work transfer here, constant volume. Second problem, a vessel is divided into three compartments by two thin partitions like this. Now, we can say this is A, this is B and this is C. A has O 2 O 2 and has a volume of 0.1 meter cube. B has volume of 0.2 meter cube and contains N 2. C has a volume of 0.05 meter cube and contains C O 2. All these parts are at the pressure of 2 bar initially and temperature 13 degrees centigrade, 286 Kelvin. Now, this is the initial condition of the three compartments where the temperature and pressure is given. Volumes are also given for these compartments. Three gases are there, O 2, N 2 and C O 2. Now, the partitions are removed and the gases are allowed to mix. Partitions are removed and then they mix. So, initially the vessel is insulated. So, all the three compartments are insulated. Determine the partial pressure of each gas and the molecular weight of the mixture. This is the first part. Second is then heat is slowly added to the system so that the mixture. So, now mixing has happened. A full uniformly mixed gas mixture is obtained. At that point, heat is slowly added so that the mixture is heated to a temperature of 100 degrees. So, initially it is 13 degrees. Then finally, it reaches 100 degrees. So, I can say T 2 equal to 100 degrees centigrade, 307 degree Kelvin. Determine the heat transfer. Specific heat constant volume C v is given for O 2, N 2 and C O 2 as 696, 747 and 750 respectively. So, this is the problem given to us. Now, total volume is what? 0.1 plus 0.2 plus 0.05 equal to 0.35 meter cube. That is the total volume. Now, when the partitions are removed and the contents mix, we can find the volume fraction or the mole fraction. So, this is the volume of oxygen, volume of nitrogen and volume of C O 2 given in this. So, now the total volume is this. Once they mix, the volume fraction can be calculated. That is volume fraction is equal to mole fraction is calculated as X O 2 equal to 0.1 by total volume 0.35 equal to 0.286. Similarly, X N 2 will be equal to 0.2 divided by 0.35 equal to 0.572 and X O 2 will be equal to 0.05 divided by 0.35 equal to 0.142. Now, mole fractions are found. Volume fraction is equal to mole fraction. So, initially individual gathers what volume the occupy is given. Finally, when they mix, the total volume is known. From that volume fraction is equal to mole fraction or calculated. Now, what is partial pressure equal to P of P is the total pressure. So, at the state only we want before the heat is added partial pressure is mole fraction of the divide ith component. This P i I can say P i equal to mole fraction into the total pressure. Total pressure is 2 bar because all the three components are 2 bar. When they mix, the pressure need not change. So, the pressure will be at 2 bar. So, the partial pressure can be calculated for each component as P O 2 will be equal to mole fraction of this 0.286 into 200 which is equal to 57.2 kilopascals. Similarly, P N 2 for nitrogen it is 0.572 into 200 which is equal to and P C O 2 will be equal to 0.142 into 228.8 kilopascals. So, these are the partial pressures. So, the first part is determine the partial pressure now molecular weight. What is molecular weight of the mixture? Molecular weight of the mixture equal to sigma x i molecular weight of i which is equal to 31.432 kg per kilo mole. The first part is now answered. Now, second part is initially it is insulated. Now, one of the insulated is removed and heat is slowly added. The mixture is heated to 100 degrees. So, how to solve that part? Since volume is constant, what happens? dv equal to 0. So, w equal to integral p dv equal to 0. So, first law is written as q equal to delta u delta ke equal to delta pe equal to 0 that is assumed here. So, now, q equal to m of the mixture cv of the mixture into t 2 minus t 1. Okay now, we have to find the mass of the mixture. So, first mass of masses of individual components. So, we should get that that will be m o 2 will be equal to m o 2 is equal to initially 2 bar now. So, 200 kilo Bascals into its volume is 0.1 divided by molecular weight is known 32. So, we can find the r that is r u by molecular weight. So, this is r of o 2 into temperature 286. So, that will give you the mass 0.269 kg. Similarly, m n 2 will be equal to 200 into 10 power 3 into here the volume is 0.2 divided by 8314 by 288 into 286. So, that will be equal to 0.471 kg. And finally, m of c o 2 will be equal to 200 into 10 power 3 into 0.05 that is the volume it occupied initially divided by 8314 minus the gas constant divided by molecular weight into 286. So, that will be equal to 0.185. So, the total mass of the mixture will be equal to 0.269 plus 0.471 plus 0.185 which is equal to 0.925. Okay, so this is the total mass. Now c v, c v of the mixture equal to sigma y i c v i. So, now what is y i mass fraction? So, now we have the mass of the mixture individual masses also we have. So, we can find this as if for example, for o 2 it is 0.269 divided by 0.925. This will be o mass fraction of o 2 into c v. c v of o 2 is 696 given plus for n 2 it is 0.471 divided by 0.925. This is y of n 2 into 743 plus 0.185 divided by 0.925. This is y of c o 2 into 750. So, that will be equal to 732.76 joule per kg Kelvin. Now we can find q as m mix is 0.925 and the c v mix is on 32.76 into T2. T2 is basically 100 degrees 100 degrees and T1 is 13 degrees. So, that if you substitute you get the answer as 500 sorry 58968.9 joules or 58.9 kilo joules that is the answer. So, the mixture problems basically a single gas which is ideal gas constitute to a mixture. So, such gases constitute to a mixture. The mixture of the gas also is considered as ideal gas, but we have to calculate the properties of the mixture based upon some weighted averages. If you have a component property given in per kg basis like joule per kg Kelvin, then you use for this property phi as a unit like this, then we have to get the mixture property as mass fraction into phi i. This is mass fraction. So, mass fraction weighted average you have to use. On the other hand, if you have joule per kilo mole Kelvin, then you have to use the mole fraction ok. So, for example, if chi equal to joule per kilo mole Kelvin, then this is actually this mixture. Now, this mixture will be calculated as sigma xi, this now this xi is the mole fraction. So, these are all weighted averages that is it. So, once you do this, then you get the mixture of gases treated as a single ideal gas.