 So, the next problem 1 kilogram of argon molecular weight 40 Cv 392 at temperature of 392 100 kilobuses contained is rigid tank that is a rigid tank argon is present ok this is 1 kg. Now, connected by a valve. So, there is a valve to another rigid tank like this where O 2 0.8 kg is present. So, here T 1 is 300 Kelvin P 1. So, this is a I will say this is a and this is B I will say T A 1 equal to T A 1 is 300 Kelvin similarly P A 1 equal to 100 kilopascals. Similarly here T B 1 is given as 400 Kelvin and P B 1 is given as 500 kilopascals. So, they are separately there in the chambers A and B as shown in the figure here. Now, the valve is open the valve is open gas is allowed to mix until they reach a equilibrium temperature of 360 degrees that is T final is 360 degrees Kelvin 360 Kelvin 360 Kelvin ok. So, now once they mix we have to calculate the heat interaction. First what is the volume of each tank? Then what is the final pressure? So, solution volume of each tank let us calculate V A is equal to what? Because M 1 R molecule of argon is 40. So, 8 3 1 4 by 40 into T is 300 divided by P is 100. So, that will be the volume which is 0.6235 5 meter cube similarly V B will be equal to mass is 0.8 0.8 into 8 3 1 4 by 32 calculate into 400 Kelvin 400 divided by pressure is 500 kilopascals 500 into 10 power 3 which is equal to 0.16628 meter cube. So, these are the initial the volumes in the two compartment A and B the individual volumes. Now, the final volume will be what? V A because this is now mixture argon goes to other side oxygen comes to this side and they mix thoroughly. It takes some time, but after the mixing both components are present in a uniform manner in both the tanks and the final temperature also is given as 360. So, the final volume that is I can consider the mixture to be in both the tanks. So, we have just add the volumes of A and B that will be equal to 0.789 83 meter cube. So, that is the final volume. So, now, what is the mass fraction? The total mass is 1.8 kg now once the mixture forms. Individual masses we know. So, from that we can calculate this. So, this will be equal to 1 divided by for argon it is 1 divided by 1.8 and molecular weight of argon is 40 plus for oxygen it is 0.8 divided by 1.8 divided by its molecular weight is 32. So, that will be the molecular weight of the mixture will be 36 kg per kilo mole and R the specific gas constant will be equal to R U divided by molecular weight of the mixture is equal to 8314 by 36 which is equal to 230.94 joule per kg Kelvin. Now, I can find the right for the final state P f is the final pressure V f is the final volume equal to molecular mass of the mixture into R of the mixture into T final. So, what is P f will be equal to M mix 1.8 kg into this is 230.94 into 360 divided by what is volume here 0.78. So, that will be equal to 189.470 Pascals or 189.47 kilo Pascals that is it. Now, consider A plus B as the system ok. Now, then Q minus W equal to delta U I can write. Now, what is now A A plus B as system in A plus B as system delta V equal to 0. So, W equal to 0. So, I can say Q equal to delta U ok. Now, is equal to U final minus U initial ok. What is U final mass of the mixture into C V of the mixture into delta V. Into T final ok. So, minus T reference some reference the same reference is there for all ok that is the U final U final minus U initial is what M argon C V argon T final minus T argon initial R 1. So, not this sorry T here T A 1 minus T ref initial initially argon has this U similarly plus M O 2 correct M O 2 minus T argon initial R 1. So, C V O 2 into T B 1 minus T ref that is the thing. So, now, you can find this a T ref will actually cancel ok. So, I can write this first of all what I want is C V mix what the Y I C V I which is equal to 1 by 1.8 into 392. Plus 0. This is for the argon 1.8 into this is C V for the argon and C V for this is 696 which is equal to 572.11 joule per kg Kelvin. So, now, I can say Q equal to delta U equal to 1.8 into 572.11 into 360. So, common T ref that will cancel. Minus 1 into 392 into 300 plus 0.8 into 696 into 400. So, that will be equal to 30407.28 joules. So, that is the heat answer ok. This is about this problem. Next problem 5 kilograms of equimolar composition of 3 gases. Equimolar mean number of moles of these gases are same with molecule moles of 16, 32 and 44 is contained in a rigid vessel ok. So, this rigid vessel let us say that is system 1. Now, 5 kg of the same mixture equimolar composition of these 3 mixture gases at the same temperature and pressure is also contained in a piston and arrangement where the piston is frictionless and free to move. Let this be system 2. It is desired to increase the temperature of the gas mixture by 20 degrees centigrade from its initial value in both the systems system 1 and system 2 from whatever be the initial temperature. What is the difference in the heat transfer given to system 2 and system 1 in kilo joules ok. So, now what is the, so it is an equimolar. That means, mole fractions of each component will be 0.333, 1 by 3. So, molecular weight of the mixture will be equal to 0.333 into I will add everything 16 plus 32 plus 44 which is equal to 0.333 which will be equal to 30.667 kg per kilo mole. Now, mass is given as 5 kg. Now, what I can do is for system 1 rigid vessel first law is q minus w equal to delta u. So, we will assume in both the cases equal to 0. So, delta e becomes delta u. So, q minus w equal delta u. Now, delta v equal to 0 for rigid vessel. That means, w equal to 0 correct. So, I can say q equal to delta u equal to m into cv minus w equal to 0. So, delta v is equal to delta v into delta t, delta t is 20 degrees, delta t equal to 20 degrees in degrees given. This is for system 1 I will say q 1. So, system 1 I will say q is q 1 ok. Now, for system 2 what is that piston cylinder arrangement? What happens here? q 2 minus w equal to delta u. So, now, w here is integral p dv equal to p into delta v ok. So, I can say which implies q 2 equal to p into delta v plus delta u which I can write as p 2 v 2 minus v 1. So, p into v 2 minus v 1 plus delta u or I can say it is delta u I can write as u 2 minus u 1. So, I can say p 2 v 2 plus u 2 minus p 1 v 1 minus plus u 1 where p 2 equal to p 1 equal to p in this case. So, this is nothing but h 2 minus h 1. I can say this q 2 will be equal to m into cp mix into delta t that is it. So, what I want basically is q 2 minus q 1 equal to m into delta t which are constants into cp mix minus cv mix ok. So, which is equal to m into delta t into r mix. So, which is equal to 5 into 20 into r mix is what? 8 3 1 4 divided by molecular weight 30 1 6 6 7. So, that is equal to 2 7 1 1 0.5 joules or 27.11 kilo joules that is the answer. So, the difference between the heat transfer to a constant pressure process in the Pistons regime arrangement and a constant volume process in a rigid vessel will be just m into delta t into r mix. Because the heat transfer to a constant volume process or in a rigid vessel will be just delta u change in the internal energy. For a constant pressure process the heat transfer will be equal to the enthalpy change ok that is the main concept here. Next problem 10 kg is of a mixture having an analysis on mass basis. Please understand mass basis 50 percent nitrogen 30 percent CO2 and 20 percent O2 is slowly compressed in an insulated Pistons regime arrangement from 100 kilopascals 280 Kelvin to 500 kilopascals. Determine the work interaction. Specific heat at constant volume cv for these 3 components are 747, 750 and 696 and the molecular masses are 28, 44 and 32 respectively. So, this is the problem. Slow compression in an adiabatic piston cylinder arrangement that is the problem here ok. Pressure temperature initially is given, final pressure is given. So, let us do this del q equal to 0 adiabatic. Then the first law can be written as minus del w equal to du or del w can be written as minus pdv equal to del u can be written as m cv dt ok. Now, m is the mass and cv is the specific heat of the mixture mass of the mixture cv of the mixture here ok. Now, equation of state pv equal to mrt or I can say p equal to mrt by v. So, I can substitute here in this. So, this becomes minus mrt dv by v ok equal to m cv dt. So, m cancels. So, I can say minus say I will say dv by v separate the variables equal to minus cv by r because r comes and dt by t ok. So, m cancels. So, I can say minus I will say dv by v separate the variables equal to minus cv by r because r comes and dt by t ok. So, this separating the variables integrating I get natural logarithm of v2 by v1 equal to minus cv by r into natural logarithm of t2 by t1 ok. So, because I want to find the final state temperature further I am developing this from the first basic principles for an adiabatic process. So, I write everything here the pdv work is only present in the form of work is not present. So, displacement because piston cylinder there will be when you compress it there will be a change in volume. So, that will cause some displacement work and at d u is written as m cv dt for the ideal gas. So, I can say v2 by v1 equal to t2 by t1 power say I will say a where a will be equal to cv by r t2 by t1 minus a ok. So, a this is cv by r is called a. Now, in terms of pressure ratio I can say v2 by v1 equal to p1 t2 divided by p2 t1 equal to which implies p1 by p2 will be equal to t1 by t2 power a plus 1 ok. Now, what is a? a equal to cv by r which can be written as cv by r is cp minus cv which is equal to 1 divided by cp by cv minus 1 or I can say this is 1 divided by gamma minus 1 a is 1 by gamma minus 1. So, that means a plus 1 will be equal to 1 by gamma minus 1 plus 1 which is equal to 1 plus gamma minus 1 divided by gamma minus 1 which is equal to gamma by gamma minus 1. So, which implies I can say p1 by p2 equal to t1 by t2 power gamma by gamma minus 1. So, this is the relation between pressure and temperature ratio in a adiabatic slow process ok. So, reversible process you can say when the slow process it is quasi static process. So, reversible process. So, reversibly adiabatic process we can say this relationship will hold good. Now, I know pressures at two states p1 and p2 temperature at one state is known. So, t2 can be found from this I find t2 equal to 4 424.5 Kelvin. So, for that gamma should be known to me ok. So, what is gamma? We will do that first. So, R mix will be equal to 8314 divided by the molecular mass 32.33. How we get molecular mass? This is nothing but 1 by sigma yi by mwi which is equal to 1 divided by 0.5 by 28 plus 0.3 by 44 plus 0.2 by 32 you understand. So, now that is molecular weight of this. So, this you find this R will come calculated as 257.16 joule per kg Kelvin. Now, C V mix equal to 0.5 into 747 plus 0.3 into 750 plus 0.2 into 696 equal to 737. So, this is the value of this 0.7 joule per kg Kelvin. Now, C P mix can be calculated as R mix plus C V mix which is equal to 994.86. Now, gamma can be calculated as gamma is now gamma mix correct gamma equal to gamma mix equal to C P mix divided by C V mix. So, this is equal to 1.3486 ok. Now, from that I can find T 2 equal to T 1 into P 2 by P 1 power gamma minus 1 by gamma which implies T 2 equal to 424.5 Kelvin that is what we have got this. Now, we can find adiabatic so, Q equal to 0. So, we can say W equal to minus delta U which is equal to minus M sorry yes. So, Q minus W equal to delta U. So, W equal to minus delta U. So, M mix C V mix into T 2 minus T 1. So, that will be equal to minus 1065976.5 joules ok. So, this is the way it is solved. So, it is a adiabatic reversible adiabatic compression process which obeys T 2 by T 1 equal to P 2 by P 1 power gamma minus 1 by gamma ok. So, I will stop here.