 A rigid container with a total volume of 10 meter cube is separated into two equal volumes by a membrane. So, let us draw this also, that is a membrane ok, this is totally v equal to 10 meter cube. So, equal volumes that means here it will be first in the first state it will be 5 meter cube and here also 5 meter cube. Initially volume on one side of the membrane contains nitrogen, so let us take left side contains nitrogen at 5 bar and 1000 Kelvin and other side is evacuated nothing is there evacuated. The membrane this is a membrane a thin membrane like aluminium foil ok, an aluminium foil and we can rupture it basically. Once the aluminium foil is ruptured what happens? There is a rapid expansion of the nitrogen from the left side to the entire volume ok. Remember ruptures and finally, there is a thermal equilibrium with the ambient. So, ambient temperature is 300 Kelvin, from 1000 Kelvin there is a thermal that means the rigid container is not insulated that is a heat transfer. Obviously there is a thermal equilibrium. So, T2 will be equal to 300 Kelvin equal to T ambient that is it. What is the final pressure and then at entropy change for the universe that is what is asked ok, entropy change of the universe. Obviously, it is irreversible process you have to understand that rapid expansion takes place correct, rapid expansion. So, irreversible. So, irreversible process occurs. So, let us say R for nitrogen will be equal to 8314 which is the universal gas constant in joule per kilo mole Kelvin divided by the molecular weight kg per kilo mole. So, that will be equal to 297 joule per kg Kelvin specific gas constant for nitrogen. Now, since gamma is given I can calculate Cv equal to what R nitrogen divided by gamma minus 1 which is equal to 742.5 joule per kg Kelvin Cp equal to 1039.5 that is nothing but what gamma R by gamma minus 1 joule per kg Kelvin ok. So, that is the expressions we have got values. Now, V1 for the nitrogen is 5 meter cube which is given in the problem P1 equal to 5 bar or 5 into 10 to the power 5 Pascal's T1 equal to 1000 Kelvin ok. Now, I can find the mass of the mass of the nitrogen will be equal to say just say m, m equal to what I can use 5 into 10 power 5 into 5 Pv by RT R is 297 into T is 1000 which is equal to 8.4175 kg. So, that is the mass contained in the left side of the chamber initially. Finally, after the membrane ruptures what happens? V2 becomes 10 meter cube the total volume. Nitrogen is now contained in total volume and T2 is given as 300 Kelvin equilibrium it is in equilibrium with ambient temperature. So, T2 also is given correct. So, the state what is the final pressure we can find by using this. So, let us apply the first law taking the rigid container as the system ok. Here what? Q minus W equal to Q minus W equal to delta U. Now, rigid container no volume change so, work equal to 0 because what since delta V equal to 0 the entire container is now taken so, Q equal to delta U that is got from the first this. So, now Q equal to delta U equal to M into Cv into T2 minus T1. T2 is known T1 is known M is known. So, we can calculate this as minus 4 3 7 5 kilojoules so, Q equal to minus 4 3 7 5 kilojoules that it is obviously from 1000 Kelvin temperature reduced to 300 Kelvin correct. So, this is 300 Kelvin this is 1000 Kelvin. So, obviously the heat has to be rejected from the whatever system you consider to the ambient so, this is negative that is done. Next is pressure P2 V2 equal to M or T2 which implies P2 what is P2 equal to mass M is 8.41 75 kg into R is 297 into T2 is 300 Kelvin divided by V2 10 meter cube. So, that will be equal to 750 kilo Bascals ok 7 particular Bascals I am just giving the units. So, now final pressure is got heat interaction is got next is entropy change. Entropy change for the universe which is equal to delta S universe equal to delta S system plus delta S surroundings ok. What is delta S system? Delta S system will be equal to M into Cp ln T2 by T1 minus R ln V2 by V1. So, this we can get from TDS relationship correct. Recall TDS equal to DH minus VDP and we know PV equal to RT ok now for the ideal gas. So, we can write DS equal to Cp DT Cp DT by T minus instead of V I can put RT by P. So, this is RDP by P. So, for the ideal gas which is perfect Cp is constant. So, I can integrate this. So, you can say delta S will be equal to Cp ln T2 by T1 minus R ln P2 by P1. So, this is the expression I have used here. So, this is actually P2 this sorry P2 by P1 P2 by P1. So, this is the expression I have used. So, now we know T1 T2 P2 is now calculated. So, P1 P2 also is known. So, from this I can calculate this as minus 5792 joule per Kelvin R minus 5792 kilo joule per Kelvin that is the systems ok. So, what is for surroundings? For surroundings why the entropy would change change for surrounding because there is a heat transfer system rejects some heat this heat is got by the surroundings. Surrounding temperature is constant at 300 Kelvin. So, Q surroundings divided by T surrounding that is the because of the heat transfer entropy is transferred from the system to surrounding. So, that is the entropy change for this. Otherwise, there is no internal irreversibility or anything here. So, this can be written as minus Q for a system divided by T surroundings. So, this is equal to 4 3 7 5 by 300 which is equal to 14.58 kilo joule per Kelvin that is the. So, what is delta S universe now? Add these two delta S system plus delta S surroundings which is equal to minus 5.792 plus 14.58 which is equal to 8.788 kilo joule per Kelvin. So, you can find this that delta S universe can be 0 or it should be only positive. So, it is no positive. So, that means that this is a highly reversible process because of the rapid expansion. So, you can see that there is a increase in entropy of the universe. Even though the systems entropy change is negative, the contribution from the surrounding entropy change is very high here. So, that is the reason. So, if delta S universe comes to be negative, then the process is not possible. So, either it should be 0, delta S universe should be 0 or it should be positive is the principle of increasing entropy. So, this is about this problem. So, nice problem where rapid expansion is there and due to which the entropy actually increases for the universe. Then the ninth problem, a perfect gas mixture. So, perfect gas means an ideal gas which has constant Cp and Cv. So, that is perfect gas containing 0.5 moles of nitrogen. The ratio of specific heat is 1.4, molecular heat is 28 and 0.5 moles of neon Cp by Cv is 5 by 3, molecular heat is 20 kg per kilo mole is expanded in an insulated turbine. Insulated turbine, so let us draw the turbine here. A gas mixture comes in that is 0.5 moles of nitrogen plus 0.5 moles of neon ok. At 8 bar, 300 Kelvin coming in this is a gas turbine ok. So, some work has to be done by the turbine. So, let us say that is this ok. Now, what is the exit pressure? So, turbine to 1 bar the expansion from 8 bar to 1 bar here is 1 bar. So, mixture is expanded here and isentropic efficiency of the turbine is given as 50 percent isentropic efficiency. So, you know that if I draw T s diagram ok this is say 1 bar constant pressure line 1 bar and this is 8 bar. So, now from 8 bar, so this is a 300 Kelvin. The expansion this state inlet state or state 1 again ok then occurs like this to 1 bar. So, this will be the T 2 s and this is the state 2 s where if there is an ideal expansion that means, see normally the turbine is adiabatic, insulated turbine is given. So, in adiabatic process there is no heat transfer that is entropy change due to heat transfer is not there. So, the entropy can change in an adiabatic process only due to the irreversibility ok. Now, if I say that the expansion is also reversible. So, adiabatic and reversible it should be isentropic. So, that is what. So, s 2 s will be equal to s 1. Why I am saying the state s 2 s? I am saying that the state 2 s now isentropically arrived at. So, 2 s means it is an isentropic state which is having the same enthalpy as the initial state 1. So, this is the exit state and inlet state also have the same entropy. Now, actual process will have some irreversibility. So, I cannot even show the process in a state diagram. So, I have to just show by dashed line like what I have shown here and this is the actual process state 2. Now, you can see that there is an increase in entropy from state 1 to state 2. Always irreversibility will cause an increase in entropy. So, now what is isentropic efficiency is the actual work developed by the turbine divided by the isentropic work W s is isentropic work developed by the turbine. So, now, this can be written as h 1 minus h 2 divided by h 1 minus h 2 s. So, this isentropic efficiency is given as 50 percent that is very pretty low 0.5 ok. So, now, what is to be got? Calculate the entropy change of the mixture across the turbine. That means, from inlet to the outlet what is the entropy change that is what is asked. It is a mixture which is entering in a gas turbine. So, that is the problem. So, after understanding the problem, we will try to solve it. Here, mole fractions we will first calculate for the mixture components for the nitrogen which is 0.5 mole divided by what? Total number of moles is 0.5 plus 0.5 ok. Equally equal moles are there initially 0.5 moles of nitrogen and 0.5 moles of neon. So, this will be 0.5 ok and the neon also will be same. So, x ne will be also 0.5 divided by 0.5 plus 0.5. So, which is equal to 0.5. These are the mole fractions. Now, for the mixture we have to arrive at the mixer mixture properties. So, individually I have molecular weight, individually I have C p by C v etcetera correct that is the ratio. But what is the molecular weight of the mixture that we can calculate molecular weight of the mixture will be equal to sigma xi m wi which is equal to 0.5 into 28 plus 0.5 into 20 which is equal to 24 kilo gram per kilo mole ok that is the molecular weight of the mixture. Now, from this I can calculate the specific gas constant of the mixture as the universal gas constant divided by the molecular weight of the mixture which is equal to 8314 joule per joule per kilo mole Kelvin divided by 24 kilo gram per kilo mole which is equal to 346.42 joule per kg specific gas constant ok. Now, I can calculate the mole mass fractions. See, there are several ways to do the mixing properties, mixer properties. I am just following one of these. The mole fractions can give the mass fractions with the help of the molecular weights. So, what is that? y n 2 will be equal to x n 2 into molecular weight of n 2 divided by molecular weight of mixture which is equal to 0.58 triple 3. Similarly, I can calculate that of neon as 0.41667. So, the molecular weight of the individual component of the mixture can be used to calculate the relationship between mole fraction and mass fraction. So, once I get y i mass fraction because I want to calculate the mixture properties like C v mix. C v mix is what? C v mix is equal to sigma y i C v i. So, now, C v i I will calculate based upon mass basis ok. So, C v for nitrogen I can calculate as what? R of nitrogen divided by gamma of nitrogen minus 1. So, now the specific gas constant joule per kg Kelvin will come. So, from this I can calculate. So, now what is R of nitrogen will be equal to the universal gas constant divided by the molecular weight that is 28. Similarly, individual C v for neon I can calculate because that is 8314 by 20 will be R divided by 5 by 3 minus 1 ok. So, I can calculate the C v mix after calculating the individual C p and C v in terms of mass that is here C v is calculated as joule per kg Kelvin. So, it is per mass basis. Whenever there is a quantity per mass basis, then mass fraction should be used for the weighted average that is the mixing property. The mixture property of the mixture will be sigma y i C v i. On the other hand, if I can calculate C v as joule per kilo mole Kelvin, then I can use x i ok. So, now, it is straight forward for me to do this in the mass basis. So, I calculate using this C v mix equal to 692.83 joule per kg Kelvin. Similarly, C p mix I can calculate as sigma y i C p i. So, C p for nitrogen will be equal to gamma of nitrogen R of nitrogen divided by gamma nitrogen minus 1. Here this is joule per kg Kelvin. So, this C p also will be in joule per kg Kelvin. So, when I do this I get this value as 1039.25 joule per kg Kelvin or you can also use C p mix will be calculated as R mix plus C v mix because C p minus C v is equal to R for the ideal gas. You can use either of this and get the value. Now, after calculating the C p mix and C v mix, now I calculate the gamma that is C p mix equal to what? C p mix divided by C v mix that is the ratio of the specific heat for the mixture that will be calculated as 1.5 ok. So, why I need this? Because, see isentropic efficiency is given. Once to apply the isentropic efficiency I need the actual or the ideal process 1 to 2 s. So, process 1 to 2 s is isentropic for an ideal gas this process is written as p v power gamma equal to constant ok. Now, equation of state is p v equal to RT. Combining these two, these two equations I can write the expression for pressure ratio p 2 by p 1 equal to T 2 s divided by T 1 power gamma mix by gamma mix minus 1. Do you understand? So, that is the relationship for the isentropic process. Please understand that in this T s diagram it is clear that the pressure ratio is same for the isentropic process which is ideal as well as the actual process. In the actual process what is p 1 8 bar p 2 is 1 bar no problem in that. But only thing is the temperature reached in the isentropic process is T 2 s and in the actual process it is T 2 which is higher than this. That means your enthalpy change which is C p delta T for the isentropic process will be more. So, more work will be delivered and for the real process less work will be delivered because of the irreversibility. So, that is why the isentropic efficiency is 0.5 of the actual process w a by w s. So, once you calculate the isentropic work done I can calculate the actual work done because of the fact that isentropic efficiency is given to me. So, now I can fix T 2 s correct. So, p 2 is 8 bar p 1 is 1 bar. So, I can find T 2 s as from this the gamma also I know now ok. So, T 1 is known T 2 s I have to calculate pressure ratio is known. So, from that I can calculate T 2 s T 2 s is 150 Kelvin ok. So, 300 2 1. So, T 1 is 300 300 Kelvin ok. So, due to expansion temperature drops to 150 Kelvin. Now, isentropic efficiency is what actual work divided by isentropic work which is equal to H 1 minus H 2 divided by H 1 minus H 2 s which is equal to C p into T 1 minus T 2 divided by C p into T 1 minus T 2 s perfect gas. So, C p constant right ok. So, now, which implies this is 0.5 0.5 is isentropic efficiency. So, I can calculate from this expression T 2 as 225 Kelvin. So, go back and see here here T 2 s is what T 2 s 150 Kelvin and T 2 is 225 Kelvin. So, you can see that the temperature difference comes down for the actual process only 75 Kelvin, but for the isentropic process is 150 Kelvin. So, you are losing the work because of the irreversibility in the process that is it. So, now, we will calculate the entropy change. So, what is asked is calculate entropy change for that I have to find the mass. So, mass of the mixture what is that m equal to total number of moles into molecular weight of the mixture. So, this is mass in kg this is kilo mole and this will be what kg per kilo mole ok. Now, what is the NT value? NT value is 0.5 moles please understand it is given as moles not kilo mole. So, 0.5 divided by 1000 plus 0.5 divided by 1000 in kilo moles now it is in kilo moles ok into molecular weight of the mixture is 24. So, which will which is mass of the mixture will be equal to 0.024 kg. So, now, what is delta s delta change of the change in entropy is equal to m into Cp mix into ln T2 by T1 minus r mix into ln P2 by P1 ok. The same type of expression what we have already used. So, now, I get this the entropy change as 10.113 Kelvin that is it because that is the sorry Joule per Kelvin Joule per Kelvin ok. So, that is the delta s ok. So, in this problem if you should understand that the components of the mixture is given you have to first arrive at the mixture properties that is very important. And first of all for that I need the molecular weight of the mixture then I can calculate the specific gas constant and now I can calculate the Cv mix and Cp mix for that I will calculate the individual Cv values. If it is mass basis then I can use mass fraction and use this mixture rule or mixing rule. If it is in mole fraction if it is per mole basis then you have to use mole fraction ok. Now, I have kept everything in the mass basis. So, I can use sigma yi Cpi etcetera. Now, gamma there is no mixing rule for that. For calculating gamma you have to use the ratio Cp mix by Cv mix only. You cannot say gamma mix is equal to sigma yi gamma a and all. So, ratio find the Cp and Cv for the mixture and take the ratio of that. Then for isentropic process undergone by an ideal gas it obeys pv power gamma equal to constant here gamma is gamma mixture and equation of state is pv equal to RT. So, combining two I can find the ratio of pressure as ratio of temperature power gamma by gamma minus 1. From that I can find the temperature which is got from an isentropic process that is T2S. Now, either of the efficiency is given using that I can find the actual temperature. Now, the entropy change is got by using the integrated form of TDS relationship ok. This is about the problem 9.