 So with our problem we were able to find the exit temperature of our mixture and now what we want to do is we want to evaluate entropy generation through this process. So I'll begin by writing out the equation for a steady flow process with multiple inputs or exits. Now we're dealing with an adiabatic mixing chamber and consequently there is no heat transfer and the process is taking place at a steady flow rate. So there is no change within the control volume with time and consequently the term on the right disappears as well. So we're left with that for the entropy generation and we have our two fluid streams that we need to look at so let's express it in terms of both the methane and the ethane. So what we'll do, we'll begin by determining the change of entropy for each of the different fluid streams and we'll begin with ethane and given that our temperatures are not changing that significantly, we will use the approximate evaluation for the change in entropy and conditions that we have. We know coming in and I've used state 1 to be the inlet of ethane and 2 being the exit. So coming in we know that we're at 200 kPa and then exiting ethane, remember it's in a mixture with methane and so consequently we do not know the pressure but we would need the mole fraction in order to determine that and then that will be multiplied by the mixture pressure at state 2 and we were told the mixture pressure leaving or for a mixing chamber it won't change, it will always be the same. So you'll have 200 kPa coming out but we need to know the mole fraction there and likewise let's take a look at the entropy change for the methane. We know the state in is 200 kPa and then conditions for the pressure leaving, again we need to know the mole fraction and then that would be multiplied by the mixture pressure at exit conditions which is 200 kPa. So in order to proceed we need to solve for the mole fractions. So that's what we will do next. So if we look at a molar flow rate, we know the mass flow rates and that would be the mass flow rate divided by the molar mass for the particular component. So we can write out the ethane molar flow rate divided by its molar mass and similarly for methane. So we get the mixture molar flow rate being the sum of both of these. So what we can do now in determining the mole fraction, we can take each of these and divide by the mixture and that will then give us the mole fraction of each individual component. So for ethane and for methane, and once we have that we can then go back a couple of steps where was it. It was right in here where we had the expressions for evaluating the change and that was changing entropy for ethane and similarly the change in entropy for methane and we said that in order to proceed from there we needed to know the mole fractions which we now have. So we can proceed and evaluate the change in entropy for both of the fluids or both of the gases coming in and exiting and notice we're using the temperature of our exit stream that we determined from part A of the problem and similarly for methane. So with that we can go back, let's see where was it. We had our expression where right here. So this is the one that we can now use to evaluate the entropy generation and so that would be the final answer for our problem roughly 3.36 kilowatts per kelvin. So that concludes an example of dealing with calculating properties of mixtures, probably the trickiest part again was dealing with the entropy change, but you just go through step by step and it's actually quite straightforward in terms of solving it. You might have to get mole fraction or mass fraction. So that concludes this lecture, the next thing that we will be doing is moving on into a different section of gas mixture where our mixture consists of both a gas and a vapor and so that will be the next section that we'll be moving into in the course. Thank you.