Skip to 0 minutes and 5 seconds So we’re going to look at a mixture of air and petrol at initial temperature of 20 degrees C contained in a rigid vessel. The mixture undergoes a series of processes, and we’re asked to evaluate the energy of the mixture after each process. It has an initial internal energy of 25 kJ and the mixture temperature is raised to 250 degrees C by a positive heat transfer of 13 kJ So I’m going to start this problem by drawing a picture and summarising all the information on it. So I’m going to draw my container here.

Skip to 0 minutes and 50 seconds It’s a rigid container, so I’m going to give it a nice, thick wall to represent the fact that it’s rigid.

Skip to 1 minute and 3 seconds And it contains an air and petrol mix. And its initial temperature T0 is 20 degrees C. And we’re going to raise the temperature to T1 at 250 degrees C. We’re told its internal energy initially, and that’s 25 kJ. And we’re being asked to find what the internal energy becomes, U1 when we transfer in a heat transfer equal to 13 kJ. So to tackle these problems, the first thing we should do is identify a system boundary. And so I’m going to use a system boundary that’s just inside the wall of my rigid container here so that it contains the entire mixture. So that dotted line represents my system boundary.

Skip to 2 minutes and 21 seconds And now I’m going to apply the first law of thermodynamics to this system here. So apply the first law to the system.

Skip to 2 minutes and 40 seconds And so in its standard form, we have a heat transfer term here, the heat transfer n minus heat transfer out plus the term representing the work n minus the work out plus the term representing the energy transfer that occurs as a consequence of mass flow. So energy is stated with mass going in minus energy associated with mass outwards. And we can equate all these energy transfers across this system boundary to the change in the internal energy of the system delta U plus the change in the kinetic energy of the system plus the change in the potential energy of the system. So our system is not moving, and therefore, there’s no change in kinetic energy. So this term goes to 0.

Skip to 3 minutes and 36 seconds So equal 0 due to no motion.

Skip to 3 minutes and 45 seconds It doesn’t change position, anyway. And so this term here, the potential energy, doesn’t change. So this is equal to 0 due to the position change.

Skip to 4 minutes and 2 seconds It’s a closed box, and so that means that this term here, the mass flow, is going to be 0. So we can set this to 0.

Skip to 4 minutes and 19 seconds And because it’s rigid, we can’t do any work with it, and so this term here also goes to 0.

Skip to 4 minutes and 33 seconds So that just leaves us with two terms here. We’ve got our heat transfer term and the internal energy. And so we can equate these and say that Q in minus Q out is equal to delta U or the final internal energy, U1, minus the initial internal energy U0. We’ve only got q going in here, so this term also can cancel. And so we can rewrite this as U1 is equal to U0 plus Q in. And we’re told this initial value is 25. And we’re putting in 13 here, and so we end up with 38 kJ. And that is the answer to the first step. So now let’s rub this out and move on to the second part.

Skip to 5 minutes and 35 seconds So now the mixture ignites and burns completely and adiabatically, and the temperature rises to 1,500 degrees C. So once again, let’s draw a picture of our container.

Skip to 5 minutes and 55 seconds And we’ll show the walls nice and thick as before, because they’re still rigid.

Skip to 6 minutes and 3 seconds But now we’re told that it burns adiabatically. So that implies there’s no heat transfer, and so they’re also insulated. So let’s just mark that on here– insulated. And we’ll shade it to emphasise that so there’s no heat transfer across these walls. And so we’ve got an air fuel mixture that is going to combust, and it will go to a carbon dioxide and water mixture. And we start this process at t1, the temperature from before, 250 degrees C, with an internal energy that we found previously. That’s 38 And we’re told that this goes to a new temperature of 1,500 degrees C. And we’re asked to find the new internal energy. That’s our unknown.

Skip to 7 minutes and 4 seconds So as before, we should draw a system boundary. And I’m going to use the same boundary as before and put it just inside the container so that all the mixture and then the combustion products are contained inside it. So this is my system boundary.

Skip to 7 minutes and 28 seconds And so to solve this, I’m going to take the same approach as before, and I’m going to apply the first law of thermodynamics to the gases inside this container. So we can say again apply the first law to the system.

Skip to 7 minutes and 56 seconds And we can write down the same expression, the heat transfer terms plus the work terms plus the energy transferred as a consequence of mass flow in minus the energy transferred as a consequence of mass flow out. And all these energy transfers across our system boundary here can be equated to the change in the internal energy plus the change in the kinetic energy plus the change in the potential energy. And as before, our system is stationary, and so the kinetic energy change is 0. So this is equal to 0 due to no motion.

Skip to 8 minutes and 50 seconds And there is no change in position, either, so this term also goes to 0.

Skip to 9 minutes and 7 seconds The walls of the container prevent any mass flow in or out, so this term here also goes to 0.

Skip to 9 minutes and 23 seconds And because the walls are rigid, we can’t have the gas do any work. And so this term here also is 0.

Skip to 9 minutes and 39 seconds And then finally, we’re told that the process is adiabatic. And I’ve shown it insulated here, and so that means there’s no heat transfer either. So this term goes to 0, too.

Skip to 10 minutes and 1 second So the only nonzero term left is this delta U And so we can rewrite this as delta U, which is equal to U2 minus U1, is equal to 0. And so that implies that u2 that we’re looking for must be equal to U1, which is 38 kJ. And that’s the answer to the second point. So although we’ve combusted our mixture here, because we’ve got a rigid insulated container, none of the energy escapes, and so there’s no change in the internal energy despite the combustion process. OK. Let’s move on to the third part.

Skip to 10 minutes and 48 seconds So in the third part of this problem, the temperature of the products of combustion is reduced to 138 degrees centigrade by heat transfer out of the system of 35 kJ. So once again, we can draw a picture to represent what’s going on.

Skip to 11 minutes and 13 seconds So we’ll have a nice, thick wall again. This time, it’s not insulated. It’s just rigid.

Skip to 11 minutes and 23 seconds And we’ve got a heat transfer outwards of 35 kJ. So Q out equals 35 kJ. And inside here, we’ve got the combustion products.

Skip to 11 minutes and 48 seconds And they’re sitting at a temperature here, that t2 we had before, 1,500, and a U2 internal energy that we just found that was 38 kJ. And we reduced the temperature to 138 degrees, and we’re asked to find the new internal energy. So we do this in exactly the same way as before. We draw a system boundary. And I’m going to put that just inside the wall of the container so that we capture all the combustion gases in that. So this is our system boundary. And we’re going to apply the first law of thermodynamics to this system again. So apply first law to this system.

Skip to 12 minutes and 59 seconds And so we have the standard terms that we’ve had previously, Q in minus Q out for the heat transfers plus the work in minus the work out to the system, and then the energy transfer as a consequence of mass flow in minus the energy transfer as a consequence of mass flow out. And all of these are equal to the change in the internal energy of the system plus the change in the kinetic energy of the system plus the change in the potential energy of the system. And as before, our system is stationary, so this term goes to 0. No change in position, so this one goes to 0 as well.

Skip to 13 minutes and 52 seconds It’s all sitting inside a container, so there can be no mass flow in or out. It’s a closed container. So this one goes to 0 as well. And then because it’s rigid, then we can’t do any work. And so this term here goes to 0 as well. And so we’re just left with the heat transfers and the change in internal energy. So we can equate those two and say that Q in minus Q out is equal to delta U. Delta u is going to be U3 minus U2. And we’ve only got a heat transfer out, so this one disappears as well. And so we can rewrite this now as U3 is equal to U2 minus q out.

Skip to 14 minutes and 50 seconds And we’re told that Q out is 35, and we’ve found U2 previously to be 38. So we get 38 minus 35 equals 3 kJ. And that’s the final internal energy left inside here after we’ve made that heat transfer.

# Combustion of petrol (worked example)

Look at the question posed below. Have a go at doing it yourself, and then watch the video. For a complete solution you can go to the PDF at the bottom of this page.

In this video Eann will walk you through the following example.

A mixture of air and petrol at an initial temperature 20°C is contained in a rigid vessel. The mixture undergoes the following processes in sequence; evaluate the energy of the mixture after each process if the initial internal energy is 25 kJ.

a) Mixture temperature is raised to 250°C by a positive heat transfer of 13kJ.

b) Mixture ignites and burns completely and adiabatically and the temperature rises to 1500°C.

c) Temperature of the products of combustion is reduced to 138°C by a heat transfer out of -35kJ.