1.7

## University of Liverpool

Skip to 0 minutes and 4 secondsSo we're going to do a worked example on an air and petrol mixture inside a rigid vessel. It's at an initial temperature of 20 degrees C. And the mixture undergoes a series of processes in sequence. And we're asked to evaluate the energy of the mixture after each process, if the initial internal energy is 25 kilojoules. So I'm going to start out by drawing the rigid box. So I'm just going to have a rectangle here. And I'll make the walls nice and thick to indicate that it's rigid. OK. So these are rigid walls. And it's an air and petrol mix in here. And we're told that it's starting temperature is 20 degrees C.

Skip to 1 minute and 0 secondsAnd it's got an initial internal energy of 25 kilojoules. And what we're asked to find is the internal energy after the first process. And the first process consists of the mixture temperature being raised to 250 degrees C. So this goes to 250 degrees C by a positive heat transfer of 13 kilojoules. So we're going to move into here, Q, in heat transfer of 13 kilojoules. And so we can solve this by using the first law of thermodynamics. And so we can write here, first law of thermodynamics.

Skip to 2 minutes and 1 secondAnd I'm going to apply it to the mixture.

Skip to 2 minutes and 8 secondsSo I'm going to show my system boundary as a dotted line just inside my box here. So the box itself is not included. But all the air mixture is. So we can write the first law of thermodynamics as the heat transfer terms. That's in and out. Plus the work in and out. Difference between them, plus the terms due to energy carried by a mass transfer in. And then also the mass transfer out. So the difference between them. And that little lot is all equal to the change in energy of the system, which will be a change in internal energy, plus the change in the kinetic energy, plus the change in the potential energy.

Skip to 3 minutes and 3 secondsSo it's a rigid box, and that means that we can't do any work. So this term here has to go to zero.

Skip to 3 minutes and 13 secondsBecause it's enclosed in a box here, then it's not possible for any of the air or the petrol to leave. So none of these mass transfers can occur. So this term will also go to zero. It's a stationary box, so there's no change in the kinetic energy. And it doesn't move in terms of its position or height, and therefore, the change in potential energy goes to zero as well. So we're just left with two terms. The change in internal energy, and the heat transfer. And nothing escapes. We're just a positive transfer in, so this one must also go to zero. And so that means that delta U is equal to the Q in.

Skip to 3 minutes and 54 secondsWe can expand this to U one minus U naught equals Q in. And so that's implies that U one will be equal to Q in plus U naught. And that's equal to the 13 kilojoules plus the 25 that we started with. So that gives the final value of 38 kilojoules. And so that's a condition at the end of the first part of the process. So we'll move now on to the second part. And we'll just clear these out of the way. So in the second process, the mixture ignites and burns completely and adiabatically, and the temperature rises to 1,500 degrees C. So we're starting from the conditions we just found in the previous case.

Skip to 4 minutes and 45 secondsSo T1 was 250 degrees centigrade, and U one that we found was 38 kilojoules. And we're going to move to a state where this is now at 1,500 degrees C. And we're asked to find the internal energy in this new state. And we're told it's adiabatic, so that means that there's no heat transfer. So effectively now, our walls are insulated. So we could shade them to indicate that that's the case.

Skip to 5 minutes and 24 secondsAnd we can apply the first law of thermodynamics again to the mixture. So I've left my boundary in place here. We're going to do the same thing. So the first law of thermodynamics applied to the mixture.

Skip to 5 minutes and 52 secondsSo just as before, we can write out the full form again. So first of all, we have the heat transfer terms, and then the work transfer terms. And then the terms due to the movement of mass across the boundary.

Skip to 6 minutes and 18 secondsAnd that's equal to the change in internal energy, and the change in the kinetic energy, and the change in potential energy. And so we look at this system here, we can see it's not moving. It doesn't change its position. So this term goes to zero, because it's stationary. This term goes to 0. It's all contained in a rigid box, so this term here goes to zero as well. And because it's rigid, there can be no work done. So this term has to go to zero. And we've insulated it, so there's no heat transfer either. So this term here goes to zero. So we're just left with delta U. And so delta U must also be equal to zero.

Skip to 7 minutes and 6 secondsAnd delta U is U two minus U one. And so that implies that U two is equal to U one, and is equal to 38 kilojoules. So there's no change in internal energy, even when we combust this air-petrol mixture. And that's because there's no way for the energy to escape. We've got an insulated box, so we can't get it out through heat. It's a rigid box as well, so it can't go out as a consequence of work. And it's all contained inside so, there can be no mass flowing across the boundary, taking anything with it. So the internal energy stays the same despite the combustion process. So now let's move on to the third part.

Skip to 7 minutes and 48 secondsSo in the last part of the process, here, the temperature of our mixture is reduced by a heat transfer out of 35 kilojoules. So this is a Q out. And, of course, we pick up the temperature from last time. So T two was 2,500 degrees C. And we're told that we cool it by this temperature transfer out, or heat transfer out, down to 138 degrees C. And we had U two from the previous part, which was 38 kilojoules. And we're asked to find what the new state is with this transfer out. So we can, again, do the first law of thermodynamics.

Skip to 8 minutes and 45 secondsand apply it to the mixture.

Skip to 8 minutes and 55 secondsSo if we write out, in its full form-- so the heat transfer terms, plus the net work transfer, plus the net energy movement as a consequence of mass flow.

Skip to 9 minutes and 17 secondsAnd all of that's going to be equal to the change in the internal energy, plus the change in the kinetic energy, plus the change in the potential energy. So our box is in the same position. So this doesn't change. And it's not moving, so this doesn't change. It's a container, so nothing escapes. So this term goes to zero. And it's rigid, so we can't do any work with the mixture. So the work terms disappear to zero. So that leaves it with just the heat transfer and the internal energy change. There's nothing coming in, so that goes to zero. And so we're left with Q out is equal to the change in the internal energy.

Skip to 10 minutes and 3 secondsAnd that's going to be U three minus U two. And so we can say that U three will be equal to Q out, plus U two. Q out is negative. So we're going to get minus 35 plus our original here, 38. So we'll be left with 3 kilojoules of energy, internal energy, left inside. And that's the end of the problem.

# 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 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.