 We're now going to work an example problem involving the adiabatic flame temperature. So what I'll begin doing is, as usual, we're writing out the problem statement. So there is our problem statement. It's a rather long problem that we're dealing with, but what we have is we're combusting methyl alcohol vapor, and we're told that it is in a container, 28 liters, originally 25 degrees C, 98 kPa, and what I want us to do is determine max pressure if combustion is constant volume. That's the first thing. The other thing that they want us to determine is max volume if combustion is constant pressure. And what we are to assume is that this is taking place adiabatically, which is specified here. So what we need to do is calculate the adiabatic temperature, and then with that we can determine these values. So let's proceed with what we know and what we don't know and try to work this problem. So that's what we're dealing with. We have combustion of methyl alcohol vapor, stoichiometric air is in the reaction, 28 liters. It's adiabatic. We know the initial conditions. We want to find max pressure if it is a constant volume combustion process, which would replicate what happens in the auto cycle. And we also want to find max volume if it was constant pressure combustion. So the first place to start with solving this problem is to do the stoichiometric balance. We don't have excess air. We have the stoichiometric amount of theoretical air. So that will simplify things a little bit. But let's begin by doing stoichiometry. So we're beginning with the carbon balance on the left and the right looking. We have one on the left and one on the right. X is then one for hydrogen. On the left we have all the hydrogen is in our methyl alcohol vapor. We have four on the left and on the right we have it in the water vapor, because that is 2y. And with that, that tells us that y is equal to 2. For diatomic oxygen, we have 0.5 in the alcohol. And given that we know x and y, we can solve for this. And therefore the theoretical amount of air here is 1.5. And finally, for diatomic nitrogen, we get this. So we can rewrite our chemical reaction as follows. So that is our reaction equation. And what we will now do is we will go ahead and we will address part A, which was maximum pressure if we have constant volume combustion. And we will use this equation in the first law for that in order to do that analysis.