00:01
So in this case, we have a rigid vessel that contains two kilomoles of carbon and two kilomoles of oxygen.
00:10
They are initially at 25 degrees c and 200 kilopascals.
00:17
The combustion occurs and the result produces one kilomole of carbon dioxide and one kilo -kilmole of carbon monoxide.
00:26
And excess oxygen at the products then are at, um, and let's see here at 1 ,000 kelvin.
00:39
So we can figure out how much excess oxygen we have, and that's a half a kilo mole.
00:47
So we know we have a constant volume process.
00:52
Now, we have, let's see here, in the input, the gas, the number of moles of gas, in the input, in the reactor, is 2 or kilomoles whatever and in the products this is all gas so we have 2 .5 kilomoles.
01:17
Now we know we can figure out the pressure in the products using the ideal gas law and again using our mole fraction from the what we had of gas in the initially and what we had in the gas in the final state and so that gives us 800 ,000, 38 .4 kilopascale so we obviously increased the pressure a lot because a lot more of this stuff became gas or a lot of this carbon all became turned into a gaseous form not a gas just form of carbon but molecules with carbon that are gas now we can again we say that this is the reference um enthalpy in our initial state here um so we have uh we want to figure out the heat transfer and so if this is the reference then we can figure out what the the we have the entope of formation of the co2 here and then the entropy above above these can be added entropy going from this state to this state and then for the let's see here for the carbon monoxide we have the entropy of formation and then the extra entropy from going from this state to this state again.
02:53
And likewise, there's no entropy of formation for the o2...