The fuel in high-efficiency natural-gas vehicles consists primarily of methane (CH₄). (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?

Video transcript

Hello everyone. So in this video we want to see the maximum amount of useful work that can be achieved by burning one mole of our ethanol under center conditions. So our key word here is going to be burning because burning is a combustion reaction. So first thing I'm gonna do is go ahead and route the combustion reaction for this problem. So that's going to be when again, starting off with our ethanol, the C two H five oh H. Which is a gas is gonna go ahead and react with three moles of our oxygen gas to give us two moles of our carbon dioxide. And that's going to also give us three moles of liquid water. So at constant temperature and pressure, the maximum work is equal to our delta G. And r delta G is going to equal to the delta G of my products, my sc delta G of my reactant. So just using this sort of table here, we're gonna go ahead and plug those values into this equation here, let's go ahead and do that. Then so again, our maximum work here is equal to, let's see here from my products, we have our carbon dioxide are two moles of carbon dioxide. So we have two moles and the corresponding delta G value for that is going to be negative 394.4 killer jules Permal. And then we also have our three moles of water. So three moles multiplied by negative 271.13 killer jewels per mole. And then we're gonna go ahead and subtract that of course with the delta G. Of my reactant putting the second role here. So we have one mole of our ethanol And the Delta G value for that is negative. 1 68.5 kill a jules Permal. And of course we're going to add that with the second starting reagents, which is going to be our oxygen gas. So we have three moles of that And the corresponding delta G is zero killer joules per mole. So once we put all that into the calculator, we get that. The maximum work is negative. 1331.69 kg joules. So we see here that the work is negative. This negative sign means that the work that's done by the system on the on the surroundings. So our answer then our final answer is that the maximum amount of work of useful work that the system can achieve is 133.1. Or let's see here. 1331.7 kg Joel's alright. And this right here is going to be my final answer for this problem. Thank you all so much for watching