Yes, this is homework.
I’m having problems with the calculations required for finding overall q, w, ∆H and ∆E for a cycle are. I understand the concept, and can do calculations for each individual step. But to me, it seems like in a perfectly efficient system, each one of those would be 0 since it’s a cycle. Google and wikipedia give me plenty of background on it, but none show me how to set up the equations. Anyone here have any other resources?
I’m not certain why you think they should be zero. Your car engine also runs on a cycle; do you expect that (absent any frictional or thermal losses) that it would pump fuel back into the fuel tank after it is done burning it?
The Carnot cycle describes an ideal heat engine in which energy is taken from a high temperature reservoir (in the case of your engine, from the high temperature and pressure of combusting fuel) and rejects it to a lower temperature reservoir (the external environment) while performing some amount of mechanical work (increase in pressure that drives the pistons) in the process. In the case of the ideal Carnot cycle the processes alternate between being isothermic and isentropic, with work equaling exactly the amount of energy transferred through the system, W = Q[sub]H[/sub] - Q[sub]C[/sub]; this also makes it reversible, i.e. by applying the same amount of mechanical energy back into the system you can pump energy so that -W = Q[sub]C[/sub] - Q[sub]H[/sub].
If your instructor isn’t explaining this to you clearly (and the tragedy of thermodynamics seems to be that for such a simple topic most instructors can’t teach it worth a damn) try H.C. Van Ness’ Understanding Thermodynamics. I think there are a couple of worthwhile discussions in *Feynman’s Lectures on thermodynamics and reversible cycles as well, though being an introductory physics course Feynman obviously doesn’t go much into applied and engineering thermodynamics in detail.
Stranger
I understand that the heat from the hot reservoir is going to cause the gas to do work on the surroundings, but why doesn’t it release an equal amount of heat when it is compressed back to the initial volume?
This is the fourth time that I have studied thermodynamic in a class, and second time I’ve tried to take pchem (and so second time with P-V work), and for some reason, I’m just now understanding the very basics of this. I’m a chem major and it’s the one concept I’ve run into I just can’t get.
Because it has lost a lot of the heat-energy in the act of expanding, transferring it to mechanical motion; a cold gas is much easier to compress than a hot one, hence why intercoolers are often used on turbochargers. This is why you want the difference between the high temperature threshold of your cycle and the low temperature to be as great as possible; all else being equal a large spread makes the cycle more efficient. The difference between them is proportional to the amount of energy that is converted to mechanical work; if you could let the gas expand to infinity the temperature would approach absolute zero; as a practical matter, you have to accept that it will be somewhat above the ambient temperature unless you have a refrigeration cycle (used in some large power production facilities, but for something like a car engine a simple heat transfer loop that uses coolant flow to reject heat to the environment suffices).
As I wrote earlier, thermo is actually pretty simple in concept, but many instructors dive right into the math without really explaining the concepts, and you end up with a bunch of students who can work problem sets without understanding thermodynamics. P-Chem, on the other hand; I found that class to be a bitch because it was such a catch all without an apparent organization. Later, after taking more specific classes in statistical mechanics, thermodynamics, some introductory quantum mechanics, et cetera that I started to see how it gelled together, but it’s a tough class.
Stranger
So without posting particulars of my homework on here, and because I for some reason can’t translate the equations on the equation sheet into anything useful for this situation, how WOULD I calculate q, w, ∆E and ∆H?
And back to theory - is there some sort of example I can think about this easier? I’m thinking about a piston in a cylinder - I can see how heating it would make the gas expand, I can see how decreasing the pressure (weight) on the top would make it expand without adding heat, I can see how it cooling would cause compression, and I can see that adding weight/pressure to the top would compress it without adding heat.
But I don’t see how this overall process does net work on the surroundings, even assuming C[sub]v[/sub] and C[sub]p[/sub] are constant and the fact that we never even mentioned the difference in compressibility between hot and cold gases in class.
Unless I’m just making this WAY too complicated, and I really just need to find q[sub]1[/sub], then multiply it by the efficiency as calculated by the temperatures of the hot and cold reservoirs and ignore all the theory behind it.
But aren’t ∆E and ∆H state functions, so they’d automatically be 0 for a cycle?
For the simple case of the Carnot cycle, you can just algebraically rearrange Carnot’s theorem. For a more complex, real world situation the math is more difficult but the principles are the same.
You say, “…I can see that adding weight/pressure to the top would compress it without adding heat.” This is incorrect; if the gas is in an adiabatic chamber and you do work on it to compress it, it will heat up (mechanical energy is converted to heat). Alternatively, when you allow the heat-energy in the gas to move a piston, you are converting heat-energy into mechanical energy via the mechanism of work and the gas cools as it expands. If you then allow the resulting gas to radiate some of its remaining heat to the environment it takes much less work to compress it back to the original volume; thus, you get net mechanical work out of the system.
Consider this illustration. Let us say that you have a pail full of sand on your deck, and you tie it to a line that runs through an overhead pulley and is attached to a bucket of beer. You kick the pail of sand off of the deck and it raises up the bucket of beer, allowing you to pull out a few longnecks for your drinking pleasure. Now you lean down and tip over the pail of sand so about half of it spills out, and now, weighing less than the bucket of beer, it automatically comes back up. In this case, net work was done proportional to the mass of the sand that you ejected from the sand pail. Add more sand to the pail to repeat the cycle.
This is the Carnot cycle except we’ve substituted height for temperature and the amount of sand in the bucket for entropy (see the T-S diagram in the above link). Obviously, the energy to do work didn’t come from nowhere; it came from the effort you expended–or, if you are clever, fooled one of your less intelligent buddies to expend–hauling bags of sand up on the deck so that you can pull off this trick. So now you learned something about thermodynamics and had a couple of brews to boot. Yet another triumphant victory for physics. Nicolas Léonard would be proud.
Stranger
So here’s an important distinction that I think I had missed before. This whole time I was trying to account for the heat throughout the whole cycle, and was thinking for some reason the heat lost by the system to its surrounds during the isothermal compression was added back to the hot reservoir. So we couldn’t give two shits where that heat’s going, we know that we are putting some known amount Q heat into the system, it’s doing work, then in the third step all the heat goes away into the surroundings never to be seen from again and gives us a fake telephone number?
To relate it to the sand analogy, we’re saying I have all kinds of burly, glistening naked men to bring me bags of sand, and that I’m just dumping the sand from the bucket into a bottomless hole somewhere, and don’t care how much sand I waste? As in, the sand that I’m kicking out of the bucket isn’t the same sand that the burly men are going to carry back up the stairs to me to refill the bucket?
(and what about ∆H and ∆E and state functions?)
Heat is just one form of energy[sup]*[/sup]. When we allow the cycle to undergo isothermal expansion the heat energy is converted to directional mechanical energy; in the case of a piston, it is the motion of the piston and the force upon whatever crank or mechanical media it interacts with. So the heat from within the system doesn’t “disappear”; it’s decrease is accounted for by the mechanical energy gained from the isothermal expansion leg of the cycle. When more heat energy escapes the system via thermal conduction to a low temperature reservoir during the isentropic expansion the energy remaining in the system drops; it is as if someone has come along and siphoned out gasoline from your gas tank even though you didn’t really go anywhere, so when it comes time to start doing compression (adding gasoline) the tank is almost empty, and you can put more fuel in it than you expected.
Okay, first of all, I did not need the image of “burly, glistening naked men”! 
The sand being hauled up is part of the energy you are going to introduce to the cycle every time at Step 1 (prior to isothermal expansion); this is what drives the cycle, and the pool from which you derive the energy to do work. However (in our analogy) your men would be hauling sand up from the bottom of the garden, below the ground under your deck, so the potential energy put into the sand by carrying it up is not completely recovered by dumping it out on the ground below the deck.
What about them? State functions merely describe the property of the state independent of path, just as mechanical work describes the amount of energy done on or by a system independent of how you got it there. (In other words, whether you took a straight route up the stairs or did one of those Billy from Family Circus runs, the only thing that matters is the difference in height between the two states). The state function you are interested in is the enthalpy differential equation. There are no overall values for the cycle; these functions have values only when you pick a specific point in the cycle.
Stranger
*Specifically, heat is randomized energy that lacks an inherent vector. It, or rather the manifestation of it, acts in all directions equally and can only be utilized by offering a lower temperature reservoir toward which it or the media it is in will flow.
I feel like everyone could use naked, burly, glistening men in their lives. if not to do any more than mow and wear censor bars.
I turned in the assignment today, and used the Carnot Theorem to calculate w from the Q that I had calculated when I had to do the (Q,w,∆H,∆E, etc for each step), then wrote down “∆H and ∆E are state functions, therefore equal 0 for a cycle” and turned it in.
And going back to the sand example, this is how I had imagined it working before our conversation, Stranger, is that I drop the bucket to get the beer, tip the bucket to get it to raise back up, then reach down, grab all the sand and put it back in the bucket, so I was understandably confused about what the hell the point was when I could just reach up and grab the damn beer.
Anyway, thanks for your help and having basically an email conversation with me to walk me through it.