First Law of Thermodynamics

In calculus, I learned that if z=xy, then z'=xy'+x'y. However, in my chemistry class, I was given this following equation and told to memorize it:

dW = -PdV

I wanted to know where this comes from, so I did some searching, and after searching it for a while, I was unable to find the answer anywhere. Where is the VdP part of the equation? And why is there a negative sign? Should VdP also be negative so that the differential equation can be integrated? (into W=-PV)

I know that this is probably a partial derivative, but I came across nothing that states why this equation must be taken as a partial derivative. Actually, I couldn't find anything other than Wikipedia that denoted it as a partial.

What if you had a pressure change but not a volume change? I'm confused.

http://en.wikipedia.org/wiki/First_law_o...modynamics

You're soo helpful.

I already looked there.

You're soo helpful.

I already looked there.

I'm sorry for linking you to a source that explains and answers all your questions. I shall refrain from doing so in the future.

I'm sorry for linking you to a source that explains and answers all your questions.

Did you bother to pineappleing read the first post or the wikipedia article you linked to? The article doesn't answer any of the questions I asked. Also, Wikipedia is such an obvious place to look that there's completely no possibility that I could miss it.

Piss off. I need legitimate help and explanation, not an impaired form of Google search.

Speaking of google search:

http://www.answerbag.com/q_view/1089309

http://en.wikipedia.org/wiki/First_law_o...ormulation

The first paragraph explains why it's a partial derivative.

Did you bother to pineappleing read the first post or the wikipedia article you linked to? The article doesn't answer any of the questions I asked. Also, Wikipedia is such an obvious place to look that there's completely no possibility that I could miss it.

Piss off. I need legitimate help and explanation, not an impaired form of Google search.

Seriously dude? Cool down.

@Fiel: That doesn't help. I already saw that. It contradicts the Wikipedia article segment that says that when the exact derivative is taken, the total is 0. Then it fails to explain why whoever wrote that "answer" derived it fully instead of taking the partial.

@Riyuran: I know it says so. It doesn't provide a good explanation. Here's the portion you linked to:



Where in this segment does it explain where the VdP part went? Where does it explain the negative sign? Where does it explain why completely differentiating it won't work? And before you link me to the thermodynamic process Wikipedia page, I already clicked it. It doesn't explain it either.

Seriously dude? Cool down.

Ultimately, I want to know how the equation is derived. Just because I made this in the Speakeasy instead of the Rubik's Cube doesn't mean that people should come in and, without fully reading the first post, link me to absolute crap and other unhelpful sources that I have already seen, especially when the sources aren't very credible to begin with (answers.com, answerbag.com). And then there's that attitude of pretending to have fully answered all of my questions when people only address a rather trivial section of my post.

If you are not interested in helping me or you don't know the answer, stay out of my thread. It's not that pineappleing hard. I'm sick of receiving piss poor replies from a forum that's supposed to contain a large userbase of intelligent individuals. At least have some decency.

Work is expressed in such a way that a change in pressure is entirely internal and U doesn't change. On the other hand, a change in volume (dV) requires work to be done on the surrounding environmetn.

Ultimately, I want to know how the equation is derived. Just because this is in the Speakeasy doesn't mean people should come in and, without fully reading the first post, link me to absolute crap and other unhelpful sources that I have already seen. And then the attitude of pretending to have answered all of my questions when you only addressed a rather trivial section of my post is moronic. If you are not interested in helping me, then stay out of my thread.

I don't know anything about deriving equations or about the first law of thermodynamics, so excuse me if this post is stupid; however, it looks like people are trying to legitimately help you. They are giving you links of stuff that, to them, explains the problem just fine. Why would they link you to stuff they don't think will help you? What, is everyone out to troll you?

Work is expressed in such a way that a change in pressure is entirely internal and U doesn't change. On the other hand, a change in volume (dV) requires work to be done on the surrounding environment.

Thank you to be the first one for trying to explain. However, I still don't understand why pressure can't change in from an external force. Why must it be entirely from an internal force? Suppose that there's a random mass of gas in space and there's entirely nothing around. (Okay, that's not entirely true because there's always gravity, but let's just assume that the gravity is negligible or 0.) Then there can only be outward expansion because the molecules that are moving must collide into each other and bounce off of each other and travel towards nothingness.

Then the volume increases and the pressure decreases, and where's the work done? Also, if dW is already work done, what does integrating it and getting W give you?

I don't know anything about deriving equations or about the first law of thermodynamics, so excuse me if this post is stupid; however, it looks like people are trying to legitimately help you. They are giving you links of stuff that, to them, explains the problem just fine. Why would they link you to stuff they don't think will help you? What, is everyone out to troll you?

I wrote in the first post that I already saw the Wikipedia article. Linking me to it again without any other comments doesn't help at all. Doing this is clear cut trolling, because it assumes that the answer is completely obvious and common sense, and it implies that I'm moronic for not seeing it. If a child tells you that he is unable to learn from Sesame Street why things don't disappear when they're hidden, telling him to go watch it over again and again isn't going to help him.

But if the answer is so obvious, then it should be easy to explain too. Then why isn't anyone willing to do it? Because they're out to troll me. If the article explains the problem(s) just fine, then do you completely understand what the article is talking about? If you do, can you explain it to me in a way that will make me understand it?

There's a negative sign because it's a convention. Work is being done on the surroundings by the system, and is therefore negative, just by definition.

I don't know anything about deriving equations or about the first law of thermodynamics, so excuse me if this post is stupid; however, it looks like people are trying to legitimately help you. They are giving you links of stuff that, to them, explains the problem just fine. Why would they link you to stuff they don't think will help you? What, is everyone out to troll you?

Harrison just generally has a stick shoved up his ass that makes him feel he's obligated to be a douchebag at every turn.

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There's a negative sign because it's a convention. Work is being done on the surroundings by the system, and is therefore negative, just by definition.

Yup. In the first law E= Q-W, you add Q most often than not, and the system does work on whatever, usually gas.[/COLOR]

It's just the definition of work. Work is force integrated over distance. Pressure is force, volume is distance. Negative sign is convention, as previously mentioned.

Stating an equation in words doesn't explain its origin.

P = F/a = F/d^2
V = d^3

P is force, but V is definitely not distance. P isn't really force either, so your explanation doesn't make logical sense.

Going from the above equations, using simple algebra, only PV will give the solution W = FD, which is what I understand to be the definition of work. However, I'm not interested in the algebra; I'm interested in the calculus behind this equation. In calculus, W = FD isn't really W. It's dW. Which leads to my original question, why the VdP term is not in the equation for dW.

And if dW is work, what is W=integral(dW)?

Force times distance is the same as pressure times volume. Pressure is force over area, and volume is volume. Overall, you still get a force times a distance.

Force times distance is the same as pressure times volume. Pressure is force over area, and volume is volume. Overall, you still get a force times a distance.

Gee thanks for repeating what I just wrote.

Pressure is the force exerted by atoms on their surroundings.

F/d^2 * d^3 = Fd

dW is change in work.