First Law of Thermodynamics

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

Pressure is a constant.

Pressure is the force exerted by atoms on their surroundings.

F/d^2 * d^3 = Fd

Gee, thanks for repeating what I just wrote.

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Gee, thanks for repeating what I just wrote.

Hey, dickasaurus rex, we are trying to help. I can't think of any other way we can. We've given why it is - multiple times, in multiple different forms. Learn to read, or rephrase your question.[/COLOR]

Gee, thanks for repeating what I just wrote.

Gee, are you always such a pompous plantain head?

If this is seriously that hard of a concept for you to wrap your head around, I suggest you ask your TA or professor to explain it to you during their office hours. I have no idea what you are trying to ask that hasn't been answered in this thread already. The derivation of the equation is very simple and only requires algebra to do. P*V is just F*D in different units. Calculus just lets you generalize W = F*D to changing pressures and volumes. Negative sign is convention. There isn't anything else to the First Law.

Where is the VdP part of the equation?

Should VdP also be negative so that the differential equation can be integrated? (into W=-PV)

What if you had a pressure change but not a volume change?

I'm pissed because you are all stating this like it's such common sense. Even your last post only uses simple algebra. I already know these backbone equations. I want to know why they work conceptually. So far, Stereo is the only person who tried to, but he didn't follow through and explain the rest:

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 posted this topic in the Speakeasy and the Rubik's Cube because I assumed that some of you must have taken at least one class that explains these concepts, and if you have, that these questions should be fairly easy to explain (and therefore not "intelligent discussion"). And what do I see? Some dude's marching band that apparently needs "intelligent discussion", and my topic being treated like it's crap from the Funhouse.

Links to wikipedia articles. Having the equations typed out in words. Wow.

If you had a pressure change, yet constant volume, you integrate with respect to pressure, not volume. The negative is only since the work is being done to the system from an external source. In the first law, it has to be there. If not, the world would ... I don't know, explode? Implode? Vdp is right after the minus sign. E=Q-Vdp.

The first law is defined within the terms PdV. I haven't seen it the other way around. I also haven't seen them both written together. What if they were changing at the same time and at different rates? Why is one part or the other ignored?

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.

W is energy, dW is work.

In an ideal gas, PV = nRT
Since in a closed system, n, R are constant, you can replace one of the other 2 variables with this equality. (eg. P = T/V) and thus eliminate one of the 3 partial derivatives (dT, dP, dV)



I think this table is another clue to the issue:
http://en.wikipedia.org/wiki/Table_of_th...d_concepts
U = TS - pV + sum(...)
H = U + pV

If you scroll down further, you see
dU = TdS - pdV + d(sum)
dH = TdS + Vdp + d(sum)

So the Vdp term is going into enthalpy for some reason.

You don't really have problems where both p and V and being changed. It doesn't even work like that irl. Ever hear of atmospheric pressure? If you do, just integrate properly.

W is energy, dW is work.

If PV=nRT, then nRT = W = energy. This part makes sense to me now. Thank you.

Since in a closed system, n, R are constant, you can replace one of the other 2 variables with this equality. (eg. P = T/V) and thus eliminate one of the 3 partial derivatives (dT, dP, dV)

In an ideal gas, PV = nRT.

Then:

-W = PV = nRT
- dW = nR(dT) = VdP + PdV.

How did you cancel only PdV or VdP, without canceling the other term, by replacement? Also, this new equation found through substitution doesn't really do anything. It basically states that total energy is the energy of all the molecules, which is really circular.

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And what about a real gas:





The equation doesn't cancel out to PV=nRT.

You can express P as a function of V, or V as a function of P. It generally makes more sense to integrate over a volume, not a pressure, as far as finding work goes.

Stereo fully explained how to cancel out an entire variable. By substitution.

This goes over a very simplistic derivation of it:

http://www.citycollegiate.com/thermodyna..._first.htm

In an ideal gas, PV = nRT.

Then:

-W = PV = nRT
- dW = nR(dT) = VdP + PdV.

How did you cancel only PdV or VdP, without canceling the other term, by replacement?

It's a bit of a cheat really, because Pressure can be treated as approximately ~constant at an infinitesimal limit, even if we know it isn't really. It leads to dP= 0 since P is a *constant* which leads to dW=-PdV

And what about a real gas:





The equation doesn't cancel out to PV=nRT.

Yes it does. They've taken n=1 there, and already substituted it in, and for an ideal gas (i.e., very dilute) V>>a,b so you can take an approximation that gives PV=RT for 1 mole of gas.

(The actual equation being nRT=(P+an^2/v^2)(V-nb)

It's a bit of a cheat really, because Pressure can be treated as approximately ~constant at an infinitesimal limit, even if we know it isn't really. It leads to dP= 0 since P is a *constant* which leads to dW=-PdV

Say you sink a cubical box in the water and it is rotating. The pressure at the bottom of the box is going to be greater than the pressure at the top. The rotational motion is going to lower the fluid pressure in a nonuniform way. Overall, there's no way dP=0.

Say you sink a cubical box in the water and it is rotating. The pressure at the bottom of the box is going to be greater than the pressure at the top. The rotational motion is going to lower the fluid pressure in a nonuniform way. Overall, there's no way dP=0.

But it can be treated as if it is for 2 very infinitesimally close volumes, which it has to be for it to be integratable in the first place. If you don't agree with it/understand it, fine, I said it was a cheat (like a lot of physics approximations), but thats exactly where it comes from.