First law of thermodynamics
Question presented for discussion in the Newsgroup sci.physics (June / 99)
Message 1
From: Alberto Mesquita Filho
Date: 16 jun 1999
Let us suppose two compartments 1 and 2 separate, one of the other, by a solid and adiabatic movable membrane. Let us suppose also that the pressures are, respectively, P1 and P2 such that P1 > P2.
Forlorn on himself, the system will develop for an equilibrium situation such that P1 = P2. As expressed notation in most of the physical-chemistry books (for instance, Castellan, for who W = integral of PopdV, where Pop = opposition pressure], we will have:
W2 = Work received by the system 2 = integral of P1dV
W1 = Work supplied by the system 1 = integral of P2dVIt is easily proven that W2 > W1. Some thing is wrong in this reasoning. Or not?
Will it be that the solution of a irreversible transformation problem as simple as this flees at the domain of the classic thermodynamics?
Note: The classic thermodynamics studies initial and final states and, in its calculations, such effects should not be considered as the friction between the movable membrane and the recipient.
Alberto
Message 2
From: Ed
Date: 16 jun 1999
Alberto: It is easily proven that W2 > W1.How so?
Message 3
From: Alberto Mesquita Filho
Date: 16 jun 1999
Ed wrote:
Ed: albmesq wrote:Alberto: It is easily proven that W2 > W1.Ed: How so?I would hope to find W2 = W1 (conservation of energy). However, if the expression dW = Popdv is correct, this doesn't happen. "Being accepted this formula as correct", easily we proved that W2 > W1.
If the formula dW=Popdv is not correct...?
Some more thing is wrong! Or not?
Alberto
Message 4
From: Theresa Knott
Date: 19 jun 1999
albmesq wrote in Message 1:
Alberto: would hope to find W2 = W1 (conservation of energy). However, if the expression dW = Popdv is correct, this doesn't happen. "Being accepted this formula as correct", easily we proved that W2 > W1.
If the formula dW=Pop*dv is not correct...?
Some more thing is wrong! Or not?Perhaps you could supply us with your easy proof?
Message 5
From: Alberto Mesquita Filho
Date: 20 jun 1999
In message 4 Theresa Knott wrote:
Alberto: would hope to find W2 = W1 (conservation of energy). However, if the expression dW = Popdv is correct, this doesn't happen. "Being accepted this formula as correct", easily we proved that W2 > W1.Theresa: Perhaps you could supply us with your easy proof?
If the formula dW=Popdv is not correct...?
Some more thing is wrong! Or not?This is, really, very easy. Notice that if the compartments C1 and C2 are separate by a movable membrane and the pressures are, respectively, P1 and P2, the opposition pressures will be respectively P2 and P1. Therefore, and as already presented in the original message:
W2 = Work received by the system 2 = integral of P1dV
W1 = Work supplied by the system 1 = integral of P2dVNow, if during the transformation P1 is always larger than P2, it results that W2 > W1. By the first law of Thermodynamics we would hope to find W1 = W2 (adiabatic process).
Some could say that the expression is just valid for reversible processes, or equilibrium transformations. In spite of this, the expression W = integral of PopdV is used in other irreversible processes that, by chance, they are compatible with the first law.
Regards
Alberto
Message 6
From: Theresa Knott
Date: 20 jun 1999
If you let the membrane move slowly, so that P1 is only infinitesimally larger than P2 at all times and the process is occurring quasistatically and reversibly then the problem doesn't occur.
If on the other hand you just let the membrane go then the membrane will undergo simple harmonic oscillation and the missing energy will be contained in the kinetic energy of the membrane. The situation is analogous to a mass vibrating on a horizontal plane that is attached to two springs.
So the answer to your question is yes pdv is the correct expression for work done and no the first law isn't violated.
Message 7
From: sbhar
Date: 21 jun 1999
In response to Message 6 "sbhar" wrote:
Well, look, if the pressures on the massless piston between chambers are equal, so also will be the forces, and it won't even move. If they are unequal the membrane or piston will accelerate, and the only thing limiting acceleration of a massless membrane/piston will be aerodynamic drag. Once the membrane hits "terminal velocity" the forces on it (pressures on it) will again be equal (since it isn't accelerating), so the work done by both chambers is equal d(PV)1 = d(PV)2. Although some of the pressure on the membrane in the direction it is moving will be aerodynamic drag pressure, which won't be the same as the pressure in the rest of the chamber, but will be exerted by a high pressure shockwave front in front of the piston. Differential heating will naturally occur in this region from the adiabatic compression of this shock/sound wave, and along with its kinetic energy, that's where the extra energy is.
The only question is what happens in the period of acceleration when the pressure must, of definition, be unequal on the piston (since otherwise it can't accelerate). I'm of two minds here. My first thought is that this means that a massless piston accelerates infinately fast, so there's no problem in a period of differential pressures (since it's infinitely short). But I don't think that's right, except where one chamber is in vacuum; in which case there is no piston terminal velocity until it gets to the gas molecule mean velocity, and the gas in the first chamber is undergoing a simple escape into vacuum (the massless piston being no impediment and making no difference) and thus undergoes no cooling (save Joule-Thompson cooling) because it's not doing any external work.
Actually, I believe that even if the piston is massless, since the piston moves through air and suffers drag, the air on the other side stands in for its mass, and during the acceleration period, the differential pressure goes into acceleraction of the gas in the shock front on the other side of the piston, just as if the piston HAD mass. Thus, the work difference goes into straight kinetic energy of this gas, not into adiabatic heating of it. I suppose the same is true, to some extent, even after the piston reaches terminal velocity, though there must be some limit as to how much gas gets moving as a shock wave once things reach steady state when a piston moves down an infinitely long tube at a given v. After that has been reached, there is no more energy stored as mass movement (kinetic energy of mass quantities of gas), and it all goes into production of sound waves (which carry energy, but not nearly as much directional energy or momentum as shock waves, which include a kinetic component, like wind. Here there is a loose analogy between wind and electrostatics, and between a shock wave and transmitter near field, and a sound wave and transmitter far field).
Message 8
From: Alberto Mesquita Filho
Date: 20 jun 1999
In message 7 sbhar wrote:
sbhar: Actually, I believe that even if the piston is massless, since the piston moves through air and suffers drag, the air on the other side stands in for its mass, and during the acceleration period, the differential pressure goes into acceleraction of the gas in the shock front on the other side of the piston, just as if the piston HAD mass. Thus, the work difference goes into straight kinetic energy of this gas, not into adiabatic heating of it. I suppose the same is [...]I agree with your explanation and I believe that there is not other plausible one. However, I don't know if the followers of a classic and orthodox thermodynamics would accept this explanation, by virtue of the use of the concept of kinetic energy. The kinetic energy, be contained in a gas or in a piston with mass, it is not usually taken in consideration in classic thermodynamics. Certainly it would be accepted by a specialist in "thermodynamics of irreversible processes"; but the existence of this discipline doesn't identify the classic thermodynamics as a "thermodynamics of reversible processes" or "equilibrium thermodynamics".
I say this because it is common, in classic thermodynamics, the study of the expansion of a gas against a piston on which a m mass is placed. In neither moment, in these cases, it is made reference about the kinetic energy of m; and this exists during the expansion. The specialists of the subject say that the gas exercises a work given by the integral of PopdV that in this case is equaled at mgdh.
In this last case, and for the same reasons that you so well marked, intermediate states exist, when a fraction of the energy is kinetic and other is potential (elevation of m). We will have, at the end of the process, a numeric coincidence among the W work provided by the gas and the integral of PopdV. This because the kinetic energy comes back for the gas.
In the system with two compartments, as I presented, it is difficult to anticipate as the kinetic energy will be redistributed; even because we could think about a temporary oscillation to happen before the establishment of the equilibrium. And, in this case, the shock waves would appear some times in a compartment, another times in other.
Regards
Alberto
Message 9
From: Ed
Date: 21 jun 1999
In response to Message 7 "Ed" wrote:
Random scratchings, meanderings & peregrinations ....
As the advancing front of molecules or their proxy, the membrane, encounters stationary (not-yet-accelerating) molecules, it pushes some of them forward, normal to the membrane and scatters others at various angles.
thus, the motion of the advancing membrane may invest it with a fore-layer of higher density molecules and these deputy-molecules, in turn, may also advance the cause by exciting others even further forward.
The net result of all the push and shove is to increase the average thermal velocity and thus the pressure and temperature.
The counter-balancing result is a corresponding decrease in pressure and temperature on the other side.
Per math, dE = d(PV) = PdV + VdP for both sides.
This continues until the pressures are equal. Then, if the chambers are thermally insulated, you are done with your homework.
And, ignoring energy dissipation to the environment due to friction heat and sound, the work balances.
If the chambers are not insulated, however then heat can flow directly through the membrane, from hot to cold, one way only, and you can't get it to go back, without providing energy to do the job.
As the heat flows, it intends to change the pressures a bit and this shifts the equilibrium position of the membrane.
dE = dE, however still applies, with no escape.
That is to say, no matter how you slice the salami, it's all baloney and the governing law is: Some Of It + The Rest Of It = All Of It.
Arguments, paradoxes, conundrums and such are really nothing very much more than evidence of thought that gives up way before it ought.
PS: Don't expect to pick up a lot of girls by showing off your Navier-Stokes equation.
Message 10
From: Ed
Date: 20 jun 1999
In Message 5 "albmesq" wrote:
Alberto:
W2 = Work received by the system 2 = integral of P1dV
W1 = Work supplied by the system 1 = integral of P2dVPero, de donde viene esta arriba?
Dice, E = PV.
Entonces, dE = d(PV) = PdV + VdP, verdad?
Message 11
From: Alberto Mesquita Filho
Date: 20 jun 1999
In message 10 "Ed" wrote:
Pero, de donde viene esta arriba?
Dice, E = P*V.
Entonces, dE = d(PV) = PdV + VdP, verdad?Sí, pero también tenemos dE = dW + dQ (primera ley) y, en el caso en discusión, "si" dW = PopdV, se deduce que PdV + VdP = PopdV + dQ.
Si dQ = 0 tendremos PdV + VdP = PopdV o entonces dE = PopdV.
Alberto
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