- In the figure, one gas is an ideal gas (dE1/dV1=0) and the other is a real gas (dE2/dV2 is not equal to 0), which can achieve heat transfer from low temperature to high temperature without consuming external energy. This is the second type of perpetual motion machine.
- Real gas (dE/dV not equal to 0). This is the simplest middle school physics knowledge.Middle school physics knowledge can defeat the second law of thermodynamics. Isn't this very funny?
- This field will produce Nobel Prizes, welcome to join.
As in your other "double chamber Carnot" thing, the basic premise dW=0 is false even if both sides are filled with ideal gases and with that the whole construct collapses.
I already described it on your other thread. If we just let both gases be ideal, you have pA=nRT/VA and identical for pB.
Therefore you have dW=nRT/VAdVA+nRT/VBdVB and if VA decreases and VB increases 1/VAdVA doesn't equal -1/VBdVB. With a real gas you get even more terms. That's not strange physical logic, that's spotting an undergraduate-level error in your derivation which collapses completely with the false premise.
Naturally, if you ignore the mechanical work that you put into a machine, everything becomes a perpetuum mobile.
Frankly, I am well aware that there is no chance of convincing you, I'm just documenting the error for other people who might come across this thread.
As I stated above, if dW is unequal 0 for ideal gases, it will be even more apart from 0 for real gases. You can take whatever real gas equation of state and insert it for pi, you will end up with an equation that has even more terms.
Unfortunately, I understood them and, as usual with your "proofs", there's a simple error in it which you don't want to be true. But anyway, I had nice entertainment.
(Isothermal) p1 * dV1+P2 * dV2=0, which is artificially set.
DQ=dE=(dE1/dV1) * dV1+(dE2/dV2) * dV2=0, the perpetual motion machine will not appear.
Thus, p1/(dE1/dV1)=p2/(dE2/dV2) is derived (3)
If the two real gases are not related, equation (3) does not hold.
DQ is not equal to 0, there is a perpetual motion machine in the picture.
If you want to prove the second law of thermodynamics is right, go ahead and prove equation (3)!
(Isothermal) p1 * dV1+P2 * dV2=0, which is artificially set.
DQ=dE=(dE1/dV1) * dV1+(dE2/dV2) * dV2=0, the perpetual motion machine will not appear.
Thus, p1/(dE1/dV1)=p2/(dE2/dV2) is derived (3)
If the two real gases are not related, equation (3) does not hold.
DQ is not equal to 0, there is a perpetual motion machine in the picture.
If you want to prove the second law of thermodynamics is right, go ahead and prove equation (3)!
If gas 1 is an ideal gas and gas 2 is a real gas, formula 3 would become p1/0=p2/(dE2/dV2). Isn't that absurd?
Yes, the artificial setting is the non-physical part in your derivation.
1) In differential form, I already wrote down above why the dW for the two chambers will not cancel out each other already for an ideal gas.
2) If you take one (or both) as real gases and insert whatever equation of state you prefer, you will get even more terms that will make dW not 0.
3) In integrated form, W for an isothermal step would be
Wiso=-nRTln((V0+VD)/V0)-nRTln((V0-VD)/V0)=-nRTln((V02-VD2)/V02)=-nRTln(1-VD2/V02) for the ideal gas couple and I'm not wasting my time integrating a real gas equation of state now. Seeing that this is not 0 is first-semester physics.
So, we're back to: ignoring half of the real existing energy terms in a derivation makes everything a perpetuum mobile. Sorry, but your Nobel prize is not happening.
*Flies away*
Using differential equations is simpler. For ideal gases, my differential equations can also reach the same conclusion. For real gases, it needs to be proven that p1/(dE1/dv1)=p2/(dE2/dv2). This formula holds and there is no perpetual motion machine. Otherwise, the opposite is true. You cannot limit your understanding to ideal gases.
Simply insert the van der Waals real gas equation of state (or Redlich-Kwong or Virial ansatz, if you prefer those) into the dW equation. You will also get a dW that's not 0. This is a trivial operation.
Arbitrarily setting dW=0 for an isothermal (or adiabatic) process here is actually a violation of the first law of thermodynamics and I got the notion that you believe at least in this one, or has that changed?
So what you have demonstrated here is actually: if we violate the first law of thermodynamics, the second one is also useless. But since the first law is valid, your "proof" is invalid.
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