# Tag Info

4

It seems you're coming at entropy from a thermodynamics standpoint. This is completely consistent with (and, at the macro scale, equivalent to) the statistical derivation of entropy, but you might find the statistical version more intuitive, if the thermodynamic version is causing you issues. I warn you, statistical physics is both math-heavy and takes some ...

3

If you look at the first law of thermodynamics, $$dU=\delta Q-\delta W=TdS - pdV$$ then consider a reversible processes ($dU=0$), then we get $$TdS=pdV$$ Then using the ideal gas law, $pV=nRT$, we find $$dS \sim \frac{dV}{V}$$ The volume considered would be the volume of the system (e.g., a gas), with its infinitesimal increase(decrease) signified by ...

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This experiment is not possible, i.e. you cannot make a glass of only protons or a glass of only electrons because of the electromagnetic repulsion, they cannot be a liquid. A liquid requires chemistry They can be a gas though, and the LHC is creating a type of proton gas to get the protons for the beams. To get hydrogen gas into a plasma phase takes ...

1

I will start by answering the second question. Let's consider the case of two species of liquid in a box, with a partition separating them. The irreversible process you describe is to remove the partition. A reversible process would be to have the partition actually composed of two separate independently movable partitions. One of these does not interact ...

1

I can't say I give this answer with great confidence, and I'll have to resort to some hand-waving. Think of the hydration shell around a hydrophobic substance. To minimize the local free energy, the water molecules will avoid any interaction with the hydrophobe and will seek to maximize attachment to the neighbouring water molecules, creating a sort of ...

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The other thing that I can think of, is when you are not interested in some parts of your system(i.e. environment), so you trace it out. Now if the environment is not separable from the rest of the system, which is usually the case; what you are left with(the reduced state) is a mixed state. Note that in this case: \rho_{AB}\ne ...

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If you call $\chi$ the exergy (availability) then $\chi = U + p_o V - T_o S$ where $p_o, T_o$ are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that $dU=TdS-pdV$ and then the exergy ...

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Here's an enlightening special case: Take $n$ bodies with temperatures $T_1,\ldots T_n$ and bring them together until they reach a final temperature $T$. The first law of thermodynamics tells you that $T$ is the arithmetic mean of the $T_i$. The second law of thermodynamics tells you that the change in entropy is $n\log(T/G)$ where $G$ is the geometric ...

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