# Tag Info

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When two identical gases mix, the state is generally indistinguisable from the previous state. If one molecule from the left of the partition changes places with a molecule from the right of the partition, does the mixture actually look any different? If the left and right molecules are identical, you would never know which ones started where. So entropy ...

9

Suppose the separated volumes of identical gas are a low-entropy state and the mixed volume is high-entropy. Imagine the reverse process from mixing. You have a single tank full of helium. You insert a partition, so now you have two half-tanks of helium. This can be done reversibly, but it takes you from the high-entropy to the low-entropy state. The entropy ...

8

Squeezing the wavefunction means confining it to a smaller space. It takes more energy to confine something within a small space than within a big one. 2, 3: These are consequences of the quantum adiabatic theorem: if you take a system in state $n$ of some system, and act on the system sufficiently slowly, it ends up still in state $n$ of the new system. ...

6

Indistinguishability There is something that is lurking in the background here, which is an important statement of physics: all helium atoms are indistinguishable. You cannot tell two helium atoms apart from each other, they are so identical, they are basically the same helium atom, twice. Now let's use that to understand what is going on. Two experiments ...

4

Why does heat added to a system cause an increase in entropy that is independent of the amount of particles in the system? Short answer: it doesn't The systems won't end up with the same entropy. Your intuition is correct that the change in entropy depends on the number of particles. The reason why you can't just reason directly from $dS = \delta Q/T$ ...

3

At the classical framework , i.e. no General relativity and astrophysical observations of the 18th century , this is a valid question. When talking of a "Universe" one must have a model , and the model depends on the state of physics knowledge at the time of the model. The second law states that entropy always increases or stays the same. One can make a ...

2

The von Neumann entropy, written in terms of the quantum mechanical density operator, is a constant of the motion if you keep track of everything (including entanglement with the environment) and don't have any collapse events (which, depending on your favorite interpretation of quantum mechanics, might not exist anyway). The thing is that this fact already ...

2

The minimal counterexample seems to me to be the following: Take two materials, placed next to each other: ____________________ | | | | Material|Material | | 1 | 2 | ____________________ E1 _ _ _ E0 _ _ They have energy levels as indicated above- both have states at E0 and E1, but one has two excited states. ...

2

As march pointed out, your reasoning is incorrect for finite $Q$ and $\Delta S$. However, the equation is true for differentials, so we have to address that. Possible 'conceptual' answers: Stat mech: the increase in $S$ is related to how much extra disorder is produced. When there are less particles, they each get a bigger share of the $dQ$, so they each ...

2

Suppose you have a heat reservoir at $300\,\mathrm{K}$, and you take $9.8\,\mathrm{J}$ of energy out of it to lift a $1\,\mathrm{kg}$ weight $1\,\mathrm{m}$ off the ground. Then the entropy of the heat reservoir has been reduced by $9.8/300 = 0.33\,\mathrm{JK^{-1}}$, but the entropy of the weight is unchanged, since it has only moved and not changed state. ...

2

TL;DR The integral in the Clausius inequality is negative the portion of the entropy change of the reservoirs that is due to energy exchange via heating with the system, i.e. $$0 \geq \oint \frac{\delta Q_{\text{sys}}}{T_{\textrm{res}}} = - \Delta S_{\text{res due to heat exchange with sys}}.$$ (I use the word "portion" because the reservoirs are allowed ...

2

If you are looking for an intuitive explanation for this, as you said in the comments above, I'll try to give you one here. We assume an heat transfert is done in a high temperature environment, which means the entropy of this environment is already high since particles are already more agitated and that the energy is dispersed in a disorded way. Now if we ...

2

Your initial "state" is not really a state of thermodynamic equilibrium. As you yourself said, you have separated hot and cold water regions which subsequently mix together. Since your initial "state" is not a true thermodynamic equilibrium state it is not describable by just the two thermodynamic variables u (energy?) and v (volume). Obviously, you need ...

1

I might see part of the problem here. There are processes in which energy is extracted via heating from a thermal reservoir, and in the process the system does positive work on the environment, and all of the energy coming in via heating gets transformed into work. There are many canonical examples in classic thermodynamics: the main one is an ideal gas ...

1

Now, why does this happen so? Both the gases, though are same, when expand, make their entropy increase, isn't it? Yes, the measure of set of accessible states increases and the "ln W" entropy increases as well. This kind of entropy however has the inconvenient property that it does not simply add when equal systems are brought together; but the entropy ...

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