# Tag Info

0

No. A low entropy "initial state" could be the result of a so-called anthropic fluctuation in a (past) eternal universe. Fluctuations about equilibrium could, fortuitously, create the initial conditions for life as we know it. This was proposed by Boltzmann and his assistant Schutz in the late 19th century, though ultimately deemed unsatisfactory by a ...

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 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. ...

-1

Total entropy of the universe is equal to the total area of the space boundary, according to holographic principle. Our universe is expanding, asymptotically approaching de Sitter space. In de Sutter space the radius of the cosmic horizon is constant and equal to the Hubble radius - the distance at which cosmological red shift becomes infinite. For ...

-1

I don't think it is wrong to think entropy is correlated to disorder specifically for the example your teacher gave provided there should not be any ambiguity in defining disorder. There is an increase in entropy when you shook the marbles in the jar, but it is really, really, really negligible. However, energy dispersal method is far more intuitive & ...

0

Making the total entropy tend to zero, makes the efficiency tend to the carnot efficiency, which is indeed the maximum efficiency for a cyclic process , according to carnot's theorem

0

Time is not like space. It is a coordinate, as the space coordinates, but that doesn't mean it is the same. Read Ben Crowell's answer here. Entropy is stochastic, it doesn't have to increase monotonously. The very low current universe entropy can make the illusion that it always does, but there is still a very low probability that it decreases for a short ...

-1

A short, alternative, and qualitative explanation: Consider the entropy of a system as not "the disorder" of a system, but the wiggle room in a system. If two gases of the same size get opened to each other, do they both experience more wiggle room? I'll ask this a different way: can individual molecules wiggle less, just as much, or more than in the two ...

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 ...

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 ...

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 ...

10

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 ...

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 ...

0

I think there is some confusion of terminology here.. A system is not at thermal equilibrium if you lock it into the most probable state. A single state is not an equilibrium, it is the complete evolution of the system which has to be observed (or known) to say if it's in equilibrium. If we take your A,B,C example, the system will be in thermal equilibrium ...

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. ...

0

The equation you have written is the change in entropy for an ideal gas. An ideal gas is a collection of small, elastic balls which do not interact with their surroundings. Thus, the potential energy of the gas is zero due to the absence of any interactive potentials and thus, the energy of the gas is just its kinetic energy. It can be shown that the ...

0

Internal Energy is a measure of the random motion of molecules. It depends only on temperature. By the definition of an isothermal process, which means than there is no change in temperature during the process the change in internal energy during an isothermal process must be zero. Note this only true for ideal gases with zero Vander Waals Forces between ...

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 ...

0

As Christoph said, entropy is a measure of microscopic freedom. This is, I think, better than "measure of energy dispersal" (a now common description), for two reasons (and each one is enough for me): 1/ freedom can be measured in bits (or nats, digits, ...). 1 bit of freedom is the freedom to choose one of 2 options. entropy measures the freedom, the number ...

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

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 ...

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 $dS = \delta Q/T$ ...

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 ...

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 ...

0

The entropy change in going from A to B in reversible case is Zero because system and surroundings are in equilibrium all along like a controlled very slow expansion of gas inside a cylinder by piston but in the irreversible case system and surrounding are not in equilibrium at all which meant even if the system goes from A to B the surrounding might go from ...

2

If a process is reversible, then the same process run backwards in time is also reversible. The system you describe is not closed because heat enters through the hot plate. This is true no matter the direction of the process. When the piston expands, heat enters the piston and the piston does work on the weight holding down the piston. The entropy change ...

Top 50 recent answers are included