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The answer is no. Once a black hole is formed you no longer have access to the information stored inside it. The information is not lost, but slowly radiated away as the black hole evaporates. So, the ideal system would be one that is almost near the density for a black hole collapse, but never reach it. Otherwise you loose "easy" access to the information ...

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I learned I can calculate the entropy $$S=-k_B\sum_jp_jln(p_j)$$ where $$p_j$$=probability at j state but I saw that the entropy is also can be calculated by $$S=-k_Bln(Z)$$ I think this equation applicable for both of isolated system and non isolated system these two equations are same ? Given the number of states $\Omega(E,\Delta ... 0 So, in a irreversible process , to calculate the variation of entropy you must subdivide all the variation is some little parts, that could be considered some isotherms and reversible(in this case, in fact, the entropy change is right zero).Obviously this passage isn't so simple, but I want to give you a simple explanation: These little parts tend to zero. ... 0 Entropy is a state function which means that the value of entropy change between 2 states depends only on the initial and final states and not the path/process taken from the initial state to reach the final state. I'm not sure about this part but I think entropy calculations are more easier for reversible processes than irreversible ones. 2 To answer your question first we need to know why do we need quantum mechanics in thermodynamics: In Quantum mechanics you can attribute a wave function(to be precise wave-packet) to a particle. . At high temperatures particles can be pictured as billiard balls because their size is much smaller compared to interparticle distance. But as the gas cools down ... 0 m includes all kinds of energies for the mass at rest, including the thermal energy, and those from the other degrees of freedom, such as the ones internal to the nucleus, as AnnaV mentioned. So you only have to compute mc^2, the HTM term is redundant. 0 I think I will go with the version from Landau and Lifshitz with quantisation of phase space: $$N = \int f \ln \frac{e}{f} \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}, \quad N = \int f \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}.$$ With the equilibrium function of $$f_{\text{eq}} = (2 \pi \hbar)^3 \, \frac{N}{V} \Bigl(\frac{1}{2 \pi m k_B ... 0 I personally like the view pesented in this paper. So, different snapshots of the universe at dfferent times, should actually be considered to be different universes which are related to each other by being correlated. While awkward from an intuitive point of view, it is a far more natural from the point fo view of quantum mechanics. 1 In classical mechanics, time is the unique parametrisation of dynamical systems. In relativity theory, one then sees that time is somewhat more than this, because there exists a global symmetry (the Poincaré-group) that involes time and space on an equal basis (called spacetime). Also, one can show that parametrisation by time is not the only way to do ... 0 "Time" is very difficult to define properly, in General Relativity time is the 4th dimension of what is known as "spacetime" and this dimension is one which you may only move "forward" and you must move through spacetime at the speed of light, which is why time slows down as you move near the speed of light. Other than this, time is not well defined; you ... 0 Let's first consider the definitions of entropy and temperature from a fundamental physical point of view The most general rigorous definition of the entropy of a system that we use in theoretical physics can be formulated without writing down any equations, it is simply the amount of information that you would need in order to specify the exact physical ... 1 I may give you an intuitive example i read before Imagine shouting in a street full of noise and shouting by the same amount in library although the shouting ( dQ ) is the same in both cases it will have a greater effect in the library ( systems with lower temperature. i hope it helps i know only that example and not expert with Thermodynamics. 1 An intuitive explanation is that entropy is a measure of disorder, so if heat is injected into two exactly identical systems except that one is held at a higher temperature, then you can imagine that the higher temperature system's disorder will increase less because it was more disordered to start with. In other words, its harder to increase the disorder in ... 0 There are a few concepts to go through regarding entropy in thermodynamics - it's a tricky subject easy to conflate with probabilistic interpretations. Temperature vs Heat Temperature refers only to the particles' speeds. Increasing the temperature of an isolated system will increase the speeds of the particles, but does not of itself affect the entropy ... 0 A quote from comments: how is a higher temp and faster moving particles not an increase in available energy and therefore a decrease in entropy? Maybe higher temperature means more available energy and more unavailable energy! It definitely sometimes means that. We have devices that decrease entropy of heat energy, in other words they make heat ... 1 Setting constants to be equal to 1 means that one measures quantities as multiples of that constant. So, for example, if one sets the speed of light, c, to 1, v=0.5 means that v amounts to half the speed of light. It may be clearer, especially to people who are not used to these kinds of conventions, to be approach this situation a bit ... 1 Consider an isentropic (idealistic adiabatic process, with no friction) process, in which the volume of a system is increased. For that to happen, the internal energy of the system would decrease because the system would be performing work. So any gain in the number of available microstates from an increase in volume is negated by the loss of internal energy ... 0 Drop a glass on the floor -> Entropy increases. Drop another glass on the floor -> Entropy increases again. From above we conclude that a big pile of broken glass contains more entropy than a small pile of broken glass. We also happen to know that a big pile of broken glass has larger mass than a small pile of broken glass. Drop a glass on the floor -> ... 0 It is perhaps best to look at the entropy of an extremely dense system in terms of information (think von Neumann entropy). Although we do not know how superdense and superenergetic matter quantizes exactly (we would need a theory of quantum gravity for that), it is clear that it must. As such, in this extreme state, particles have likely been squeezed into ... -1 Calling entropy "disorder" is somewhat misleading, it can also be described as the amount of area containing an amount of energy or even information. A black hole will contain a significant amount of quantum data in an incredibly small space, making them objects with high entropy. However this is a fixed amount of entropy and to continue the propagation of ... 0 I did the following :$$K_s=\frac{C_v}{C_P} K_T\frac{-1}{V}\Big(\frac{\partial V}{\partial P}\Big)_{s} = \frac{\Big(\frac{ \delta Q}{d T}\Big)_{V}}{\Big(\frac{ \delta Q}{d T}\Big)_{P}} \frac{-1}{V}\Big(\frac{\partial V}{\partial P}\Big)_{T}\Big(\frac{\partial V}{\partial P}\Big)_{s} = \frac{\Big(\frac{ \delta Q}{d T}\Big)_{V}}{\Big(\frac{ ... 0 Since at the infinitesimal singularity there was less space, there were less microstates. If you consider a cup of cold water in a pressurised, insulated vessel at absolute zero, the system has zero entropy because the atoms cannot move, there is no energy and thus less possible states the system can be in. Here we have the opposite, but the same situation: ... 1 Let me show you that there is no contradiction by pointing out e.g. that for ordinary expansion periods (that is away from first order phase transitions, decouplings...) the total entropy is actually constant in time while the universe is getting bigger and cooler. Or, going back in time, the universe is getting hotter while S is kept constant. How is this ... 0 Entropy is not the existence of heat or energy, but is more accurately described as the spread of energy. A universe with high heat and low matter density has very low entropy, the same way that a cup of hot water has low energy distribution when compared to a cold pool. If you throw the hot water into the cold pool the heat will spread throughout the pool ... 4 Many materials contain microscopic paramagnetic moments which are free to randomly orient themselves in zero magnetic field. If you apply a magnetic field, then the magnetic moments are able to lower their energy by entering a lower entropy state where they are magnetized. In such a material the entropy tends to decrease as magnetic field increase, at least ... 0 1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy? Pair of values$P,V$may not be enough to specify the state of the system, there may be other thermodynamics state variables. If so, it is possible system returns to the same values$P,V\$, but it is in a ...

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It's an example of adiabatic expansion. If you have a container full of gas and you expand the container, the gas cools. Entropy is preserved. Adiabatic processes preserve entropy. Any decrease in entropy due to lowered energy, and correspondingly fewer possible velocities for the particles, is offset by an increase in entropy due to the expanding volume, ...

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