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A short answer for a specific system might be helpful here. Imagine a system of two containers linked by a valve. One container is full of gas - the other is empty - it contains no gas - it has a vacuum. Answer to question 1 If the valve is opened the gas will fill both containers - and the entropy will increase, but temperature remains constant as no ...


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If we want to examine gravitational collapse from a statistical mechanics point of view, we find that there's a tradeoff between the fact that a more spread-out collection of matter has more possible position states, whereas a more concentrated collection has more possible momentum states (because more of the system's potential energy has been converted to ...


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When textbooks write $dS=dQ/T$, quasi-static process with no friction is considered. $S$ is entropy of the system, $dQ$ is heat accepted by the system from the environment and $T$ is temperature of the system. You can add indices to $dQ$, but they are not necessary since $dQ$ already means total heat accepted from the environment. There are no other kinds or ...


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Yes this setup violates the second law of thermodynamics .The law states that - "It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work." Now you know the answer


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This isn't a complete answer, but John Baez gave a pretty good treatment of this in a series of blog posts (part 1, part 2, part 3, part 4; arXiv paper with some more stuff). Basically, he defines what he calls the "quantropy", which is just the classical entropy formula with $\beta$ replaced by $-i/\hbar$ and the energy replaced with the action. (Note ...


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To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...


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This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $ dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...


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Since you have the expression of $\Omega$ your work is almost done. First remember that the entropy for a micro-canonical (fixed energy) system at thermal equilibrium is given the very famous Boltzmann's formula : $$S=k_B\,ln(\Omega)$$ Then, simply use the Stirling's approximation to evaluate $ln(N!)\approx Nln(N)-N$ (because $N>>1$, i.e. very long ...


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You have to consider entropy to answer this question. The idea of the Boltzamnn brain presupposes an expectation of a universe at thermal equilibrium, or maximally high entropy. In order to create the initial state from which the natural processes you describe (evolution, development etc.) require to operate, the universe has to first start in a state of ...


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ΔS universe = ΔS system +ΔS surroundings . hotosystem I of plants has a quantum efficiency of 0.99 and an energy efficiency of >0.96. The system may be considered a canonical ensemble which consists of a Thermal bath in which the particles are suspended and monochromatic light emitted ny an incandescent lamp with a definite Tr. Please note that the lamp ...


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This definition is not correct. A thermodynamic state $A$ of a particular system is said to be more ordered than another thermodynamic state $B$ of the same system if $A$ is invariant under less symmetry operations than $B$. The entropy of a thermodynamic state of a particular system, on another hand, quantifies the number of microstates which are ...


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The expression $$ I=-\sum_i p_i\ln p_i $$ is a function of probabilities $p_i$, also called information entropy. It can be calculated for any values of $p_i$, even those that are not appropriate for an equilibrium state. In such case it has nothing to do with equilibrium thermodynamic entropy $S$. When some equilibrium state $X$ is prescribed, however, the ...


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The Gibbs entropy is the generalization of the Boltzmann entropy holding for all systems, while the Boltzmann entropy is only the entropy if the system is in global thermodynamical equilibrium. Both are a measure for the microstates available to a system, but the Gibbs entropy does not require the system to be in a single, well-defined macrostate. This is ...


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Let's consider just one cycle of the Szilard engine. Aside from discussion of energy free state polling, one of the main points of Bennett paper (if you mean Charles Bennett, "The Thermodynamics of Computation: A Review", Int. J. Theo. Phys., 21, No. 12, 1982) here is that you must build a finite state machine (a very simple three-state machine) as a minimal ...


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I've been reading a text which states quite clearly that $\frac{\partial\log{\Omega}}{\partial U} \approx -\beta$ so the rate at which accessible microstates become available would approximately fall with increased energy. I'm trying to get a handle on why the rate of increase would typically fall with higher values of $U$, and by implication $T$. The ...


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Position yourself on the surface of body A and shoot a ray in any direction. After a certain number of reflections from the mirror surfaces (possibly none) it will hit either body A or body B. Now expand the ray into a very narrow cone, such that all the rays in the cone finish on the same body. Looking into that cone you will see a radiant intensity ...


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Ingenious. A and B are small, but they cannot be points.The image of B is magnified at A. Therefore if A and B are the same size, some of the light from B will miss A.


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For this process, we found that the entropy change of the universe was negative. You found nothing of the sort. You instead found that a tiny, tiny fraction of the photons emitted by the Sun hit the Earth, and that a tiny fraction of those photons that do hit the Earth trigger an endothermic reaction. The second law of thermodynamics does not prohibit ...


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In most cases in thermodynamics it's safe to assume the heat bath is the same temperature as the system. Usually the system is assumed to be in contact with the heat bath, and at equilibrium, in which case it will be at the same temperature. However, this need not always be the case, necessarily. In writing that answer I was careful to emphasise that it's ...


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Reversibility implies that the entropy change of the universe is 0. An isentropic process need not necessarily be reversible, provided that the entropy of the surroundings is increased. For an irreversible process, heat can be removed from the system in order to make it isentropic (since $dQ < TdS$. As heat is removed, you will have an isentropic ...


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Entropy and irreversibility are related to each other. Irreversibility is the one because of which process takes place and due to this energy degradation takes place that is exergy (availability) decreases. And basically irreversibility(I) is defined as I = To * ( del s ) That is when you multiply change in entropy of universe with the reference ...


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Yes you are right. Entropy of universe or isolated system is always generated (for irreversible process) or equal to zero ( for reversible process) Entropy of system increase because of heat addition and that of surrounding decreases ( -del Q / T ).



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