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## New answers tagged entropy

1

The entropy of a gas does not simply depend on the number of ways to arrange the particles that make up the gas but also on the number of ways of distributing the available energy between those particles. In an adiabatic expansion there is no heat transfer but he gas does do work on its surroundings. This reduces the internal, and so reduces then number of ...

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The equation: $$dS = \frac{dQ}{T}$$ only applies to reversible processes. For an irreversible process $dS \gt dQ/T$. To see this start with the expression for the change in internal energy: $$dU = dQ - dW$$ The internal energy is a state function, so this equation always applies whether the process is reversible or irreversible. So for a reversible ...

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The change of entropy is Q/T only if the heat transfer is reversible if the process is irreversible you can't obtain the change of entropy through the formula Q/T After the heat transfer the change of entropy o the whole system is $\Delta S>Q/T-Q/T_0>0$

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Yes. Just for the other system, which is gaining heat. Entropy is lowered for the system loosing heat (negative $Q$), but is increased (even more than what was lowered) for the external system gaining heat (positive $Q$). $$\Delta S \geq \int \frac{dQ}{T}$$ Usually I have seen it denoted as $$\Delta S = \int \frac{dQ}{T}+\sigma_{gen}$$ where the ...

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$$s=k_B \ln(v_2/v_1)^N$$ so if you are able to measure all the quantities present in this equation, you will be able to measure entropy. Here, $k_B$ is the Boltzmann constant and $N$ the number of molecules in the system.

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It is true that there is no correct a priori way of deriving the "correct" entropy as the notion of "correct" will depend upon what we mean by this word. However, if statistical mechanics introduces a quantity that has the same name as the most important quantity of thermodynamics, I suspect it has, at the very least, to carry a meaning very close to the ...

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From a purely thermodynamical point of view, why does that entropy have to be a maximum at equilibrium? Say there is equilibrium, i.e. no net heat flow, why can the entropy not be sitting at a non-maximal value? The statement that entropy has maximum possible value in equilibrium means this: given imposed constraints (volume, total energy, molar ...

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In your case, heat conduction is considered to be quasi-static for the subsystems. Let $\Delta Q$ be positive number, energy transferred as heat. Then, change of entropy $S_2$ of the subsystem 2 is $\Delta Q/T_2$, change of entropy $S_1$ of the subsystem 1 is $-\Delta Q/T_1$. For the whole system containing both subsystems, the change of entropy is ... 0 Entropy of a system can increase either by entropy transfer or due to entropy generation. If we live in a world where every process is reversible then entropy will never be generated only transfered from one system to other. Entropy generation is due to irreversibilities in a process.Lets consider water taken inside an adiabatic container we do some work on ... 1 You are correct in your proof. However the only situation in which both inequalities are satisfied is when the latter one is equal to zero. In this way you are saying that it is only zero for the reversible path. If the Clausius statement is violated then the limit of the composite system of Carnot engines is violated. This then comes back to why is the ... 4 Consider what happens to the total energy of the gas as it shrinks. By the virial theorem K=-P/2, we can write the total energy E of the gas cloud as E=K+P=-P/2+P=P/2. Additionally, according to equation (2) from the webpage, P\sim -N^2/V^{1/3}, so E\sim -N^2/V^{1/3}. Thus, as volume the volume decreases, the total energy becomes more and more ... 1 The Baez article is strongly misleading in that it applies simplified concepts and arguments appropriate for gases with short-range interactions to a system with many particles interacting purely gravitationally. Systems where short-range forces dominate (Lenard-Jones, van der Waals forces... ) such as rarified gases can be described in such simplified ... 0 Modern work by Jarzynski, Crooks, and others shows that a more fundamental statement of the second law is \langle e^{-\beta\Delta S}\rangle = 0. If we expand this to first order we find a path to the conventional second law \begin{align}1 &\geq \langle e^{-\beta\Delta S}\rangle \\ &\geq 1-\langle\beta\Delta S\rangle \end{align} which implies ... 0 Space and Time is fundamental. We still don't know if spacetime is a part of some deeper entity. Einstein's equation do tell us that space exists, but what it is, no body knows. What is space made of or is time a single entity, this has been a subject of discussion. 2 There is also, a gas of photons in the right side. It was trapped there when you assembled your box, and since you are assuming a perfectly zero emissivity, these photons must be perfectly reflected from all surfaces. That means they are blue shifted if the wall moves toward the right and red-shifted if the wall moves toward the left. Result: If you ... 0 A closed system with fixed entropy does not imply an isolated system. Fixed entropy means that dS = 0, the entropy of the system does not change. A closed system is one in which no mass can be transferred in or out of the system but heat and/or work can be exchanged in and out of the system. The change in energy of a closed system can then be described as ... 0 2nd law of thermodynamics has many almost equivalent formulations. The traditional ones always assume closed system, isolation is not needed - heat and work transfer are assumed to be allowed. One formulation: When thermodynamic system goes from equilibrium state 1 to equilibrium state 2, the entropies of these states obey the relation S(2) - S(1) \geq ...

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