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Although entropy change is defined in terms of a reversible differential transfer of heat divided by the temperature at which the heat is transferred, you can have entropy change without heat transfer....

The free expansion isn’t reversible because the gas flows down a pressure gradient (that arises when you remove the piston). Any energy flow down a gradient generates entropy.In contrast, during the (...

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What am I missing ?Entropy can be generated without there being heat transfer, i.e., when $Q=0$. That's the case for a free expansion into a vacuum. The classic example given is an ideal gas located ...

If you have an irreversible adiabatic process between two thermodynamic equilibrium end states of a system, there exists no possible reversible adiabatic process between these same two end states. So ...

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The point made by Zwiebach in those notes is true even for time-independent Hamiltonians.Let $\psi:\mathbb R\rightarrow \mathscr H$ be a trajectory through the Hilbert space $\mathscr H$, where $\psi(...

Yes there is. Naturally these degeneracies should be motivated by symmetry considerations or else they would typically destroyed by perturbations. In particular, to have non trivial irreducible ...

For an ideal gas $$PV=RT$$Since $$dU=dQ-dW$$For adiabatic process$$C_v dT = -{PdV}$$Substituting $R dT = VdP+PdV$$$VdP = -\frac{(R+C_v)}{C_v}PdV$$Since $C_p -C_v =R$ and $\gamma= \frac{C_p}{...

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I solved the issue. I will explain in detail how the existence of an operator conserved by the adiabatic evolution allows us to make sense of the spectral flow argument.Let's start with the problem ...

in $B \rightarrow C$, the working substance is isolated, and thus $Q => 0$, so $\Delta E_{int} = W$ where $W$ is the work done by theenvironment to the system, not work done by the environment on ...

If you expand a gas adiabatically using a piston, the process isisoentropic.An adiabatic process is not isentropic unless it is also reversible. To be reversible, it must be carried out quasi ...

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In an adiabatic process, that is a process in which mathematically$dQ=0$, $PV^γ=\text{constant}$.First of all, the equation only applies to an ideal gas undergoing a reversible adiabatic process. ...

But that implies that when work is done with no heat flow (i.e. reversible adiabatic expansion), the entropy of the system doesn't change.Reversible processes are by definition the processes which ...

If friction is allowed to occur in the gas (because its movement isn’t infinitesimally slow), then it ends up warmer than it would otherwise.A warmer amount of the same gas has a higher temperature, ...

Assume the initial volume on either side of the piston to be $V$.After the piston is released and we allow everything to come to thermal equilibrium (the temperature is then $T$ on both sides), the ...

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Recently, I have also came across this definition. Somehow, the concept is poorly explained in recent literature. Therefore, I went back to the original Zener's paper: Proc. R. Soc. Lond. A 1932 ...

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In Thermodynamics parlance, an adiabatic process is one in which there is no exchange of heat between the system and its surroundings. One way of accomplishing this is to have perfect insulation ...

Rubber is pretty damn complex!On a macroscopic level, you do work to the rubber and it heats. When the rubber does work back, it cools. Just like when you compress a gas and then the gas expands.The ...

Imagine a Hamiltonian of the form $\hat H = \hat H_0 + \lambda \hat V$ where $\hat V$ is not assumed to be small. If $\lambda = 0$, then we can find a set of energy eigenstates $|\psi_{n}\rangle$ ...

I think the easiest way to see this is with an example. Consider a Hamiltonian that changes suddenly at $t=0$\begin{align}H(t) = \begin{cases} H_1 & t \le 0 \\ H_2 & t > 0 \end{cases}\...

During constant volume heating, the gas's volume is fixed, and hence it does no work on its surroundings. During constant pressure heating, the gas is allowed to expand against its surroundings, ...

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The partial and total derivatives are different things, but they are related via the Chain Rule:For $f(x,y,z)$, the differential of $f$ is:$$df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\...

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Why does internal energy end up higher in an irreversible adiabaticprocess, even though the work done is less?It is precisely because the work is less that the decrease internal energy in an ...

What is meant byIt was then noted that Callen's argument, which was repeated by Leff,could not be correct since the equilibrium condition was derived fromthe first law, rather than the second law....

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To answer your first question, yes two (different) adiabatic paths can intersect in a plot. See State A in Fig 2. However, for this to be possible, at least one of them has to be irreversible. For ...

No, that is obviously wrong. It's trivial to construct a hamiltonian on a two-level system which maintains a constant gap as it swaps the ground state with the excited state. As an explicit example:...

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It's important to distinguish between several things:A process which does not change the entropyA process which involves no transfer of heatA reversible processBlundell is correct in that (2) and (...

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This is a particularly complicated problem to resolve. We encountered a similar problem on Physics Forums at the end of 2019 and into 2020, and another member named Andrew Mason worked with me to ...

J. Murray already gave you a detailed (very nice) answer.Intuitively: adiabatic processes are (by definition) something that live in the limbo between dynamics and statics.In quantum mechanics the &...

In an adiabatic irreversible process, entropy is generated within the system, and there is no transfer of entropy to the surroundings because heat cannot flow out. Thermally, the surroundings do not ...

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The second law leads to (or can be stated)$$\Delta S_{\rm tot} \ge 0$$not $\Delta S_{\rm tot} > 0$ (where 'tot' refers to everything that undergoes some change during the given process). The ...