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1

You can wait three or four minutes and let the pressure drop or..... If you feel lucky you can try tapping the can a dozen times, as is shown on this video https://youtu.be/NQYO3Dp8lCA

6

The temperature appearing the the Clausius inequality is definitely the temperature of the "boundary interface (with the surroundings)", or simply the temperature of the sources. One of the best places I have seen this discussion is in Fermi's book, chapter 5, section 11. He is explicit about it. To see this you have to recapitulate the steps in obtaining ...

0

Take gas in a container with a piston connecting to its surrounding as an example. When, at the moment you release the piston, the pressure difference between the surrounding and the container is very small, the piston will move slowly because there is not a lot force to accelerate it. In this situation, the system's response (temperature, pressure change) ...

1

A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a gas enclosed by a freely moving piston in a cylinder. Let us say it is in mechanical equilibrium with the atmosphere, that is, the pressures on the piston match. If you increase the external pressure infinitesimally the piston ...

0

That calculation has restrictions, but, one in particular should be mentioned, that master equation is supposed to be connected to this entropy, but is not necessarily, the master equation can be connected to general entropic form, and that is a fundamental idea for a more complete proff.

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It doesn't seem possible for the original premise to be correct. If you start at a certain state, and carry out a reversible Carnot cycle ending up at the same original state, you can design a great big Carnot cycle and you can design a tiny little Carnot cycle. Certainly the reversible work for these two cycles will not be the same.

3

Since the system - engine plus sources - is isolated and reversible, we have $\Delta S = 0$. The entropy change of the system after a complete cycle of the engine is just $$\Delta S=\Delta S_{\mathrm{sources}},$$ since $\Delta S_{\mathrm{engine}}=0$ for a cycle. Consider the hot reservoir at temperature $T_1$. To calculate its change of entropy we imagine a ...

6

To get the entropy change for a system experiencing an irreversible process, the first step is to forget entirely about the actual irreversible process and, instead, devise a reversible process that takes the system between the same initial and final equilibrium states. That is what is meant by $dq_{rev}/T$. The reversible process that you devise does not ...

10

The bottom line is that hot water loses heat at high temperature, giving a small negative entropy change while the cold water gain heat at low temperature resulting in a high entropy change. The net entropy change is positive. We can explicitly see this: At any instant, the infinitesimal change in the entropy of the system is dS=\frac{dQ_H}{T_H}+\frac{...

3

Entropy is a tricky concept at first but with some rigorous thought it becomes more and more intuitive. When you mix hot and cold water they become inseparable in a closed system (closed to energy....we can still change the volume). Imagine putting a label on every single water molecule of the cold water and then picking them all out after you mix them with ...

3

Mixing the water at different temperatures is an irreversible process, in that once you mix them, you get an average temperature for the system and you can't undo the process, without doing work on the system. Entropy Postulate: If an irreversible process occurs in a closed system, the entropy S of the system always increase. This is because before you ...

6

From a strictly by-the-formulas point of view, entropy change is heat transfer divided by the temperature over which the heat transfer occurs. The heat transfer is clearly the same for both volumes but positive for the cold volume and negative for the hot volume (heat flowed out of the hot volume and into the cold volume) but the average temperature over ...

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