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My Thermodynamics book discusses three types of expansions of an ideal gas and the change in entropy $\Delta S$ associated with each expansion, all of which take an ideal gas in a piston from $V$ to $V+\Delta V$:

  1. Slow Isothermal: $\Delta S = \frac{q}{T}$, which is sensible since the spatial part of multiplicity $\omega$ increases (momentum part of $\omega$ doesn't increase since $U(T)=3/2 NkT$ for an ideal gas stays constant for isothermal process)

  2. Fast Adiabatic: $q=0$ for an Adiabatic Process. But since $S$ is a state function and the endpoints of the process ($V$ to $V+\Delta V$) are the same as above (Slow Isothermal), the change in entropy $\Delta S$ must be the same. Therefore, in this case, $\Delta S > \frac{q}{T}$. Combining these two scenarios, $\Delta S \geq \frac{q}{T}$ in general, where the equality applies for a slow, reversible process and the inequality for a fast one. Conceptually, the author justifies this as "In the expansion into a vacuum, the molecules would strike only stationary walls and hence no energy would be lost from the gas." -- which I do not understand. The walls are of course moving outwards as the volume expands, so what does he mean?

  3. Slow Adiabatic: This is the one I'm stuck up against. The author argues that since $q=0$ and this is a slow process, we can apply the equality that $\Delta S=\frac{q}{T}=0$. But $S$ is a state function, and the endpoints of this process is the same as the above two. Thus, why isn't the change in entropy $\Delta S$ the same?

I have one final question. The book calculates the entropy change when water is heated from $0 °C$ to $70 °C$ by integrating $S=\int \frac{q}{T} dT$ instead of simply calculating $q/T$. Is this because the change in temperature is significant and cannot be treated to be constant throughout the process?

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  • $\begingroup$ Which thermodynamics book? $\endgroup$ Mar 7, 2023 at 16:39
  • $\begingroup$ Ralph Baierlein - Thermal Physics $\endgroup$
    – DarkRunner
    Mar 7, 2023 at 16:46

1 Answer 1

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The walls are of course moving outwards as the volume expands, so what does he mean?

It’s assumed that the moving walls are moving so fast—thus the description, fast adiabatic—that no or a negligible proportion of gas particles strike them during expansion.

S is a state function, and the endpoints of [slow adiabatic expansion] is the same as [those of isothermal and free expansion]. Thus, why isn't the change in entropy ΔS the same?

The endpoints aren’t the same. The gas did work without receiving compensating heat, so it’s colder in this example.

Is this because the change in temperature is significant and cannot be treated to be constant throughout the process?

Yes.

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  • $\begingroup$ Finally. Finally I understand. Good lord, thank you. $\endgroup$
    – DarkRunner
    Mar 7, 2023 at 18:11

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