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Well, I am a little bit confused about this question. I learn that reversible adiabatic processes are isentropic. So $\Delta S=0$. Through $\Delta S=\frac{ q}T$, we can say that $q=0$. But if you take an irreversible adiabatic process, due to friction, $\Delta S\gt0$. Which makes:

$$\begin{align} \Delta S=\frac qT&\gt0\\ q&\gt0 \end{align}$$

So is it safe to say $ q\gt0$ for the irreversible adiabatic process? If not, where I am wrong?

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    $\begingroup$ What do you mean by "Δq"? There is q, and δq, but Δq is (probably) not what you are referring to. $\endgroup$
    – Buck Thorn
    Commented Apr 2, 2023 at 17:55
  • $\begingroup$ For most cases, using Delta q is wrong, as just q should be used, at q means system energy change, like W or Delta U or Delta H. The only case where Delta q is justified, with proper legend, is in very specific scenarios the difference of two different q values. (I have it wrong myself in one of my comments - habits die hard.) $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 18:10
  • $\begingroup$ Are you saying that $\Delta S=\frac{\Delta q}{T}$ both for reversible and irreversible processes? $\endgroup$ Commented Apr 2, 2023 at 18:16

3 Answers 3

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The definition of an isentropic process is $\mathrm{d}S=0$.
The definition of an idiabatic process is $\delta q=0$.

Be aware that $\mathrm{d}S \ge \frac{\delta q}{T}$. So even if $\delta Q=0$ as the adiabatic process requirement, there is still possible $\mathrm{d}S \gt 0$.

Imagine isolated system as a water reservoir. There is done mechanical non-PV work on this system by mixing water by the blender screw - like the classical Joule experiment about heat/work equivalence.

The process is adiabatic with irreversible friction - but no heat exchange, so $\delta q = 0$.

The process is not isentropic, as it is irreversible and entropy increases, with $\mathrm{d}S \gt \frac{\delta q}{T} = 0$

If the irreversible action like friction occurs out of the system, like at the piston with friction or mechanical machinery behind, the system may do non-PV work on the surrounding via friction, which converts work to heat.

$$\mathrm{d}S_\mathrm{sys} = \frac{\delta q_\mathrm{sys}}{T} = 0$$

$$\mathrm{d}S_\mathrm{surr} \ge \frac{\delta q_\mathrm{surr}}{T} \gt 0$$


Formally, process can be isentropic even if $\delta q <> 0$, if $\mathrm{d}S_\mathrm{sys}= \mathrm{d}S_\mathrm{sys,irrev} + \frac{\delta Q}{T} = 0$, so $\mathrm{d}S=\frac{\delta q}{T}=0$ is not correct definition of isentropic process.

E.g. Assume a system consisting of 2 reservoirs with temperatures $T_1$ and $T_2$ and there is system surrounding of temperature $T_3$, when $T_1 \gt T_2 \gt T_3$. Entropy increase term of system will be caused by the spontaneous balancing of temperature in both system reservoirs, while the entropy decrease term will be caused by passing heat to the surrounding.

With properly chosen temperatures, there would be ongoing isentropic process for the system, with $Q \ne 0$.

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    $\begingroup$ It is just good to emphasize that generated heat is equal to the work done by the friction force and isn't connected to the PV work. This is to avoid confusions by using the same terms. $\endgroup$ Commented Apr 2, 2023 at 14:35
  • $\begingroup$ ........Made fix $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 14:37
  • $\begingroup$ Note that by IUPAC Gold book, both q and Q are used for heat. $\endgroup$
    – Poutnik
    Commented Apr 3, 2023 at 6:24
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If process is irreversible than adiabatic process between states 1 and 2 will generate entropy in the surroundings. Entropy is a state function which means that if system goes from state 1 to state 2, entropy change of the SYSTEM is the same regardless of process being irreversible or not.

For adiabatic process this change is equal to zero. However, if adiabatic state change is irreversible, entropy will be generated outside the system due to irreversibility causing entropy of the surroundings to increase.

Generally for irreversible process, 2nd law is:

$$ dS_{tot} = dS_{sys} + dS_{sur} > 0 $$

By definition of isentropic process:

$$dS_{sys} = 0 $$

This leaves us with: $$ dS_{sur} > 0 $$

Last inequality shows that irreversible state change generated entropy in the surroundings. This can be work done by friction when piston is moved, for example.

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    $\begingroup$ The way to do that is to exchange heat with the surroundings causing its entropy to increase. For adiabatic process, $\Delta q_\mathrm{sys}=0$ by the definition. But, the system may do work on the surrounding via friction, which converts work to $\Delta q_\mathrm{surr} \gt 0$. $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 13:12
  • $\begingroup$ That is true if process is reversible. Irreversible state changes will generate heat which may be generated inside the system. If that is the case, system needs to discard it to be able to get to the state 2. $\endgroup$ Commented Apr 2, 2023 at 13:20
  • $\begingroup$ But then it is not an adiabatic process. // OTOH, if it is an idiabatic process, but irreversible due not being in equilibrium during the transition, there is still $\Delta q=0$, but in absolute values, $W_\mathrm{irrrv,comp} \gt W_\mathrm{rev, comp}$ or $W_\mathrm{irrrv,exp} \lt W_\mathrm{rev, exp}$ $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 13:29
  • $\begingroup$ That holds, if we don't allow any heat exchange with the surroundings (which is how adiabatic process can be defined). However, than we can't say that entropy change of the system is zero, it must be positive and surroundings have entropy change zero. This discussion shows that we need to be careful what exactly we mean by adiabatic. Whether no heat exchange or constant entropy of the system. In thermodynamics and state changes, usually it is meant constant entropy, but that only works if system can discard generated entropy by heat due to the irreversibility. $\endgroup$ Commented Apr 2, 2023 at 13:29
  • $\begingroup$ Not "can be defined", but "is defined". Is says nothing about entropy change, that may or may not be zero and adiabatic process may or may not be isentropic. $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 13:31
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First of all, there is no such thing as $\Delta q$ unless you are referring to the difference in the heat evolved in two different processes. For an adiabatic process, q=0 by definition, and therefore the difference in the heat evolved in two adiabatic processes is and will always be, $\Delta q=0$, irrespective of whether these are reversible or irreversible.

Second, the second law of thermodynamics relates the heat exchanged in a process to the entropy change, with a reversible process as a limiting case for which the equality case in the following equation holds: $$\Delta S \geq \frac{q}{T}$$

In other words,

$$\Delta S = \frac{q_{\textrm{reversible}}}{T}$$

That "reversible" condition is very important. The equality does not hold otherwise. The 2nd law of thermodynamics states that for any irreversible process at constant T between the same states, q is less than $T\Delta S$. This just guarantees that the entropy change of the system and surroundings offset properly to generate an overall positive entropy change for the universe. For instance, in a spontaneous exothermic process the change in the entropy of the system cannot decrease by more than $|\frac{q}{T}|$.

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