9
$\begingroup$

If my understanding is correct, neither reversible nor adiabatic processes are necessarily isentropic.

But are reversible adiabatic processes always isentropic?

$\endgroup$
2
  • $\begingroup$ Related: physics.stackexchange.com/q/52231/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 1, 2014 at 13:22
  • $\begingroup$ Just to alert you, the term "adiabatic" is now somewhat in a state of change. In thermal physics it was first widely used to mean "without heat transfer" but it has become very common to use it to mean "reversible and without heat transfer" and that is how it is usually used in areas outside thermal physics (e.g. basic quantum mechanics). $\endgroup$ Commented Mar 30, 2021 at 23:08

5 Answers 5

6
$\begingroup$

Yes.

From Clausius theorem the following inequality can be deduced: $$\delta Q \le TdS$$ where the equality holds in the reversible case.

So, a reversible adiabatic process is necessarily isentropic, but irreversible adiabatic processes are not so.

To put it in another way, in an irreversible process, according to the above inequality, either entropy changes, or heat must be somehow removed from the system to make it possible to have zero change in entropy. So an irreversible isentropic process can not be adiabatic.

$\endgroup$
0
6
$\begingroup$

Yes. For a reversible process, we have the relation \begin{align} dS = \frac{\delta Q}{T} \end{align} and for an adiabatic process, we have (by definition) \begin{align} \delta Q = 0, \end{align} which implies that \begin{align} dS=0. \end{align}

$\endgroup$
2
$\begingroup$

I want to explain from fundamentals, rather than invoking the Clausius theorem. Here goes:

STEP 1: REVERSIBLE --> No entropy creation or destruction during the process

Entropy cannot be created or destroyed in any part of any reversible process whatsoever. The reason is, first, entropy can never be destroyed under any circumstance (second law of thermodynamics), and second, if entropy was created in (part of) the process, it would be destroyed in the corresponding part of the reversed-version-of-the-process. But again, entropy cannot be destroyed.

STEP 2: ADIABATIC --> No entropy flow during the process (into or out of the region in question)

If no heat is flowing into or out of the region (definition of "adiabatic"), and no matter is flowing into or out of the region (also part of the definition of "adiabatic" I think), then there's no way that entropy could be flowing into or out of the region.

CONCLUSION: The entropy of the region is fixed if the process is both reversible and adiabatic.

$\endgroup$
1
  • $\begingroup$ +1 This is a nice conceptual explanation. Nothing at all wrong with using formulae, but this is more satisfying in my view. $\endgroup$
    – Noldorin
    Commented Oct 11, 2021 at 0:43
1
$\begingroup$

To answer this question, lets start with combined first and second law of thermodynamics, \begin{equation} dU = TdS-PdV \end{equation} For ideal gas, $dU=c_{v}dT$. Thus $$TdS = c_{v}dT+PdV \qquad$$ Change in entropy in terms of exact differentials (by using the equation of state for an ideal gas $PV=RT$), $$\tag{1} dS = c_{v}\frac{dT}{T}+ R\frac{dV}{V}$$

By taking log on both sides of the ideal gas equation $PV=RT$ and taking differentials on both sides yields \begin{equation} \frac{dP}{P}+\frac{dV}{V} = \frac{dT}{T} \end{equation} Using the above equation in Equation (1), and making use of the relations $c_{p}-c_{v}=R; \gamma=c_{p}/c_{v}$, we find

$$dS= c_{v}\left[\frac{dP}{P}+\frac{dV}{V}\right]+R\frac{dV}{V}$$ or \begin{equation} \frac{dS}{c_{v}} = \frac{dP}{P} + \gamma\frac{dV}{V} \end{equation}

Integrating between two states 1 and 2, gives \begin{equation} \frac{\Delta S}{c_{v}} = \ln\left(\frac{P_{2}}{P_{1}}\right)+\gamma\ln\left(\frac{V_{2}}{V_{1}}\right) = \ln\left[\frac{P_{2}}{P_{1}}\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}\right] \end{equation} Using both sides of the last expression as exponents we obtain

$$\tag{2} \boxed{\frac{P_{2}V_{2}^{\gamma}}{P_{1}V_{1}^{\gamma}} = e^{\Delta S/c_{v}}}$$

Equation (2) describes a general process. For the specific situation in which entropy is constant, i.e. $\Delta S = 0$, we recover the expression $PV^{\gamma}=\text{constant}$. Which is applied for reversible adiabatic process. We now see, through the use of the second law of thermodynamics, a deeper meaning to the expression, and to the concept of a reversible adiabatic process, in that both are characteristics of a constant entropy, or isentropic process.

So in general an adiabatic process is not necessarily isentropic-only if the process is reversible ($\Delta S =0$) and adiabatic ($dQ =0$), we call it isentropic.

$\endgroup$
0
$\begingroup$

Every reversible adabatic process is isentropic. But the converse is not true.Every isentropic process may be reversible adabatic or may not be reversible adabatic.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.