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I was reasoning on the criterion for spontaneity, using the internal energy as indicator. In a closed system with only P-V work, dq = dU + PdV so that at constant volume dqV = dU. Consequently from $dS \ge dq/T$

$dS - dU/T \ge 0$

which can be rearranged to

$TdS \ge dU$

At either constant internal energy ($dU = 0$) or constant entropy ($dS = 0$), this expression becomes, respectively,

$dS_{V,U} \ge 0$

$dU_{V,S} \le 0 \quad \mathbf{1.0}$

Equation 1.0 says that is a spontaneous isentropic and isochoric process sees its internal energy decrease. Since the entropy of the system is unchanged, there must be an increase in entropy of the surroundings, which can be achieved only if the energy of the system decreases as energy flows out as heat, but this seems to contradict the fact the process is isentropic for the system.

Can you devise a concrete example of such a process, in the absence of non-PV work?

gioretikto
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2 Answers2

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An adiabatic and reversible process is always isoentropic. However, the opposite is not unconditionally true. It is perfectly possible to have irreversible isoentropic processes. Isoentropic, in such a case, refers to the entropy of the system. Of course, such transformations are not adiabatic, i.e. some heat must be exchanged with the surrounding universe. This exchange of heat does not contradict the condition that the system is closed, which refers to the absence of mass exchanges.

The key point to clarify how processes of this kind may be set up is to remember that a system may change its entropy in two ways:

  1. by exchanging heat $q$ with surroundings at temperature $T$, ($q$ is positive if the heat goes into the system, negative if it is extracted from the system): $$ \Delta S_{exch} = q/T; $$
  2. by an internal generation of entropy, $\Delta_{gen}$, due to internal irreversible processes.

The total change of entropy of the system is $$ \Delta S = \Delta S_{exch} + \Delta S_{gen}. $$ While $\Delta S_{gen} \geq 0$, $\Delta S_{exch}$ may be positive or negative. In the latter case ($\Delta S_{exch} \lt 0$), it may compensate exactly a positive $\Delta S_{gen}$, to make the process isoentropic ($\Delta S = 0$), but still irreversible since the universe entropy is increased by the positive amount $-\Delta S_{exch}$.

As a concrete example of a process of this kind, one can think of a rigid and closed vessel, able to exchange heat with the surroundings, containing two equal volumes of two different perfect gases separated by a wall. By removal of the internal wall, there will be a production of entropy $\Delta S_{mix}$, (entropy of mixing). However, the total entropy of the system may be kept constant if heat equal to $T \Delta S_{mix}$ is transferred to surroundings at temperature $T$. The total transformation is irreversible but isoentropic for the gas in the vessel. Of course, the internal energy of the system decreases in such a process, and the equilibrium condition will correspond to the minimum of the internal energy at constant total entropy of the system. At the same time, the entropy of the universe will be increased by $\Delta S_{mix}$.

GiorgioP-DoomsdayClockIsAt-89
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By definition an isentropic process is both adiabatic and reversible. But a reversible process cannot be spontaneous. Since you have stipulated the system is both adiabatic (=0) and constant volume (=0), the system is not closed but isolated and $\Delta U=0$.

Moreover, if a spontaneous process occurs in the system as you specified, entropy is generated in the system. Since the system cannot transfer that entropy to the surroundings in the form of heat, $\Delta S_{sys}>0$. An example of such a process is the free expansion of a gas into a vacuum occurring in a rigid, perfectly insulated container.

Hope this helps.

Bob D
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