My sir told me entropy change due to two reasons

  1. entropy created

  2. entropy due to heat exchange

Prof also said that entropy produced is zero for reversible process but not for irreversible process.

But I do not understand how they both are different. I am getting confused.

Can anybody explain me this with an example each?


Your professor is using a framework in which the entropy change of a closed system is equal to the sum of the entropy created within the system (by irreversibilities, such as viscous dissipation) plus the entropy entering and leaving the system through its boundaries. The entropy entering through each boundary of the system is given by $Q/T_\textrm{boundary}$ where $Q$ is the heat passing through that part of the boundary and $T_\textrm{boundary}$ is equal to temperature at the boundary through which the heat is flowing. So, in this framework, $$\Delta S=S_\textrm{created}+\sum{\frac{Q}{T_\textrm{boundary}}}$$ If the process is reversible, then $S_\textrm{created}=0$ and $T_\textrm{boundary}=T$, where T is the (uniform) temperature of the system.

For more on this powerful approach, see Fundamentals of Engineering Thermodynamics by Moran et al.

  • $\begingroup$ What do you mean by (by irreversibilities, such as vicsous dissipation) . Can you explain it little more $\endgroup$ – user123733 Dec 8 '16 at 2:17
  • $\begingroup$ Are you familiar with the physical property known as viscosity and, if so, are familiar with the situation in which a fluid (gas or liquid) is sheared between two infinite parallel plates? $\endgroup$ – Chet Miller Dec 8 '16 at 11:33
  • $\begingroup$ yes I am familiar , the tendency of not to flow $\endgroup$ – user123733 Dec 8 '16 at 11:36
  • 1
    $\begingroup$ In the simple situation I described (as well as all other fluid dynamic situations), viscosity causes mechanical energy to be converted into internal energy. You are doing work to deform the fluid, but not getting any mechanical benefit. This is one way by which entropy is generated within a system. The rate of entropy generation is proportional to the square of the deformation rate, and the constant of proportionality is related to the viscosity. $\endgroup$ – Chet Miller Dec 8 '16 at 11:50
  • $\begingroup$ @Chester Miller: Just two comments. 1. Entropy can increase not only via surface heating but also via bulk heating (as it happens in a microwave). 2. It's possible to have a reversible process (no entropy production) with inhomogeneous temperature throughout the system. These two possibilities are contained in the full formula $$\frac{\partial s}{\partial t} \geqslant \nabla\cdot\left(\frac{\boldsymbol{q}}{T}\right) - \frac{Q}{T},$$ valid for a closed system, where $\boldsymbol{q}$ and $Q$ are the surface and flux heating, and the equality holds for reversible processes. $\endgroup$ – pglpm Mar 31 '18 at 13:37

Entropy change can also occur due to mass exchange in addition to heat exchange. That entropy of universe can never be destroyed, but can either remain constant or simply increase, is the second law of thermodynamics. Second law is a postulate and ought to be accepted as such; it cannot be "explained" any further (at least not within classical equilibrium thermodynamics). The idealized textbook processes in which entropy remains constant are given the name "reversible processes". Real processes almost always end up increasing entropy of universe. To understand how entropy is created is the goal of the emerging field of non-equilibrium thermodynamics, and trust me, you don't want to go there. Read Thermodynamics by Fermi.


Not sure exactly how your prof distinguished them, but maybe the first distinction is talking about that entropy does increase even if thermal energy exchanged is zero.

Consider, the case of Free-Expansion, where in an enclosed chamber an ideal gas of $n$ moles is confined to a smaller volume $V_i$ by a partition.

When the gas is allowed to get dispersed over the whole volume $V_f$ by puncturing the partition, it does no work on the partition; neither does any thermal energy come in from outside nor does any get out to outside.

The internal energy remains the same.

It can be easily found out using the First Law and that change in the internal energy is zero (since the gas performed no work and no thermal energy is exchanged),

$$\Delta S_\textrm{free-expansion}~=~ n\mathcal R\ln\frac{V_f}{V_i}~~\gt 0 ~~~~~~~~[\because ~V_f\gt V_i]\,.$$

So, entropy change is positive even though no thermal energy is exchanged.

  • $\begingroup$ Prof also said that entropy produced is zero for reversible process but not for irreversible process. $\endgroup$ – user123733 Dec 7 '16 at 15:53
  • $\begingroup$ Change in entropy is zero for reversible processes provided the system is completely isolated. $\endgroup$ – user36790 Dec 7 '16 at 16:20
  • $\begingroup$ yes , for isolated $\endgroup$ – user123733 Dec 7 '16 at 16:26
  • $\begingroup$ What do you mean by puncturing the partition. And how does this perform no work . $\endgroup$ – user123733 Dec 7 '16 at 16:31
  • $\begingroup$ Why internal energy remains same $\endgroup$ – user123733 Dec 7 '16 at 16:32

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