2
$\begingroup$

While deriving the condition for spontaneity, $\Delta$G$\leqslant$0, we start by saying that

$\Delta S_{tot}$ $\ge$0 $\Rightarrow$ $\Delta S_{sys}$ + $\Delta S_{surr}$ $\ge0$

If $Q$ is the heat transferred to the system from the surroundings, then $−Q$ is the heat lost by the surroundings, so that $\Delta S_{ext}$ = - ${Q \over T}$, corresponds to the entropy change of the surroundings.

$\Delta S_{int}$ - ${Q \over T}$ $\ge 0 $,
But, isn't $\Delta S_{int}$ = ${Q \over T}$ , since $Q$ is the heat gained by the system? Consequently, wouldn't we always get $\Delta S_{int}$ - ${Q \over T}$ = ${Q \over T}$- ${Q \over T}$=0 ?

$\endgroup$
  • $\begingroup$ You're assuming that the external and internal temperature are the same $\endgroup$ – John Rennie Oct 5 '13 at 9:40
  • $\begingroup$ @JohnRennie that assumption is pretty ubiquitous though, at least in the context of calculating the Gibbs energy change of a chemical reaction. $\endgroup$ – Nathaniel Oct 5 '13 at 9:45
3
$\begingroup$

There is an asymmetry between the system and the surroundings. In fact, at constant temperature $T$, we are in a canonical or grand-canonical situation (depending on cases), where the "surrondings" is a heat reservoir, with Energy, Number of Particles,etc... much greater than the system. When we modify some external data (ex : volume), the set (sytem + heat reservoir) is evolving until it reaches a thermodynamic equilibrium, and this is done by a positive variation of the total entropy (sytem + heat reservoir) and a diminution of the (Helmholtz, Gibbs) free energy of the system.

The heat reservoir is so big, than we may consider it approximatively in thermodynamic equilibrium, during this evolution, with a temperature T,so, if we consider the heat gained by the heat reservoir $Q_R$, the variation of entropy of the heat reservoir is approximately $\Delta S_R = \frac{Q_R}{T}$.

On the other hand, during the evolution, the system itself cannot be considered in thermal equilibrium, so the notion of temperature is not well defined, so we cannot express a variation of entropy of the system using variation of heat and temperature like $\Delta S = \int dS$ with $dS = \frac{\delta Q}{T}$.

For instance, for simplicity, consider Helmotz free energy, and consider that no work is exchanged between the system and the reservoir. The conservation of energy is written :

$$\Delta U + \Delta U_R=0 \tag{1}$$ where $U$ and $U_R$ are the internal energies of the system and the heat reservoir.

We have :

$$ \Delta U_R = \Delta Q_R = T \Delta S_R\tag{2}$$ The first equality comes from the hypothesis of no work exchange, and the second equality comes from the approximated thermodynamic equilibrium of the heat reservoir.

The total entropy (system + heat reservoir) is increasing during the evolution, so we have :

$$\Delta S + \Delta {S_R} \ge 0 \tag{3}$$

Now, from $(1)$ and $(2)$, we get : $\Delta {S_R} = - \frac{\Delta U}{T}$, so, finally : $\frac{1}{T} (T\Delta S - \Delta U) >0$, and noting the Helmoltz free energy $F = U- TS$, we may write :

$$\Delta F \le 0 \tag{4}$$

$\endgroup$
  • $\begingroup$ Isn't Delta S for system a state property? So, whether the system is evolving or has evolved, won't the change in entropy be the same = delta Q over T? $\endgroup$ – user30602 Oct 5 '13 at 13:42
  • 2
    $\begingroup$ During the evolution, the system is not in thermal equilibrium, so it is not correct to define a temperature for this system. At the opposite, it is correct to say that $T$ is the temperature of the heat reservoir. If there were not asymetry between the system and the heat reservoir, we will be in the situation described in the text of your answer, that is : the variation of the total entropy would be ssystemically zero, which is not correct. $\endgroup$ – Trimok Oct 5 '13 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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