Another statement of second law of thermodynamics can be formulated in terms of system properties and not properties + surroundings (isolated system). For a closed system at constant temperature and pressure:
$$dG_{T,p,n_i} \leq 0$$
Are $n_i$ constant during a process? We know that during a chemical reaction Gibbs free energy take its minimum value at equilibrium. But during the reaction the number of molecules $n_i$ change. So would it better the inequality to take the form of:
$$dG_{T,p} \leq 0$$
?
Interpretation of the differential In mathematics the differential of a function $f(x)$ can be interpreted as the change in $f(x)$ for an infinitesimal change in the argument $x$, that is:
$$df(x)= f(x+dx)-f(x)$$
I have trouble in the interpretation of $dG \leq 0$. When I was introduced in thermodynamics I always thought of the spontaneity inequalities in the following manner.
Suppose that the system is at equilibrium state (state $1$) at time $t$. Something happens (e.g. a chemical reaction takes place) so the system will reach a new equilibrium state (state $2$). In that case $G$ can be thought only as function of $ξ$ (the extent of reaction). Because the Gibbs energy must take its minimum value at equilibrium I interpreted the $dG \leq 0$ as:
$$dG(t)= G(t+dt)- G(t) \leq 0$$
In other words Gibbs free energy decreases until it takes its minimum value. And because we can plot $G$ as function of $ξ$ the reaction will proceed in the direction where $G$ is minimized as shown in the following figure (the left figure):
Although the above interpretation seems intuitive I have some problems.
$1$st Problem With the above interpretation is not clear if the equality should be interpreted as that the function have a constant value over an inteval or that it has reached a minimum value. For example if we were monitoring $G$ over time and we noticed that $G$ has a constant value would this imply that the system have reached equilibrium? Look at the following figure:
In the mid region of the function (where it is constant) the differential is zero. So how $dG \leq 0$ should be interpreted? In general we can't find a function $φ$ so that $G=φ(t)$ and this should be expected as Thermodynamics doesn't deal with how the state of the system varies with time.
$2$nd Problem In the first scheme I have drawn an alternative path (right figure) that the system could take in order to minimize $G$. In that case there is an abrupt change in the extent of the reaction but still the systems keeps decreasing its $G$. To be honest I haven't seen any reaction reaching equilibrium in that way but I can't find a reason why it should not behave this way. It should be noted that the line representing the path depicts the change in $ξ$ in an abrupt manner and it should not be confused with a continuous change (think it like a $5$ step process).
In summary I want to make clear how to interpret the differential expression of the second law. Is a smooth incease of entropy (of surroundings + system) over time assumed?
