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In the case of the Ehrenfest classification for the first order phase transition it is said:

If the first derivative of the free energy is discontinuous then we have a first order phase transition.

Now I know that for the free energy we have : $dF= -pdV - SdT + \mu dN$.

From here we get : $-(\partial F/\partial T)_P=S$.

This is easy to understand. But it can be also found by taking the derivative of the chemical potential wrt Temperature:

$-(\partial \mu/\partial T)_P=S$. Where does this equation comes from?

And an additional question regarding phase transitions.

If we are observing the liquid-gas phase transition. In our lectures the professor said that the system will always choose the state with the smaller chemical potential. So if for a fixed temperature value we have the system in the gas phase and we increase the pressure beyond the Vapor pressure then the system will jump into the liquid state because the chemical potential in the liquid phase has a smaller value. My question is the following:

In the graph done by the professor (which I don't know how to illustrate here) it looks as if the chemical potential for the liquid phase decreases in value as pressure increases. But for the chemical potential the following equation was derived in the lecture:

$\mu = \frac G N = \frac F N + \frac {PV}N = f + P \nu$

where: G is the Gibbs energy, F is the free energy, N the number of particles in the system, f the free energy for particle, and $\nu$ inverse of density. The point is that the chemical potential is proportional to the pressure. Which means that regardless of the phase in which the system is found, an increase in pressure means an increases in chemical potential for the gas and the liquid state and that the graph done by the professor is not correct,meaning for the liquid phase the chemical potential increases, but the one for the gas increases more, hence the system changes phase (into liquid) as pressure increases.

Am I correct about the part in bolt?

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From your first formula (for the differential of Helmholtz free energy $F$) one can derive $$ -\left(\frac{\partial{F}}{\partial{T}}\right)_V = S. $$ Notice that the derivative is at constant volume, since the differential form clearly shows that $T,V$, and $N$ are the independent variables $F$ depends on. The partial derivative of the chemical potential wrt temperature is $$ -\left(\frac{\partial{\mu}}{\partial{T}}\right)_P = \frac{S}{N}=s. $$ Notice the division by N. The formula in the question, without division by $N$, cannot be correct because chemical potential and temperature are intensive quantities, while $S$ is extensive. Such formula can be easily obtained by recognizing that the Gibbs free energy $G=F+PV$ is a homogeneous function of degree one of its extensive variable $N$ and then, from the Euler theorem, $$ G = \mu N. $$ Therefore, every statement about the Gibbs free energy per particle is equivalent to a statement about the chemical potential as a function of $P$ and $T$. The derivative with respect to pressure of the chemical potential, at constant temperature, is $$ \left(\frac{\partial{\mu}}{\partial{P}}\right)_P =\frac{V}{N}=v $$ the volume per particle. By definition, this is a strictly positive quantity, therefore $\mu$ must be a strictly increasing function of the pressure.

Moreover, in a one-phase system, the Gibbs free energy is a strictly concave function of the pressure (for the stability of the thermodynamic equilibrium). This implies that the chemical potentials of two homogeneous phases in the neighbor of a first-order phase transition should behave as in the following figure. enter image description here

The most stable phase corresponds to the lower curve (minimum Gibbs potential). On the left of the crossing point, the stable phase (purple) has a higher slope, thus corresponding to the phase with the higher volume per particle (lower density),i.e. the gas phase. On the right of the crossing (at high pressures) the stable phase has a lower slope and then a lower volume per particle (higher density).

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  • $\begingroup$ Ok, so both chemical potentials increase, it's just that the one for the gas has higher value, and this is the reason the system goes into liquid phase. The graph made by professor was horrible and implied that the chemical potential of the liquid was decreasing with Pressure increase. Thanks $\endgroup$
    – imbAF
    Jul 1 at 18:39

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