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The water boiling process can be represented by the equation: $$ \ce{H2O(l) → H2O(g)} \tag{1} $$ Since free energy depends on pressure $$ \left(\frac{\partial G}{\partial P}\right)_T = V \tag{2} $$ and the molar volumes of water are related by the inequality $V_\ce{H2O}(\ce{l}) \ll V_\ce{H2O}(\ce{g})$, evaluate how the boiling temperature of water will change with increasing pressure.

I understand that the boiling temperature will increase, because with an increase in pressure an equilibrium will shift in the direction of smaller amount of molar volumes, according to Le Chatelier's principle. Unfortunately, I cannot find out how to predict an increase in the boiling point of water referring to the statement that $\mathrm{d}G = V_\mathrm{m}\,\mathrm{d}P$ (as far as I understand, Gibbs free energy will experience an increase as well).

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    $\begingroup$ Have you looked at the P-T diagram for water? $\endgroup$
    – Jon Custer
    Jan 3 at 16:03

2 Answers 2

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It is hard to explain! Boiling point is an artifact! it depends on having a flexible container of an inert component, an atmosphere. [Altho a constant force stopper might also work.] For a one component system in equilibrium the Gibbs Phase rule, F = C -P +2, states F = 3-P. This means when 3 phases are present there are no degrees of freedom, a triple point of fixed T, P and chemical activities notice not amounts of the phases just their activities. When 2 phases are present there is one degree of freedom, 1 phase-2 degrees of freedom. Assuming physical possibility, for a liquid -vapor equilibrium this means if one controls either Temperature or Pressure the equilibrium will adjust to establish equilibrium at the new T or P. If both are controlled either one of the phases will vanish or an accidental equilibrium point will be established [the point will fall on the liquid-vapor line in the phase diagram].

Addition of an inert gas to a pure material means additional components. In the case of air it can be considered one component for physical changes: F = 2 -P +2 or F = 4 - P. A This means for 3 phases, a triple point, there is a triple point for each inert gas pressure an envelope of phase diagrams in the 4th dimension. A liquid-vapor equilibrium with 2 degrees of freedom responds to the temperature or the inert gas pressure. Increasing either of these puts more stress on the liquid phase, raising its chemical potential. This means that increasing the temperature OR increasing the inert gas pressure will raise the vapor pressure of the liquid because the chemical potential of the liquid becomes higher than that of the vapor. It seems that this means increasing air pressure increases the vapor pressure and should reduce the boiling point. True the vapor pressure is increased but the external pressure is also increased. This is where the artifact part enters! For a liquid to boil the vapor pressure must be increased enough to push back the higher atmospheric pressure. This means that the boiling temperature must be increased. The net effect is that increased inert gas pressure increases the boiling point. This sort of explains why the effect of changing pressure is less for boiling pt elevation than for freezing point depression; there are two opposing factors: increased inert gas pressure increases the vapor pressure, and the temperature must be increased to raise the vapor pressure enough to push back the atmosphere.

A disclaimer: This analysis is my own. It is based on passages in Daniels and Alberty, Physical Chemistry That I cannot recombine upon rereading the book, and my pondering the phase equation. This analysis describes what I call the 4th dimension of phase equilibria. The pure substance phase behavior is transformed into an envelope of almost parallel diagrams separated in the 4th dimension.

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Expressing Eq 2 in differential format as $$\left( \mathrm{d} \Delta G\right)_T = \Delta V \mathrm{d}P $$ where $\Delta V = V_\ce{H2O(g)}-V_\ce{H2O(l)} > 0$ leads to the obvious conclusion that increasing the pressure increases the Gibbs' free energy of the gas relative to liquid, at constant T. This would suggest that the liquid is preferred at higher pressure (with T constant).

The question however is specifically how the boiling T will change. We know it will increase, but need to show this. In essence we need to derive the Clapeyron equation.

Start from the differential form of G:

$$dG = \left(\frac{\partial G}{\partial P}\right)_T dP + \left(\frac{\partial G}{\partial T}\right)_P dT$$

For the equilibrium system $$d \Delta G = 0 = \left(\frac{\partial \Delta G}{\partial P}\right)_T dP + \left(\frac{\partial \Delta G}{\partial T}\right)_p dT$$ or $$ \frac{\mathrm{d}T}{\mathrm{d}P} = -\left(\frac{\partial \Delta G}{\partial P}\right)_T / \left(\frac{\partial \Delta G}{\partial T}\right)_p\\ \frac{\mathrm{d}T}{\mathrm{d}P} = -\Delta V / \left(\frac{\partial \Delta G}{\partial T}\right)_p$$ From the definition of dG ($= VdP -SdT$), $$\left(\frac{\partial G}{\partial T}\right)_P = -S$$ so that $$ \frac{\mathrm{d}T}{\mathrm{d}P} = \frac{\Delta V}{ \Delta S}$$ Since $\Delta S>0$ for evaporation it follows that $$ \mathrm{d}T = \frac{\Delta V}{ \Delta S}\mathrm{d}P >0$$

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  • $\begingroup$ you confuse pressures! for a 1component system at T there is a vapor pressure if this vapor pressure is increased the chemical potential of the gas is increased forcing condensation and a new equilibrium at a higher T[heat is released]. Adding an inert gas at constant volume increases the potential of the liquid not the gas this raises the vapor pressure and lowers the T [heat is required] An inert gas or applied force n a non constant volume needs an additional pressure to push back the atmosphere or plunger so inceased external pressure a higher T to supply the vapor pressure. $\endgroup$
    – jimchmst
    Mar 5 at 23:23
  • $\begingroup$ A phase diagram involves liquid and vapor, liquid and solid, solid and vapor, and at one point liquid, solid and vapor. There is no inert gas present. Addition of inert gas overlays a new phase diagram for each inert gas pressure. $\endgroup$
    – jimchmst
    Mar 5 at 23:32
  • $\begingroup$ @jimchmst What inert gas? There is no mention of an inert gas either in the problem or in my solution afais. Don't see what point you are trying to make. Have you heard of the Clapeyron equation? It describes the relationship between T,P points on a two-phase coexistence curve. $\endgroup$
    – Buck Thorn
    Mar 6 at 7:12
  • $\begingroup$ The atmosphere is the inert gas. the Clapeyron equation describes the relationship between T and the vapor pressure with one component not the boiling point, boiling is evaporation into an unconstrained volume against an opposing pressure, usually the atmosphere tho it can be a vacuum pump.The vaporpressure curve is monotonic to the critical point, there is no boiling point without an external pressure[inert gas]. $\endgroup$
    – jimchmst
    Mar 6 at 10:53
  • $\begingroup$ @jimchmst I still don't see your point. The question is about the boiling temperature. The boiling Ts are the points on the coexistence line described by the CC equation. I think you attempt to propose a definition for "boiling" as a particular type of event involving inert gases etc, but I don't think that is pertinent here. $\endgroup$
    – Buck Thorn
    Mar 6 at 13:20

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