ml>Phase transitions
* 221 Notes
1 Gases2 Microscopic Energies3 First Law4 2nd & third Law5 Phase transitions6 Mixtures7 Phase Diagrams8 Equilibrium9 Molecular Interactions
Phase transitionsPhase DiagramsThermodynamics considerationsLet's develop some relations that assist us to account for the features of the phase diagrams. We will certainly usage Gibbs energies in this discussion. First, we need to build an knowledge of a concept of soimg.orgical potential, m. The soimg.orgical potential is actually described, making use of Gibbs energy concepts.

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Recallequation 4.39, the general equation for Gibbs energy adjust expressed as a duty of T and p.


This equation contains the presumption that no variables other than temperature and also pressure are varied. In the instance of phase changes, we carry out have moles of the two (or more) phases transforming so a more basic equation could be


The latter 2 terms in the equation would certainly additionally be supplied in soimg.orgical reactions, one such term for eextremely reactant or product. We currently specify the soimg.orgical potential as the price of change of Gibbs energy through a readjust in number of pshort articles so,



. 5.4

In the case of one mole of a pure substance, we can substitute m directly forGm. Similar to Gibbs energies, a spontaneous process will certainly result in a lower soimg.orgical potential.

We are currently prepared to check out the phase diagrams.

We recontact from our conversation leading to the Maxwell relation (equation 4.40) that the price of readjust of Gibbs power through temperature is ssuggest the negative of the entropy. For a pure substance, we have the right to substitute soimg.orgical potential for Molar Gibbs power so


From this equation we can view that as the temperature goes up, the soimg.orgical potential decreases (S is constantly positive).

This Figure mirrors the change in soimg.orgical potential for a device that is being warmed up from the solid state via the liquid state and also ultimately into the gaseous state. Note that the slope is steeper for the gas than for the liquid, which is steeper than for the solid. This comes from the reality that the molar entropy is greatest for the gas and lowest for the solid.

Now follow the diagram from the solid phase right into the liquid phase area. If the substance doesn't melt at Tf, it's soimg.orgical potential would follow the light colored line. However before, as lengthy as tbelow is no kinetic barrier to the shift, tright here is a various state (liquid) easily accessible to it in this area and the solid will certainly spontaneously adjust to the liquid, for this reason lowering its soimg.orgical potential.

We can perform the exact same logic for a phase adjust in the other direction a liquid is cooled listed below Tf, without freezing. It's soimg.orgical potential is plainly higher than that for the solid phase so the spontaneous phase change in the region listed below Tf, is from liquid to solid.

We can use this logic for all phase readjust procedures. Consider a solid that is heated so much that it is currently over Tb. Tbelow are two phases (liquid and also gas) easily accessible to it that have actually a reduced soimg.orgical potential. The liquid is just metasteady at best so we view that the transition directly to the gas phase is the a lot of most likely transition.

We can additionally discover the response of the mechanism to press alters using the equation 4.40, making the exact same substitutions as we did to obtain to equation 5.5, we arrive at


This shows that a plot of m versusp returns a slope equal to the molar volume. Since molar quantities are always positive, the slope will be positive and also steepest for gases.

The inset phase diagram shows the course adhered to (consistent T).

If we trace the soimg.orgical potential as we boost p at this temperature we see that the device passes via gas, liquid and also ultimately solid says at the preferred temperature

This chosen temperature is reduced than the triple allude. At this temperature, the soimg.orgical potential of the liquid is constantly greater than that of the solid or the gas so liquid is never thermodynamically secure.
As the temperature is collection at the triple allude temperature, we view that all three phases have the same soimg.orgical potential once press is additionally at the triple point.
In this graph, the temperature is collection above the crucial temperature. This graph of a superinstrumental fluid mirrors that the slope shifts from gas-like slopes to liquid prefer slopes smoothly, tright here is no disaffix in the slope, i.e., no equilibrium suggest through two phases
We recognize that many materials experience an elevation in melting allude as soon as tright here is a rise in exterior pressure. In various other words, as we squeeze on a substance, it will certainly tend to freeze at a greater temperature. This functions for the majority of materials except water, where the reverse trfinish happens. This is as a result of the molar volume distinction between the liquid and solid phases. Generally, the liquid has a bigger molar volume. In water, the solid has actually the bigger molar volume (much less dense) compared to the liquid. Figure 4.10 (From Atkins) illustprices the impact this has on the soimg.orgical potential versus temperature .


As the device experiences a boost in push, the soimg.orgical potential will readjust, depending on the molar volume. Thus, in (a), the molar volume of the liquid is higher than that of the solid and hence, the rise in molar volume is better for the liquid than for the solid. This outcomes in an boosted melting suggest.

In (b), the solid's soimg.orgical potential increases more, resulting in a lower melting allude.

Example: calculate the change in soimg.orgical potential for ice and also for water at 0ºC as the push is changed from 1.00 bar to 2.00 bar. the thickness of ice is 0.917 g/cm3 and of water is 0.999 g/cm3.

Using equation 5.6 we can quickly combine to get Δm = Vm×Δp . The molar volume is calculated from the density and the molar mass, therefore Vm = M/rand also for this reason, Δm = M/r×Δp

Thus, if the mechanism was at equilibrium at 1.00 bar then it is no much longer at equilibrium at 2.00 bar and the spontaneous direction is to adjust the ice right into water, so as to lower the in its entirety soimg.orgical potential.

Notice that as the pressure transforms, the liquid's soimg.orgical potential changes. This implies that the liquid's vapour push will certainly readjust when exterior press is applied. For instance, if a closed device containing a liquid and vapour at equilibrium is pressurized by including an inert gas, then the additional push exerted on the liquid will readjust it's soimg.orgical potential, as we calculated over yet the gas (assuming ideal) press will certainly not change. Hence, a mechanism that was previously at equilibrium will certainly no much longer be and the vapour push will rise. This goes against assumptions made in first year that the vapour pressure depfinished only on the temperature of the liquid.

Following the phase boundaries


The entire mechanism at equilibrium must have actually unicreate soimg.orgical potential throughout. Otherwise, tbelow would be a flow of pwrite-ups from area of high potential to low potential until equilibrium is reached. Therefore, if two phases are in equilibrium as depicted here, along the phase change line, then both phases have actually the same soimg.orgical potential. If a press is used, which shifts the system out of equilibrium then the temperature will certainly readjust (as a result of some pwrite-ups moving from one phase to the other) until equilibrium is re-establimelted.

The slope of the phase boundary is dp/dT. We can readjust equation 4.39 by substituting soimg.orgical potential for molar Gibbs power (along with molar other parameter too) to get

Vm,a dp – Sm,adT = Vm,b dp – Sm,bdT

(Vm,a – Vm,b) dp = (Sm,a– Sm,b)dT

, this is the Clapeyron equation


The Clapeyron equation is a specific equation describing the slope of any phase readjust boundary.

Solid-liquid boundary

If the 2 phases in question are the solid phase and the liquid phase then the transition in question will certainly be fusion. We deserve to rewrite our Clapeyron equation particular to this equilibrium. in this situation, the molar entropy of fusion is the same as the molar enthalpy separated by the change temperature.

given that the enthalpy of melting is positive, we check out that the slope of this transition will certainly depend on the relative molar volumes of the solid and also liquid If the molar volume of the solid is less than that of the liquid then the slope of the line will certainly be positive. This is the many prevalent situation. If the molar volume of the solid is greater than the molar volume of the liquid, as in water, then the slope of this phase boundary will be negative.If we wish to determine a macroscopic adjust, we must integrate:
This is the equation of a straight line via a very steep slope.

Liquid-vapour boundary

If the 2 phases in question are the liquid phase and also the vapour phase then the shift in question will certainly bevaporization. We deserve to recreate our Clapeyron equation specific to this equilibrium. As in the solid/liquid equilibrium, the molar entropy of vaporization is the very same as the molar enthalpy split by the change temperature.

This time, we have a instance wright here the molar volume of the liquid phase is incredibly small compared to the molar volume of the gas. So, we will assume that the readjust in volume is simply the molar volume of the gas If the gas behaves ideally, then Vm = RT/p. to give us
Example: What is the slope of the solid/liquid curve at the normal melting allude for water? ΔfusH0 = 6010 J/mol.

Use equation 5.8. Because this we're sitting on the equilibrium line, tbelow is no change in interior power so ΔU 0 = 0, for this reason, ΔfusS0 = ΔfusH0/T

Note the incredibly large value for the slope (incredibly steep) and it's negative since the adjust in volume as water goes from solid to liquid is negative.

Order of phase transitions

First-order phase transitions

Consider two phases , α and also β, in equilibrium. Most phase transitions involve alters in enthalpy and in volume. These alters number right into the differences in the slope of the soimg.orgical potential curve on either side of the transition suggest.

In words, the distinction in the slope of soimg.orgical potential versus push is simply the distinction in the molar volumes of the 2 phases.

The difference in the slopes of the soimg.orgical potential versus temperature is the difference in entropy or the difference in enthalpy over temperature for the change in question. The slope is different on either side of the transition allude. It is disconstant at the shift.

This is a first-order phase shift. Remember that at the change suggest, we are changing the enthalpy of the system yet not its temperature. Thus, the warmth capacity Cp at the shift point is unlimited. In other words, as we warmth the device that is at the shift point, no temperature readjust happens bereason all the warm is going into the phase shift.


Fig.4.16 (from Atkins) The alters in thermodynamic properties accompanying (a) first-order and (b) second-order phase transitions.
Second-order Phase transitions

In a second-order phase transition, the initially derivative of the slope of m is not disconstant yet it's second derivative is. In other words, tright here is an inflection suggest at the phase transition. This type of transitions occurs between conducting and superconducting phases of metals at low temperatures.

l transitions

These transitions are not initially order yet their warmth capacity goes to infinity at the transition point. We experienced such a shift between the 2 various liquid phases of helium at very low temperature. These transitions tfinish to incorporate order/disorder transitions, paramagnet/ferromagnetic transitions and the fluid/superliquid change of He.

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Fig.4.17(Atkins) The λ-curve for helium, wbelow the warmth capacity rises to infinity. The shape of this curve is the beginning of the name λ-change.

Last updated:21-Oct-2014

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