ABSTRACT
VpdTdSdU −≤ (10.2) Therefore the quantity 0V ≤−+ TdSpddU (10.3) cannot increase, so that, for a given temperature and pressure of the surroundings, the condition for equilibrium is given by 0V =−+ TdSpddU (10.4) Simpler forms of this general relation can be derived under the condition that two of the thermodynamic coordinates remain constant. constant: Eq.(10.4) reduces to VandS ,0=dU so that the conditions for equilibrium are
0,0V,0 === dUddS (10.5) constant: Eq.(10.3) becomes pandS ,0V ==+ dHpddU and the set of conditions reads 0,0,0 === dHdpdS (10.6)
V constant: Eq.(10.3) can be written as andT ( ) 0==− dFTSUd and the conditions on the system are
0,0V,0 === dFddT (10.7)
constant: Eq.(10.3) reduces to pandT ( ) 0V ==−+ dGTSpUd and the appropriate set of conditions is 0,0,0 === dGdpdT (10.8)
The four sets of conditions for equilibrium are equivalent. The thermodynamic potentials involved are always coupled through their specified variables. Equilibrium is defined as the minimum of the appropriate potential function depending on which variables are held constant. The analogy with mechanical potential energy shows that we deal with conditions for stable equilibrium. When a system consists of more than one phase, the thermodynamic functions of the whole may be constructed out of those of the component phases. For a system of two phases, denoted as and ,α β we can then write G m g m gα α β β= + (10.9)
where the specific Gibbs functions depend on p and T only, while G is a function of p,T and the masses ,m mα β of each phase. A unique condition for phase equilibrium holds whatever the external constraints, as we shall demostrate below. At constant from Eqs.(10.8) and (10.9) the condition for equilibrium becomes
pandT
0dG g dm g dmα α β β= + = (10.10) As the system is closed, mass is conserved according to 0,dm dmα β+ = so that Eq.(10.10) reduces to
g gα β= (10.11) Therefore the condition for equilibrium between phases is that their specific Gibbs functions are equal. The result also holds when more than two phases are present. For constant T and V the proper condition for equilibrium is given by Eq.(10.7) and we can write F m f m fα α β= + β so that
(10.12)
0dF f dm f dm m df m dfα α β β α α β β= + + + =
We can express the constant volume V in terms of the specific volumes of the two phases, so that
v v const or v v v vm m m d m d dm dmα α β β α α β β α α β β+ = + + + =
Multiplying the last equation by p and taking into account the condition for equilibrium, we get ( ) ( ) ( ) ( ) 0vvvv =+++++++ βββααβββαα pddfmpddfmdmpfdmpf αα where ,0v =−=+ sdTpddf since T is assumed constant. Using again it follows that
0=+ βα dmdm
v v orα αf p f p g gα β β+ = + = χ (10.13)
that is the same condition for equilibrium as derived in Eq.(10.11) under different constraints If S and p are constant we may use the enthalpy H m h m hα α β β= + and the condition (10.6) for equilibrium
0dH h dm h dm m dh m dhα α β β α α β β= + + + =
under the constraint of constant entropy, which reads
const or 0m s m s s dm s dm m ds m dsα α β β α α β β α α β β+ = + + + = Subtracting the last relation, multiplied by T, from the condition for equilibrium we obtain ( ) ( ) ( ) ( ) 0=−+−+−+− βββααβββαα TdsdhmTdsdhmdmTshdmTsh αα Using ,0V ==− dpTdsdh since p is assumed to be constant, together with the conservation of total mass, the equation reduces to orαh Ts h Ts g gαα β β− = − = β (10.14) which is also identical to Eq.(10.11). Finally, the internal energy U m u m uα α β β= + is used if S and V are constant. Then, by means of Eq.(10.4), one obtains
0dU u dm u dm m du m duα α β β α α β β= + + + =
and the constant constraints are expressed as
v v const or v v v vm m dm dm m d m dα α β β α α β β α α β β+ = + + + = 0 and
const or 0m s m s s dm s dm m ds m dsα α β β α α β β α α β β+ = + + + = Multiplying the two differential relations by respectively, and subtracting them from the condition of equilibrium, we may write
Tp and
( ) ( )v vαu p Ts dm u p Ts dmα α α β β β β+ − + + − = 0
since Taking into account the conservation of total mass, we obtain again that the condition (10.11) for equilibrium holds
v 0.du pd Tds+ − =
v v orα αu p Ts u p Ts g gα α β β β+ − = + − = β (10.15) 10.2. FIRST ORDER PHASE TRANSITIONS In a one component system, a phase transition can be best represented in a pT diagram, as it usually occurs at given p and T. At any point on the equilibrium boundary between two phases α and β, illustrated in Figure 10.1, the specific Gibbs functions must be equal. We may consider two phase transitions at p,T and dTTdpp ++ , respectively, for which we can write
,α α αg g g dg g dgβ β β= + = + so that αdg dgβ= (10.16) Since we have
v , vαdg dp s dT dg dp s dTα α β β β= − = − it follows that v vdp s dT dp s dTα α β β− = − (10.17)
T
p
phase β
phase α
dT
dp
Figure 10.1. Phase equilibrium boundary in a pT diagram Hence, one obtains
(v v )dp s s dT α β α β
− = − or V V (V V S Sdp
dT T α β
−= =− − (10.18) Equation (10.18) where λ is the latent heat transferred during the phase transition is called the Clapeyron equation. It gives the gradient of the phase boundary in a pT diagram and holds for first order phase transitions, in which there are changes of entropy and volume.
