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When a Substance is in equillibrium between two phases at a given temperature and pressure, the molar free energy is the same in each phase i.e. \({ G }_{ 1 } = { G }_{ 2 }\).
Consider a pure single component in phase A in equillibrium with another phase B, at a given temperature (T) and pressure (P).
If \({ G }_{ A }\) and \({ G }_{ B }\) are the free energies per mole of the component in two phases A and B respectively. $$ \therefore { G }_{ A } = { G }_{ B } \qquad .... (1) $$ $$ \therefore { G }_{ B } - { G }_{ A } = 0 \ or \ \Delta G = 0 $$ If the temperature is raised from \(T\) to \(T+dT\), to maintain the eqilibrium pressure will change from \(P\) to \(P+dP\) and new free energies will be \({ G }_{ A }+d{ G }_{ A }\) and \({ G }_{ B }+d{ G }_{ B }\). As the system is in new equilibrium free energies are equal, $$ { G }_{ A }+d{ G }_{ A } = { G }_{ B }+d{ G }_{ B } \qquad .... ( 2 ) $$ Since, \({ G }_{ A } = { G }_{ B }\) from equation (1)
Therefore equation (2) becomes, \(d{ G }_{ A } = d{ G }_{ B } \qquad .... ( 3 ) \)
Gibbs free energy is defined as, \( G = H - TS \qquad .... ( 4 )\)
Where H is enthalpy and S is entropy of the system respectively.
Also \(H = E + PV\) where E is internal energy of system.
On substituting this in equation (4) we get, $$ G = E + PV - TS $$ On differentiating this equation $$ dG = dE + \left( PdV + VdP \right) - \left( TdS + SdT \right) $$ $$ dG = dE + PdV + VdP - TdS - SdT \qquad .... ( 5 ) $$ According to first law of thermodynamics heat exchanged dq is , $$ dq = dE + dw \qquad .... \left( 6 \right) $$ also for reversible process \(dq = TdS\) and \(dw = PdV\) Lets put these values in equation (6) $$ \therefore TdS = dE + PdV $$ Hence replacing \(dE + PdV\) in equation (5) we get, $$ dG = TdS + VdP - TdS - SdT $$ or $$ dG = VdP - SdT \qquad .... ( 7 ) $$ This equation gives change of free energy when system undergoes reversible change.
For phase A equation (7) can be, \(d{ G }_{ A } = { V }_{ A }dP - { S }_{ A }dT\) and for phase B it is \( d{ G }_{ B } = { V }_{ B }dP - { S }_{ B }dT\) from equation (3) $$ d{ G }_{ A } = d{ G }_{ B } $$ $$ \therefore { V }_{ A } dP - { S }_{ A } dT = { V }_{ B } dP - { S }_{ B } dT $$ i.e. $$ { S }_{ B }dT - { S }_{ A }dT = { V }_{ B }dP - { V }_{ A }dP $$ i.e. $$ \left( { S }_{ B } - { S }_{ A } \right) dT = \left( { V }_{ B } - { V }_{ A } \right) dP $$ $$ \therefore \frac { dP }{ dT } = \frac { \left( { S }_{ B } - { S }_{ A } \right) }{ \left( { V }_{ B } - { V }_{ A } \right) } $$ or $$ \frac { dP }{ dT } = \frac { \Delta S }{ \Delta V } \qquad .... ( 8 ) $$ If q is the heat exchanged reversibly per mole during the phase transition at temperature T, then change of entropy (\(\Delta S\)) of the system is given by, $$ \Delta S = \frac { { q }_{ rev } }{ T } = \frac { \Delta H }{ T } $$ where \(\Delta H\) is the molar latent heat per mole of the phase change taking place at temperature T. On putting this value in equation (8), we get $$ \boxed { \frac { dP }{ dT } = \frac { { q }_{ rev } }{ T\Delta V } } $$ or $$ \boxed { \frac { dP }{ dT } = \frac { { q }_{ rev } }{ T\left( { V }_{ B } - { V }_{ B } \right) } } $$ or $$ \boxed { \frac { dP }{ dT } = \frac { \Delta H }{ T\left( { V }_{ B } - { V }_{ B } \right) } } $$ This equation is known as \(\underline { Clapeyron Equation }\) .
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