By Sunil Bhardwaj

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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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