By Sunil Bhardwaj

3572 Views


Gibbs free energy is defined as, $$ G = H - TS \qquad .... ( 1 ) $$ 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 (1) we get, $$ G = E + PV - TS $$ On differentiating this equation $$ dG = dE + \left( PdV + VdP \right) - \left( TdS + SdT \right) $$ $$ dG = dE + PdV + VdP - TdS - SdT \qquad .... ( 2 ) $$ According to first law of thermodynamics heat exchanged dq is , $$ dq = dE + dw \qquad .... ( 3) $$ also for reversible process dq = TdS and dw = PdV Lets put these values in equation (3) $$ \therefore TdS = dE + PdV $$ Hence replacing \(dE + PdV\) in equation (2) we get, $$ dG = TdS + VdP - TdS - SdT or dG = VdP - SdT \qquad .... ( 4) $$ This equation gives change of free energy when system undergoes reversible change.

At constant temperature: \({ dT = 0 } \)

Therefore equation (4) becomes, $$ dG = VdP $$ or $$ { \left( \frac { dG }{ dP } \right) }_{ T } = V $$ On differentiating w.r.t \({ n }_{ i }\) $$ { \frac { \partial }{ \partial { n }_{ i } } \left( \frac { dG }{ dP } \right) }_{ T, N } = \overline { { V }_{ i } } $$ where \(\overline { { V }_{ i } }\) is partial molal volume of component i or $$ \frac { d }{ dP } { \left( \frac { \partial G }{ \partial { n }_{ i } } \right) }_{ T, N } = \overline { { V }_{ i } } $$ or $$ \boxed { { \left( \frac { d{ \mu }_{ i } }{ dP } \right) }_{ T, N } = \overline { { V }_{ i } } } \qquad ....( 5) $$ where \({ \mu }_{ i }\) is the chemical potential and $$ { \mu }_{ i } = \frac { \partial G }{ \partial { n }_{ i } } $$ Therefore eqauation (5) signifies that the rate of change of chemical potential w.r.t. pressure of any component of the system is equal to partial molal volume of that component at constant temperature and composition.

At constant pressure: \(dP = 0\)

Therefore equation (4) becomes, $$ dG = -SdT $$ or $$ { \left( \frac { dG }{ dT } \right) }_{ P } = -S $$ On differentiating w.r.t \({ n }_{ i }\) $$ { \frac { \partial }{ \partial { n }_{ i } } \left( \frac { dG }{ dT } \right) }_{ P, N } = \overline { -S_{ i } } $$ where \(\overline { S_{ i } }\) is partial molal entropy of component i, or $$ \frac { d }{ dT } { \left( \frac { \partial G }{ \partial { n }_{ i } } \right) }_{ P, N } = \overline { -S_{ i } } $$ or $$ \boxed { { \left( \frac { d{ \mu }_{ i } }{ dT } \right) }_{ P, N } = \overline { -S_{ i } } } \qquad.... ( 6) $$ where \({ \mu }_{ i }\) is the chemical potential and $$ { \mu }_{ i } = \frac { \partial G }{ \partial { n }_{ i } } $$ Therefore eqauation (6) signifies that the rate of change of chemical potential w.r.t. temperature of any component of the system is equal to partial molal entropy of that component at constant pressure and composition.