Answer:
According to Gibbs Helmholtz
equation:
\[\Delta G=\Delta
H-T\Delta S\]
\[\Delta G=\Delta U+P\Delta V-T\Delta
S\,\,\,\,\,\,\,\,\,\,.....(i)\]
From second law of thermodynamics:
\[\Delta S=\frac{q}{T}\]
(Reversible process) ... (ii)
From first law of thermodynamics:
\[\Delta
U=q-{{w}_{rev}}\]
\[{{w}_{rev}}\] = Reversible work done
by the system
From (ii) and (iii)
\[\Delta S=\frac{\Delta
U+{{w}_{rev}}}{T}\]
or \[T\Delta S=\Delta
U+{{w}_{rev}}\]
Putting the value of \[T\Delta S\] in
(i) we get
\[\Delta
G=\Delta U+P\Delta V-[\Delta U+{{w}_{rev}}]\]
\[=P\Delta
V-{{w}_{rev}}\]
\[\mathbf{-\Delta
G=}{{\mathbf{w}}_{\mathbf{rev}}}\mathbf{-P\Delta V}\]
Thus, decrease in Gibbs free
energy is measure of reversible work done by the system exclusive of work of volume
expansion.
From (i). If\[\Delta H=+ve\], and \[\Delta
S=+ve,\] then process will be spontaneous when \[T\Delta S>\Delta H\] ; in
this case \[\Delta G\] will be negative.
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