Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (pi,Vi) to (pf,Vf). With the help of a pV plot compare the work done in the above case with that carried out against a constant pressure pf.


Gibbs free energy is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.
Mathematically, this results may be derived as follows
The relationship between heat absorbed by a system q, the change in its internal energy, ∆U and the work done by the system is given by the equation of the first law of thermodynamics, therefore,
q=∆U+Wexpansion+Wnon-expansion........(i)
Under constant pressure condition, the expansion work is given by p∆V,
∴q=∆U+p∆V+Wnon-expansion(∵∆U+p∆V=∆H)
=∆H+Wnon-expansion...............(ii)
For a reversible change taking place at constant temperature,
∆S=qrevTorqrev=T∆S...........(iii)
Substituting the value of q from Eq. (iii) in Eq. (ii), we get
T∆S=∆H+Wnon-expression
or∆H-T∆S=-Wnon-expression.......(iv)
For a change taking place under conditions of constant temperature and pressure,
∆G=∆H-T∆S
Substituting this value in equation (iv), we get
∆G=-Wnon-expansion............(v)
Thus, free energy change can be taken as a measure of work other than the work of expansion. For most changes, the work expansion can not be converted to other useful work, whereas the non-expansion work is convertible to useful work.
Rearranging equation (v), it may write as
-∆G=Wnon-expansion=Wuseful
As -∆G=Wuseful therefore, ∆G has the same units as those of work i.e., joule
∆G=∆H-T∆S
If ∆H=positive and ∆S=positive, then
∆G will be negative i.e., process will be spontaneous only when T∆S>∆H in magnitude, which will be so when temperature is high.