Discuss the variation of free energy change with Temperature and Pressure.

The variation of free energy change with change in temperature and pressure is discussed below:

Consider the following equation:
G = H – TS …………………………………….……. (1)
As H = U + PV
Substituting the value of H in equation (1) we will get:
G = U + PV – TS …………………………………… (2)

Now differentiating the equation (2), we will get:
dG = dU + pdV + VdP – TdS – TdS – SdT ……………………………… (3)

The first law equation for an infinitesimal change may be given as:
dq = dU – dw

If the work done is only due to expansion, then –dw = PdV
Therefore, dq = dU + PdV
For a reversible process,
dS = dq / T
TdS = dq = dU + PdV …………………… (4)

Combining equations (3) and (4) we will get:
dG = VdP – SdT …………………… (5)

This equation gives the change of free energy when a system changes reversibly i.e. a change in pressure as well as in temperature.

At constant pressure i.e. dP = 0, then equation (5) will be as:
dG = – SdT …………….. (6)
Or (∂G / ∂T) _P = -S
At constant temperature i.e. dT = 0, then equation (5) will be as:
dG = VdP ……………….. (7)
Or (∂G / ∂P) _T = V

Let the free energy of a system in the initial state be G₁ and in the final state be G₂ when there is change in pressure take place but at constant temperature. Then integrating the equation (7), the free energy change ∆G will be given as:

∆G = G₂ – G₁ = (P²∫ P₁) VdP

Where P₁ and P₂ are initial and final pressures, respectively.

For one mole of an ideal gas, PV = RT
Therefore,
∆G = G₂ – G₁ = RT (P²∫ P₁) dP / P
= RT ln (P²/ P₁) = RT ln (V¹ / V₂)
Where V₁ and V₂ are initial and final volumes, respectively.

For n moles of the gas:
∆G = G₂ – G₁
= nRT ln (P²/ P₁) = nRT ln (V¹ / V₂)