Question

Which of the following thermodynamic relations is correct?
a) $dG = VdP -SdT$
b) $dE = PdV + TdS$
c) $dH = -VdP + TdS$
d) $dG = VdP + SdT$

Answer 1

Hint : Gibbs Energy is the highest work that a thermodynamic system can operate at a fixed pressure and temperature. The reversible work in thermodynamics indicates a unique method in which work is taken out such that the system persists in perfect equilibrium with all its surroundings.

Complete step-by-step solution:
The relation between Gibbs free energy and enthalpy is given by:
$G = H - TS$
The relation of enthalpy is given by:
$H = E + PV$
Put the value of enthalpy in Gibbs free energy formula:
$G = E + PV - TS$
Differentiate above formula-
$dG = dE + PdV + VdP – TdS - SdT$……($1$)
As we know this relation,
$dq = dE - dW$…...($2$)
And work done is $dW = -PdV$ ..….($3$)
For reversible process,
$TdS = dq$ ……($4$)
Combining ($2$), ($3$) and ($4$)-
$TdS = dE + PdV$
$\implies dE + PdV – TdS = 0$……($5$)
From equation ($1$) and ($5$);
$dG = VdP -SdT$
Option (a) is correct.

Note: The Gibbs free energy estimate is the maximum number of non-expansion work obtained from a thermodynamically closed system. This maximum can be achieved only in a completely reversible manner. When a system changes reversibly from an initial position to a final position, the drop in Gibbs free energy equals the work performed by the system to its surroundings and subtracts the pressure forces' work.