Introduction

For a pure substance at constant pressure, the correct variation of Gibbs free energy G with temperature T is shown in option (C), where the slopes follow
(∂G/∂T)_P = −S and increase in magnitude from solid to liquid to gas, so that all three straight lines decrease with T and intersect at the melting and boiling points.

Concept: Gibbs Free Energy vs Temperature

For a single phase at constant pressure, the temperature derivative of molar Gibbs free energy is
(∂G/∂T)_P = −S, where S is the molar entropy.

Because entropy is always positive, the slope of G vs T is negative, so G decreases linearly with increasing T for each phase over small temperature ranges.

Typical entropies follow S_solid < S_liquid < S_gas, so the magnitude of the negative slope is smallest for the solid and largest for the gas.

At the melting point and boiling point, the lines for the two phases that coexist intersect (equal G), giving kinks in the overall G(T) curve.

Why Option (C) Is Correct

• All three lines (solid, liquid, gas) have negative slopes, so G decreases with T for every phase, consistent with (∂G/∂T)_P = −S < 0.
• The slopes obey |slope_solid| < |slope_liquid| < |slope_gas|, matching S_solid < S_liquid < S_gas.
• The solid and liquid lines intersect at the melting point, and the liquid and gas lines intersect at the boiling point, which is exactly how the standard G–T diagram for a pure substance looks.

Therefore, option (C) is the only option consistent with fundamental thermodynamic relations for Gibbs free energy versus temperature at constant pressure.

Why Options (A), (B) and (D) Are Wrong

Option (A)

• The gas line has a positive slope, implying (∂G/∂T)_P > 0 and thus negative entropy for the gas, which is impossible for a real phase.
• The relative slopes do not follow the entropy order; the gas should have the most negative slope, not a positive one.

Hence option (A) violates the basic relation between Gibbs free energy, entropy and temperature.

Option (B)

• Both solid and liquid lines appear with positive slopes, again implying negative entropy for these phases, which contradicts the requirement that entropy is positive.
• Because all slopes are positive, G increases with T for every phase, directly contradicting the observed behavior of G–T plots and the equation (∂G/∂T)_P = −S.

Thus option (B) is thermodynamically inconsistent.

Option (D)

• The solid line has a positive slope, which would mean negative entropy for the solid phase, contradicting thermodynamics.
• The gas line has the smallest magnitude of slope instead of the largest, so the entropy ordering S_solid < S_liquid < S_gas is not respected.

Therefore option (D) also fails to represent the correct Gibbs free energy vs temperature behavior.

Short Exam‑Style Summary

• Use (∂G/∂T)_P = −S; slopes are negative and proportional to −S.
• Entropy order gives slope magnitudes: solid (least negative), liquid, gas (most negative).
• Lines for coexisting phases intersect at phase transition temperatures, giving kinks in the composite G(T) curve.
• Only graph (C) satisfies all these conditions, so the correct answer is (C).