Question

The change in Gibbs free energy for vaporisation of a pure substance is

• A. Positive
• B. Negative
• C. Zero
• D. May be positive or negative

Hint:

For any phase change at equilibrium, \(\Delta G=0\). This includes melting, boiling, and sublimation at equilibrium.

Answer 1

The Correct Option is C

For a pure substance undergoing vaporisation at equilibrium, the liquid phase and vapour phase coexist. At equilibrium: \[ G_{\text{liquid}}=G_{\text{vapour}} \] This means there is no net driving force for phase change. The change in Gibbs free energy is: \[ \Delta G=G_{\text{vapour}}-G_{\text{liquid}} \] Since both phases are in equilibrium: \[ G_{\text{vapour}}=G_{\text{liquid}} \] Therefore: \[ \Delta G=0 \] So, for vaporisation of a pure substance at its boiling point under equilibrium condition, the change in Gibbs free energy is zero. Hence, the correct answer is: \[ \text{Zero} \]