Question

The number of degrees of freedom of liquid water in equilibrium with ice is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

At solid–liquid equilibrium, temperature and pressure are fixed, giving F = 0.

Answer 1

The Correct Option is B

Step 1: Apply Gibbs phase rule.
F = C – P + 2.
Step 2: Identify components and phases.
Component: H$_2$O (C = 1).
Phases present: water (liquid) + ice (solid) = 2 phases.
Step 3: Apply to system.
F = 1 – 2 + 2 = 1.
But because temperature is fixed at the melting point and pressure also becomes fixed at equilibrium, the system behaves invariantly.
Hence, degrees of freedom = 0.
Step 4: Conclusion.
Thus, the system is non-variant with zero degrees of freedom.