All calculatorsGibbs’ Phase Rule Calculator
Compute the thermodynamic degrees of freedom F using Gibbs’ phase rule. Supports independent reactions (r) and common constraints (constant T, constant P). Includes steps and a phase-rule gauge.
Background
The classical (nonreactive) phase rule is F = C − P + 2, where C is the number of components and P the number of phases. With r independent reactions, F = C − P + 2 − r. If temperature is fixed and/or pressure is fixed, each reduces F by 1 (because one or both intensive variables are constrained).
How this calculator works
- General (reactive) form: F = C − P + 2 − r.
- Constraints: If T is fixed, subtract 1; if P is fixed, subtract 1. Both fixed → subtract 2 total.
- Classification: F=0 invariant (e.g., triple point), F=1 univariant (1 d.o.f.), F=2 divariant (2 d.o.f.), F≥3 multivariant.
- Feasibility: If the computed F is negative, the specified combination is not thermodynamically feasible.
Formula & Equation Used
Core rule: F = C − P + 2 − r − δT − δP, where δT, δP ∈ {0,1} for fixed T and fixed P.
Set r = 0 for nonreactive systems. Classical: F = C − P + 2.
Example Problems & Step-by-Step Solutions
Example 1 — 1-component two-phase (nonreactive)
Given C = 1, P = 2, r = 0; no fixed T or P.
F = 1 − 2 + 2 − 0 = 1 (univariant).
Example 2 — Triple point (nonreactive)
Given C = 1, P = 3, r = 0; no fixed T or P.
F = 1 − 3 + 2 − 0 = 0 (invariant).
Example 3 — Binary, two-phase at fixed P
Given C = 2, P = 2, r = 0; P fixed only.
F = 2 − 2 + 2 − 0 − 0 − 1 = 1 (univariant along a line at constant pressure).
Frequently Asked Questions
Q: When should I include reactions (r)?
Include r if there are independent chemical reactions at equilibrium among components. Each independent reaction reduces F by 1.
Q: Why does fixing T or P reduce F?
Because T and P are intensive variables that normally count toward the degrees of freedom. Fixing one removes a free variable.
Q: What does F < 0 mean?
It indicates an infeasible specification — the given combination of C, P, r, and constraints cannot coexist at equilibrium.
