Question

Consider four minerals P, Q, R and S in a three-component chemical system (A-B-C) as shown in the figure. For a crossing tie-line relationship, the variance (degrees of freedom) of the equilibrium mineral assemblage at X is .........
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Hint:

Variance in a system with multiple components and phases is calculated using the formula: \( \nu = C - F + 2 \), where \( C \) is the number of components and \( F \) is the number of phases.

Answer 1

Correct Answer: 1

Step 1: Understanding variance in a three-component system.
In a three-component system (A-B-C), the formula for variance (\( \nu \)) is given by: \[ \nu = C - F + 2 \] Where:
- \( C \) is the number of components (3 in this case, A, B, and C),
- \( F \) is the number of phases (in this case, four minerals P, Q, R, and S, so \( F = 4 \)).
Step 2: Applying the formula.
Substituting the values into the formula: \[ \nu = 3 - 4 + 2 = 1 \] Step 3: Conclusion.
Therefore, the variance (degrees of freedom) of the equilibrium mineral assemblage at X is 1.