ρ V l V g ρ l V l ρ g V g ρ l ρ V l ρ ρ g V g V l V g ρ ρ g ρ l ρ 21 Hamza J

# Ρ v l v g ρ l v l ρ g v g ρ l ρ v l ρ ρ g v g

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ρ ( ¯ V ( l ) + ¯ V ( g ) ) = ρ ( l ) ¯ V ( l ) + ρ ( g ) ¯ V ( g ) ( ρ ( l ) ρ ) ¯ V ( l ) = ( ρ ρ ( g ) ) ¯ V ( g ) ¯ V ( l ) ¯ V ( g ) = ρ ρ ( g ) ρ ( l ) ρ (2.1)

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Hamza J. Magnier LIQUID SOLID PRESSURE TRIPLE POINT CRITICAL POINT TEMPERATURE VAPOR Figure 2.1: Pressure-temperature diagram for a general one-component system. where ρ = N/ ( ¯ V ( l ) + ¯ V ( g ) ) is the overall density of the system. Equation (2.1) is known as the lever rule. It allows us to compute the relative amounts of the two-coexisting phases. The dashed-line represents the triple point. Anywhere along the dashed-line, the vapor, liquid, and solid phases can simultaneously exist. 2.3 Conditions for phase equilibrium Now let’s derive the mathematical conditions for equilibrium between two, coexisting phases. We consider an isolated system that is separated into two phases, which we label A and B . The volume occupied by each phase can change; in addition, both phases can freely exchange energy and material with each other. Because the system is isolated, the total energy ¯ U , the total volume ¯ V , and the total number of moles N in the system must remain constant. This leads to the following relations: ¯ U ( A ) + ¯ U ( B ) = ¯ U = δ ¯ U ( A ) = δ ¯ U ( B ) (2.2) ¯ V ( A ) + ¯ V ( B ) = ¯ V = δ ¯ V ( A ) = δ ¯ V ( B ) (2.3) N ( A ) + N ( B ) = N = δ N ( A ) = δ N ( B ) (2.4) where ¯ U ( i ) is the total energy of phase i , ¯ V ( i ) is the total volume of phase i , and N ( i ) is the total number of moles in phase i . The total entropy of an isolated system at equilibrium is maximized. Thus, any displacement from equilibrium must correspond to a decrease in entropy δ ¯ S 0 (2.5)
Hamza J. Magnier Figure 2.2: Temperature-density diagram for a general pure substance. The change in the total entropy of the system is given by the sum of the entropy change in phase A and the entropy change in phase B . That is δ ¯ S = δ ¯ S ( A ) + δ ¯ S ( B ) = 1 T ( A ) δ ¯ U ( A ) + p ( A ) T ( A ) δ ¯ V ( A ) μ ( A ) T ( A ) δ N ( A ) + 1 T ( B ) δ ¯ U ( B ) + p ( B ) T ( B ) δ ¯ V ( B ) μ ( B ) T ( B ) δ N ( B ) (2.6) Substituting the relations given in Eq. (2.4) into Eq. (2.6), we find δ ¯ S = 1 T ( A ) δ ¯ U ( A ) + p ( A ) T ( A ) δ ¯ V ( A ) μ ( A ) T ( A ) δ N ( A ) 1 T ( B ) δ ¯ U ( A ) p ( B ) T ( B ) δ ¯ V ( A ) + μ ( B ) T ( B ) δ N ( A ) 0 µ 1 T ( A ) 1 T ( B ) δ ¯ U ( A ) + µ p ( A ) T ( A ) p ( B ) T ( B ) δ ¯ V ( A ) µ μ ( A ) T ( A ) μ ( B ) T ( B ) δ N ( A ) (2.7) The quantities δ ¯ U ( A ) , δ ¯ V ( A ) , and δ N ( A ) on the right-hand side of Eq. (2.7) can be chosen arbitrarily.

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• Fall '19
• Salvatore Ziccone

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