A phase diagram is drawn for a mixture of two substances at a fixed pressure p with a liquid phase and a gaseous phase (Fig. 6.5). The substances are labelled 1 and 2 and the diagram is shown as a function of the concentration c1 of substance 1. There is a range of temperature for which there is

Question

A phase diagram is drawn for a mixture of two substances at a fixed pressure p with a liquid phase and a gaseous phase (Fig. 6.5). The substances are labelled 1 and 2 and the diagram is shown as a function of the concentration [latex]c_1[/latex] of substance 1. There is a range of temperature for which there is coexistence of two phases. Answer the following questions, treating the concentrations [latex]c^A_1[/latex] and c^B_1[/latex] as given values.
a) Apply the Gibbs phase rule (6.62) to find the number of degrees of freedom when two phases coexist at a fixed pressure p.

f = 2 + m(r − 1) − r (m − 1) = r − m + 2
b) We distill a substance 1 with an initial concentration [latex]c^A_1[/latex] by heating the liquid up to the temperature [latex]T_C[/latex]. Determine the final concentration of substance 1 after distillation.
c) A mixture with a concentration [latex]c^C_1[/latex] of substance 1 is put in a container. The mixture is brought to a temperature [latex]T_C[/latex] while the pressure remains at p. Establish that,

[latex] N_\ell (c^C_1 - c^A_1) = N _g (c^B_1 - c^C_1) [/latex].

where [latex]N_\ell[/latex] the amount of mixture in the liquid phase, and [latex]N_g[/latex] that in the gas phase. This is known as the lever rule.

Accepted Answer

a) According to Gibbs phase rule (6.62) r = 2 and m = 2 imply that f = 2. If the pressure p is fixed then there is only one degree of freedom. If we choose [latex]c_1[/latex] as the variable, then the temperature where the two phases coexist can be read on the phase diagram for each value of [latex]c_1[/latex].

f = 2 + m(r − 1) − r (m − 1) = r − m + 2

b) When the liquid mixture reaches the temperature [latex]T_C[/latex], a vaporisation takes place. After condensation, the concentration of substance 1 is [latex]c^B_1[/latex].

c) The concentration [latex]c^A_1 , c^B_1[/latex] and [latex]c^C_1[/latex] are defined as,

[latex]c^A_1 = \frac {N_{\ell 1}}{N_{\ell }}[/latex] and [latex]c^B_1 = \frac {N_{g 1}}{N_{g }}[/latex] and [latex]c^C_1 = \frac{N_{\ell 1 }+ N_{g1}}{N_{\ell }+ N_{g}} [/latex].

where [latex]N_{\ell} = N_{\ell 1} + N_{\ell 2}[/latex] and [latex]N_g = N_{g1} + N_{g2}[/latex] are the amounts of substances 1 and 2 in the liquid phase [latex]\ell [/latex]and the gaseous phase g. Thus,

[latex](N_\ell + N_g) c^C_1 = N_\ell c^A_1 + N_g c^B_1[/latex].

which implies that

[latex]N_\ell ( c^C_1 - c^A_1 ) = N_g ( c^B_1 - c^C_1) [/latex].

This result would be obtained if we had two ‘weights’ [latex]N_\ell[/latex] and [latex]N_g[/latex] hanging from the ends of a mechanical lever of length AB at equilibrium around its axis at C.

Question 6.12: A phase diagram is drawn for a mixture of two substances at ...

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