Question

Suppose that at a pressure of 1 atm and temperature 300 K, a given material has two possible phases: liquid (L) and solid (S). We start the system with - NL​=5 moles in the liquid phase and - NS​=2 moles in the solid phase. For these values of pressure and temperature, - μL​=5×10−21 J - μS​=9×10−21 J. What is the equilibrium value of NL​ ? (a) 2 moles (b) 5 moles (c) 7 moles (d) 0 moles (e) 1.49 moles

Answer 1

The equilibrium value of
`$N_L$` is 5 moles. This means that at equilibrium, the system will have 5 moles of material in the liquid phase and 2 moles in the solid phase.

At equilibrium, the chemical potentials of the two phases must be equal. This means that the free energy of transferring a mole of material from one phase to the other must be zero. Mathematically, we can express this as:

`N_L \mu_L+N_S \mu_S=\left(N_L+N_S\right) \mu_(a v g)`

where:

`$N_L$` and
`$N_S$` are the number of moles in the liquid and solid phases, respectively

`$\mu_L$` and
`$\mu_S$` are the chemical potentials of the liquid and solid phases, respectively

`$\mu_{\text {avg }}$` is the average chemical potential of the system

We are given that:

`\begin{aligned}& N_L=5 \text { moles } \\& N_S=2 \text { moles } \\& \mu_L=5 * 10^(-21) \mathrm{~J} \\& \mu_S=9 * 10^(-21) \mathrm{~J}\end{aligned}`

We can solve the equation for
`N_L` as follows:

`\begin{aligned}& \mu_{\text {avg }}=(N_L \mu_L+N_S \mu_S)/(N_L+N_S) \\& \mu_{\text {avg }}=\frac{\left(5 * 10^(-21) \mathrm{~J}\right)(5)+\left(9 * 10^(-21) \mathrm{~J}\right)(2)}{5+2} \\& \mu_{\text {avg }}=6.67 * 10^(-21) \mathrm{~J}\end{aligned}`

`\begin{aligned}& N_L=(\mu_(a v g)-\mu_S)/(\mu_L-\mu_S)\left(N_L+N_S\right) \\& N_L=\frac{\left(6.67 * 10^(-21) \mathrm{~J}\right)-\left(9 * 10^(-21) \mathrm{~J}\right)}{\left(5 * 10^(-21) \mathrm{~J}\right)-\left(9 * 10^(-21) \mathrm{~J}\right)}(7) \\& N_L=5 \text { moles }\end{aligned}`