1. Derivation of the Clapeyron Eq. Please recall that earlier this semester, we began with the Clapeyron Eq. and derived the Clausius-Clapeyron Eq. At the time we said we said that we would derive the Clapeyron Eq. from First Principles. ». Now, please consider two phases of a single, pure substance that are in equilibrium with each other. Wewill designate the phases a-phase and p-phase. Although the system is 2-phase, each phase is singlephase. Therefore, we can write the 4th Fundamental Property Relation for each phase. \text { Equilibrium means that } T=T^{\text {sat }} \text { and } P=P^{\text {sat }} \text {, } \text { Recall that at in equilibrium } d \bar{g}^{\alpha}=d \bar{g}^{\beta} \text { which leads to } \bar{v}^{\alpha} d P^{s a t}-\bar{s}^{\alpha} d T^{s a t}=\bar{v}^{\beta} d P^{s a t}-\bar{s}^{\beta} d T^{s a t} Divide through by dTsat and collect terms by volume and entropy. Let's consider the transition from a-phase and p-phase, that is the "aß" phase transition. (We could just as easily consider the "Ba" phase transition. The result would be the same.) Please show that \underbrace{\left(\bar{v}^{\alpha}-\bar{v}^{\beta}\right)}_{=-\Delta \bar{v}^{\alpha \beta}} \frac{d P^{s a t}}{d T^{s a t}}=\underbrace{\bar{s}^{\alpha}-\bar{s}^{\beta}}_{=-\Delta \bar{s}^{\alpha \beta}} . With an algebraic rearrangement, please show that \frac{d P^{s a t}}{d T^{s a t}}=\frac{\Delta \bar{s}^{\alpha \beta}}{\Delta \bar{v}^{\alpha \beta}} We will come back to this. h. Write the 2nd Fundamental Property Relationship and integrate from the a-phase to the B-phase. Recall that for a phase change T ' = Tsat = constant and dP = 0. From the integration, solve for A5%³.R \text { i. Substitute the result for } \Delta \bar{s}^{\alpha \beta} \text { in Part } h \text { into Part g to show that } \frac{d P^{s a t}}{d T}=\frac{\Delta \bar{h}^{\alpha \beta}}{T \Delta \bar{v}^{\alpha \beta}} a. Please write the 4th Fundamental Property Relation for a single phase, pure substance.