Derivation of the Clapeyron Eq. Please recall that earlier this semester, we began with the Clapeyron Eq. andderived the Clausius-Clapeyron Eq. At the time we said we said that we would derive the Clapeyron Eq. fromFirst Principles. a. Please write the 4th Fundamental Property Relation for a single phase, pure substance. b. Now, please consider two phases of a single, pure substance that are in equilibrium with each other. We will designate the phases a-phase and ß-phase. Although the system is 2-phase, each phase is single phase. Therefore, we can write the 4th Fundamental Property Relation for each phase. \text { c. Equilibrium means that } T=T \text { sat and } P=P_{\text {sat }} \text {. } \text { d. 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} . e. Divide through by dTsat and collect terms by volume and entropy. f. Let's consider the transition from a-phase and ß-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 \Delta \bar{s}^{\alpha \beta}} With analgebraic rearrangement, please show that1

Write the 2nd Fundamental Property Relationship and integrate from the a-phase to the p-phase. Recall that for a phase change T = Tsat = constant and dP = 0. From the integration, solve for A5%³.