(Solved): 1. The Clausius-Clapeyron relation for the slope of a phase boundary in the PT plane of a si ...

1. The Clausius-Clapeyron relation for the slope of a phase boundary in the $P?T$ plane of a simple substance is given by $\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{?S}{?V}$ where $P,T,S$ and $V$ have their usual meanings. (a) Sketch the P-T phase diagram of a simple pure substance. Identify all phases, the triple point, and the critical point on the diagram. [ 4 ] (b) At atmospheric pressure, the fusion temperature of mercury is $?38.8^{?}\mathrm{C}$. Use the Clausius-Claperyon relation to find the pressure required to solidify mercury at room temperature, $T=20^{?}\mathrm{C}$. The latent heat of fusion of mercury is $L_f=2.30\mathrm{kJ}/\mathrm{mol}$. The densities of liquid and solid mercury are ${?}_l=13.7\mathrm{g}/{\mathrm{cm}}^3$ and ${?}_s=14.2\mathrm{g}/{\mathrm{cm}}^3$ respectively. Assume that $L,{?}_l$ and ${?}_s$ are constant over the relevant range. [ 6 ] (c) For a gas-liquid transition at low pressure, a common approximation is to neglect the volume of the liquid phase compared to the vapour. Using this approximation show that the Clausius-Clapeyron equation can be written as $\frac{dP}{dT}=\frac{L_{e}P}{R{T}^2}$ where $L_e$ is the molar latent heat of evaporation. Assuming that $L_e$ is independent of temperature, show that $P(T)=P_0\exp \left[\frac{L_e}{R}\left(T_0^{?1}?T^{?1}\right)\right].$ (d) The vapour pressure of mercury at $20^{?}\mathrm{C}$ is $160\mathrm{mPa}$. Use the result derived in (c) to find its vapour pressure at $50^{?}\mathrm{C}$ given that the latent heat of evaporation is $L_e=59.3\mathrm{kJ}{\mathrm{mol}}^{?1}$. Discuss briefly whether this change has a significant effect on the reading of mercury barometers. [ 3 ]