The vapor pressure of an unknown solid is given by ln(P/Torr)=22.413−(2211/T), and the vapor pressure of the liquid phase of the same substance is approximately given by ln(P/Torr)=18.352−(1736/T). Determine the enthalpy of vaporization, the enthalpy of sublimation, and the enthalpy of fusion. Then calculate the triple point temperature and pressure

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Step 1Enthalpy of vaporization and enthalpy of sublimation can be calculated as follows:\begin{array}{l}\ln (P)=22.413-\frac{2211}{T} \\\ln (P)=\text { constant }-\frac{\Delta \mathrm{H}_{\text {sublimation }}}{\mathrm{R}}\end{array}For the given solid,\begin{array}{l}\frac{\Delta \mathrm{H}_{\text {sublimation }}}{\mathrm{R}}=2211 \\\Delta \mathrm{H}_{\text {sublimation }}=2211 \times 8.314 \mathrm{~J} / \mathrm{molK} \\\Delta \mathrm{H}_{\text {sublimation }}=18.38 \times 10^{3} \mathrm{Jmol}^{-1}\end{array}For the liquid.\begin{array}{l}\frac{\Delta H_{\text {vaporization }}}{\mathrm{R}}=1736 \\\Delta \mathrm{H}_{\text {vaporization }}=1736 \times 8.314 \mathrm{~J} / \mathrm{molK} \\\Delta \mathrm{H}_{\text {vaporization }}=14.43 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\end{array}Step 2Enthalpy of fusion can be calculated as\begin{aligned} \Delta \mathrm{H}_{\text {fusion }} & =\Delta \mathrm{H}_{\text {sublimation }}-\Delta \mathrm{H}_{\text {vaporization }} \\ & =18.38 \times 10^{3} \mathrm{Jmol}^{-1}-14.43 \times 10^{3} \mathrm{Jmol}^{-1} \\ & =3.95 \times 10^{3} \mathrm{Jmol}^{-1}\end{aligned}Step 3Triple point temperature and pressure can be calculated as\begin{array}{l}22.413-\frac{2211}{\mathrm{~T}_{\mathrm{tp}}}=18.352-\frac{1736}{\mathrm{~T}_{\mathrm{tp}}} \\ 4.061=\frac{475}{\mathrm{~T}_{\mathrm{tp}}} \\ \mathrm{T}_{\mathrm{tp}}=116.97 \mathrm{~K}\end{array}\begin{array}{l}\ln \frac{P_{t p}}{\text { Torr }}=22.413-\frac{2211}{116.97} \\ \ln \frac{P_{t p}}{\text { Torr }}=3.511 \\ \mathrm{P}_{\mathrm{tp}}=33.48 \text { Torr }\end{array} ...