Solution to a physicochemical problem on entropy
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It is known that the saturated vapor pressure of solid benzene at -5°C is 2.28 kPa. When 1 mole of supercooled liquid benzene at -5°C freezes at a pressure of 101.325 kPa, ΔSm = -35.46 J·K-1·mol-1, with 9860 J·mol-1 of heat being released. Find the saturated vapor pressure of liquid benzene at -5°C. Assume benzene vapor to be an ideal gas. The answer is 2.67 kPa. How to solve it?
The normal melting point of benzene is 278.65 K. The solution should follow these steps: Benzene (liquid) at 278.65 K ---------→ Benzene (solid) at 278.65 K ↑ ↓ ↑ ↓ Benzene (liquid) at 268.15 K---------→ Benzene (solid) at 268.15 K
Reply to 3# dalin123: Thank you. This can be used to calculate the enthalpy change of L-S. Can its purpose be to determine the saturated vapor pressure of liquid benzene at 268.15 K? What is the next approach?