Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 405 torr?

• A. 251.9 K
• B. 720.7 K
• C. 924.2 K
• D. 333.2 K

Answer: (D)

Verified step by step guidance

1. Identify the known values: the heat of vaporization (ΔHvap) is 30.72 kJ/mol, the normal boiling point (T1) is 80.1°C, and the normal atmospheric pressure (P1) is 760 torr. The external pressure (P2) is given as 405 torr.
2. Convert the normal boiling point from Celsius to Kelvin by adding 273.15 to the Celsius temperature: T1 = 80.1 + 273.15 K.
3. Use the Clausius-Clapeyron equation to relate the change in pressure to the change in temperature: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1), where R is the ideal gas constant (8.314 J/mol·K).
4. Rearrange the Clausius-Clapeyron equation to solve for T2, the boiling temperature at the new pressure: 1/T2 = 1/T1 - (R/ΔHvap) * ln(P2/P1).
5. Substitute the known values into the rearranged equation and solve for T2. Remember to convert ΔHvap to J/mol by multiplying by 1000, and ensure all units are consistent.