Determine the vapor pressure (in mm Hg) of a substance at 29°C, whose normal boiling point is 76°C and has a ΔHvap of 38.7 kJ/mol.

• A. 96 mm Hg
• B. 48 mm Hg
• C. 21 mm Hg
• D. 13 mm Hg

Answer: (C)

Verified step by step guidance

1. Identify the known values: the normal boiling point (T1) is 76°C, the vapor pressure at this temperature (P1) is 760 mm Hg (since it's the boiling point), the temperature of interest (T2) is 29°C, and the enthalpy of vaporization (ΔHvap) is 38.7 kJ/mol.
2. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. T1 = 76 + 273.15 K and T2 = 29 + 273.15 K.
3. Use the Clausius-Clapeyron equation to relate the vapor pressures and temperatures: \( \ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \), where R is the ideal gas constant (8.314 J/mol·K).
4. Rearrange the equation to solve for P2, the vapor pressure at 29°C: \( P2 = P1 \times \exp \left( -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \right) \).
5. Substitute the known values into the equation and calculate P2. Remember to convert ΔHvap from kJ/mol to J/mol by multiplying by 1000.