Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0℃. The normal boiling point of isopropanol is 82.3℃ and the heat of vaporization (ΔH_vap) is 39.9 kJ/mol.

• A. 1.35 × 10⁻³ atm
• B. 5.67 × 10⁻² atm
• C. 9.97 × 10⁻¹ atm
• D. 17.6 atm
• E. 43.1 atm

Answer: (B)

Verified step by step guidance

1. Identify the given values: the normal boiling point of isopropanol (T1 = 82.3°C), the heat of vaporization (ΔHvap = 39.9 kJ/mol), and the temperature at which you need to find the vapor pressure (T2 = 20.0°C).
2. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. This gives T1 = 355.45 K and T2 = 293.15 K.
3. Use the Clausius-Clapeyron equation to relate the vapor pressures at two different temperatures: \( \ln \left( \frac{P_2}{P_1} \right) = \frac{-\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \), where R is the ideal gas constant (8.314 J/mol·K).
4. Rearrange the equation to solve for \( P_2 \), the vapor pressure at 20.0°C: \( P_2 = P_1 \cdot \exp \left( \frac{-\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \right) \).
5. Assume the vapor pressure at the boiling point (P1) is 1 atm, as it is the normal boiling point. Substitute the values into the equation and solve for P2.