Chem 162 - CH 11 - Vapor Pressure (Clausius-Claperyon Equation)

Definition: Vapor pressure is the pressure of a gas above a liquid.
Conceptual Overview: Explained in relation to intermolecular forces; gaseous water molecules above liquid water determine the vapor pressure.

Heat of Vaporization: The energy required to transform a liquid into a gas; it influences vapor pressure.
Temperature: As the temperature increases, molecular activity increases, leading to a higher vapor pressure since more molecules evaporate into the gaseous phase.

Equation: The formula that relates vapor pressure to temperature:
- [ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) ]
Variables:
- P: Vapor pressure (units can vary, must match for P1 and P2)
- ( \Delta H_{vap} ): Heat of vaporization (in kilojoules per mole)
- R: Universal gas constant; use 8.314 J/(mol K) for energy-related calculations.
- T: Temperature (must be in Kelvin)

Important to convert between joules and kilojoules based on the context of the problem.
Temperature Conversion: Always convert Celsius to Kelvin for calculations-
[ T(K) = T(°C) + 273.15 ]

Given:
- Vapor pressure of diethyl ether at 20.0 °C (P1): 439.8 mmHg
- Heat of vaporization: 29.2 kJ/mol
- Find vapor pressure at 34.0 °C (P2).
Provided Data:
- T1 = 20.0 °C = 293 K
- T2 = 34.0 °C = 307 K
- Convert (29.2 \text{ kJ/mol} = 29200 \text{ J/mol})
Plug values into the Clausius-Clapeyron equation:
[ \ln \left( \frac{P_2}{439.8} \right) = \frac{29200}{8.314} \left( \frac{1}{293} - \frac{1}{307} \right) ]
Calculate the right-hand side to find (P_2)
Resulting P2 ≈ 760 mmHg.

The calculated vapor pressure at 34 °C indicates that this is also the boiling point of ether, as it equals atmospheric pressure (approximately 760 mmHg).
Relates to previous discussions on boiling points: vapor pressure equals atmospheric pressure at the boiling point, confirming that this is where the phase change occurs.