How To Calculate Enthalpy Of Vaporization From Vapor Pressure And Temperature

Clausius-Clapeyron Equation:

\[ \Delta H_{vap} = -R \times \frac{d(\ln P)}{d(1/T)} \]

Enthalpy of Vaporization (ΔH_vap):

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1. What is Enthalpy of Vaporization?

Enthalpy of vaporization (ΔH_vap) is the amount of energy required to convert a liquid into a gas at constant pressure and temperature. It represents the energy needed to overcome intermolecular forces during the phase transition from liquid to vapor.

2. How Does the Calculator Work?

The calculator uses the Clausius-Clapeyron equation:

\[ \Delta H_{vap} = -R \times \frac{d(\ln P)}{d(1/T)} \]

Where:

\( \Delta H_{vap} \) — Enthalpy of vaporization (J/mol)
\( R \) — Gas constant (8.314 J/mol·K)
\( \frac{d(\ln P)}{d(1/T)} \) — Slope of ln P vs 1/T plot (K)
\( P \) — Vapor pressure (Pa)
\( T \) — Temperature (K)

Explanation: The equation relates the temperature dependence of vapor pressure to the enthalpy of vaporization through the slope of ln P versus 1/T.

3. Importance of ΔH_vap Calculation

Details: Calculating enthalpy of vaporization is crucial for understanding phase transitions, designing distillation processes, predicting boiling points at different pressures, and studying intermolecular forces in liquids.

4. Using the Calculator

Tips: Enter the slope obtained from plotting ln P versus 1/T. The slope should be determined from experimental vapor pressure data at different temperatures. A negative slope value will yield a positive ΔH_vap value.

5. Frequently Asked Questions (FAQ)

Q1: Why is the negative sign in the equation?
A: The negative sign accounts for the fact that vapor pressure increases with temperature, resulting in a negative slope for ln P vs 1/T, which when multiplied by -R gives a positive ΔH_vap value.

Q2: What are typical ΔH_vap values?
A: Typical values range from 20-50 kJ/mol for most liquids. Water has a relatively high ΔH_vap of 40.7 kJ/mol due to strong hydrogen bonding.

Q3: How accurate is this method?
A: The method assumes ΔH_vap is constant over the temperature range and the vapor behaves as an ideal gas. Accuracy improves when measurements are taken over small temperature intervals.

Q4: Can this be used for any liquid?
A: The equation works best for pure liquids that don't associate or dissociate. It may be less accurate for associating liquids or near the critical point.

Q5: What if I have vapor pressure data at only two temperatures?
A: For two data points, you can use the integrated form: ΔH_vap = -R × (ln(P₂/P₁))/(1/T₂ - 1/T₁)