Calculating Enthalpy Of Vaporization

Clausius-Clapeyron Equation:

\[ \Delta H_v = R \times \frac{T_1 \times T_2}{T_2 - T_1} \times \ln\left(\frac{P_2}{P_1}\right) \]

Gas Constant (R):

J/mol·K

Temperature 1 (T₁):

Temperature 2 (T₂):

Pressure 1 (P₁):

Pressure 2 (P₂):

Enthalpy of Vaporization (ΔH_v):

Unit Converter ▲

Unit Converter ▼

From:	To:

1. What is the Clausius-Clapeyron Equation?

The Clausius-Clapeyron equation describes the relationship between vapor pressure and temperature for a substance. It's used to calculate the enthalpy of vaporization (ΔH_v), which is the energy required to vaporize one mole of a liquid at constant temperature and pressure.

2. How Does the Calculator Work?

The calculator uses the Clausius-Clapeyron equation:

\[ \Delta H_v = R \times \frac{T_1 \times T_2}{T_2 - T_1} \times \ln\left(\frac{P_2}{P_1}\right) \]

Where:

\( R \) — Gas constant (8.314 J/mol·K)
\( T_1, T_2 \) — Temperatures in Kelvin
\( P_1, P_2 \) — Vapor pressures at T₁ and T₂ respectively (in Pascals)
\( \Delta H_v \) — Enthalpy of vaporization (J/mol)

Explanation: The equation assumes constant enthalpy of vaporization and ideal gas behavior over the temperature range considered.

3. Importance of Enthalpy of Vaporization

Details: Enthalpy of vaporization is a fundamental thermodynamic property that indicates the strength of intermolecular forces in a liquid. It's crucial for designing distillation processes, predicting evaporation rates, and understanding phase behavior in chemical engineering applications.

4. Using the Calculator

Tips: Enter the gas constant (typically 8.314 J/mol·K), two different temperatures in Kelvin, and the corresponding vapor pressures in Pascals. Ensure T₂ ≠ T₁ to avoid division by zero.

5. Frequently Asked Questions (FAQ)

Q1: Why must temperatures be in Kelvin?
A: The Kelvin scale is an absolute temperature scale required for thermodynamic calculations where zero represents absolute zero.

Q2: What if I have pressures in different units?
A: Convert all pressures to Pascals (Pa) before calculation. 1 atm = 101325 Pa, 1 mmHg = 133.322 Pa, 1 bar = 100000 Pa.

Q3: How accurate is this calculation?
A: The accuracy depends on how well the assumptions (constant ΔH_v and ideal gas behavior) hold for your substance over the temperature range.

Q4: Can I use this for any substance?
A: This works best for substances that follow the ideal gas law and have relatively constant enthalpy of vaporization over the temperature range.

Q5: What if T₂ equals T₁?
A: The equation requires T₂ ≠ T₁ to avoid division by zero. Use two different temperature points.