Clausius Clapeyron Calculator

Clausius-Clapeyron Equation:

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Initial Pressure (P1):

Final Pressure (P2):

Enthalpy (ΔH):

J/mol

Initial Temperature (T1):

Final Temperature (T2):

Equation Balance:

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1. What is the Clausius-Clapeyron Equation?

The Clausius-Clapeyron equation describes the relationship between vapor pressure and temperature for a substance. It's particularly useful for phase transitions and calculating how vapor pressure changes with temperature.

2. How Does the Calculator Work?

The calculator uses the Clausius-Clapeyron equation:

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Where:

\( P_1, P_2 \) — Initial and final vapor pressures (Pa)
\( \Delta H \) — Enthalpy of vaporization (J/mol)
\( R \) — Gas constant (8.314 J/mol·K)
\( T_1, T_2 \) — Initial and final temperatures (K)

Explanation: The equation calculates how vapor pressure changes with temperature for a given substance, assuming constant enthalpy of vaporization.

3. Importance of Clausius-Clapeyron Calculation

Details: This calculation is crucial for understanding phase equilibria, predicting boiling points at different pressures, and designing distillation and evaporation processes in chemical engineering.

4. Using the Calculator

Tips: Enter all values in appropriate units (pressures in Pa, enthalpy in J/mol, temperatures in Kelvin). All values must be positive and non-zero.

5. Frequently Asked Questions (FAQ)

Q1: What assumptions does this equation make?
A: It assumes constant enthalpy of vaporization and ideal gas behavior, which may not hold for all substances or temperature ranges.

Q2: Can this be used for other phase transitions?
A: Yes, with appropriate enthalpy values, it can be adapted for sublimation and other phase changes.

Q3: Why use natural logarithm in the equation?
A: The natural logarithm form provides a linear relationship between ln(P) and 1/T, making it easier to determine enthalpy from experimental data.

Q4: What are typical enthalpy values?
A: Enthalpy of vaporization typically ranges from 20-50 kJ/mol for common liquids at their boiling points.

Q5: How accurate is this equation?
A: It provides good approximations for many substances over moderate temperature ranges, but may deviate for polar molecules or near critical points.