Clausius Clapeyron Calculator

Vapor Pressure & Heat of Vaporization Solver

Input Parameters

Closed phase transition cell Vaporizing
Final Vapor Pressure (P2)
1.9482 atm
Liquid vapor pressure increases exponentially with heating.
Mathematical Working & Substitutions
ln(P2 / P1) = −(ΔH_vap / R) × (1/T2 − 1/T1)

Understanding the Clausius-Clapeyron Equation

What is the Clausius-Clapeyron Equation?

The Clausius-Clapeyron Equation is a key relationship in chemical thermodynamics. It describes the exponential boundary between two phases of matter (specifically liquid and vapor) by relating vapor pressure directly to absolute temperature and the liquid's enthalpy of vaporization.

The Mathematical Expression

The standard integrated form of the Clausius-Clapeyron equation is written as:

ln(P2 / P1) = −(ΔHvap / R) × (1 / T2 − 1 / T1)

An alternative representation that eliminates the negative sign is:

ln(P2 / P1) = (ΔHvap / R) × (1 / T1 − 1 / T2)

Where the variables represent the following parameters:

Physical Significance & Dynamic Equilibrium

Inside a closed container at a given temperature, a liquid and its vapor exist in **Dynamic Equilibrium**. The rate of liquid molecules escaping into the gas phase (evaporating) is exactly equal to the rate of gas molecules returning to liquid (condensing). The pressure exerted by the gas under these conditions is the **Vapor Pressure**.

As the temperature increases:

  1. Molecules gain kinetic energy, increasing their speeds.
  2. A greater fraction of molecules overcome the intermolecular forces holding them in the liquid phase.
  3. The number of gas molecules increases, which causes the vapor pressure to rise **exponentially** with temperature, as predicted by the Clausius-Clapeyron equation.

Solved Examples

Example 1: Calculating Vapor Pressure of Water

Problem: The normal boiling point of water is 100°C (where its vapor pressure is 1.0 atm). Given its heat of vaporization is 40.7 kJ/mol, calculate its vapor pressure at 120°C.

Solution:

  1. Define variables:
    • P1 = 1.0 atm, T1 = 100°C = 373.15 K
    • T2 = 120°C = 393.15 K
    • ΔHvap = 40.7 kJ/mol = 40,700 J/mol
    • R = 8.314 J mol−1 K−1
  2. Apply equation: ln(P2 / 1.0) = (40700 / 8.314) × (1 / 373.15 − 1 / 393.15)
  3. Calculate reciprocal difference: (1/373.15 − 1/393.15) = 0.0026798 − 0.0025435 = 0.0001363 K−1.
  4. Evaluate product: ln(P2) = 4895.357 × 0.0001363 = 0.6672.
  5. Solve for P2: P2 = e0.6672 = 1.95 atm.
Example 2: Calculating Heat of Vaporization

Problem: Ethanol has a vapor pressure of 40.0 mmHg at 19°C and 100.0 mmHg at 35°C. Estimate its enthalpy of vaporization.

Solution:

  1. Given values:
    • P1 = 40.0 mmHg, T1 = 19°C = 292.15 K
    • P2 = 100.0 mmHg, T2 = 35°C = 308.15 K
  2. Rearrange equation for ΔHvap: ΔHvap = [ R × ln(P2 / P1) ] / (1 / T1 − 1 / T2)
  3. Calculate terms:
    • ln(100.0 / 40.0) = ln(2.5) = 0.9163
    • (1/292.15 − 1/308.15) = 3.423 × 10−3 − 3.245 × 10−3 = 1.78 × 10−4 K−1
  4. Solve: ΔHvap = (8.314 × 0.9163) / (1.78 × 10−4) = 7.618 / (1.78 × 10−4) = 42,797 J/mol = 42.8 kJ/mol.
Example 3: Calculating Boiling Point at Altitude

Problem: At an altitude where atmospheric pressure is 0.80 atm, what is the boiling point of water? (Boiling point is the temperature where vapor pressure equals external pressure. Normal BP of water = 100°C at 1.0 atm; ΔHvap = 40.7 kJ/mol).

Solution:

  1. Given: P1 = 1.0 atm, T1 = 373.15 K, P2 = 0.80 atm.
  2. Apply equation: ln(0.80 / 1.0) = (40700 / 8.314) × (1 / 373.15 − 1 / T2)
  3. −0.2231 = 4895.36 × (0.0026798 − 1 / T2)
  4. Divide by 4895.36: −0.00004557 = 0.0026798 − 1 / T2
  5. Solve for 1/T2: 1/T2 = 0.0026798 + 0.00004557 = 0.0027254 K−1.
  6. T2 = 1 / 0.0027254 = 366.92 K = 93.8°C. (Water boils at a lower temperature at high altitudes).

Frequently Asked Questions

What assumptions are made in the integrated Clausius-Clapeyron equation?
The integrated Clausius-Clapeyron equation makes three major simplifying assumptions:
  • Constant heat of vaporization: It assumes ΔHvap is constant and does not vary with temperature over the studied range.
  • Ideal gas behavior: The vapor is assumed to behave like an ideal gas.
  • Negligible liquid volume: The molar volume of the liquid phase is assumed to be negligible compared to the molar volume of the gas phase.
Why does the boiling point of a liquid decrease at high altitudes?
A liquid boils when its vapor pressure equals the surrounding atmospheric pressure. At high altitudes, atmospheric pressure is lower. Therefore, the liquid needs to be heated less to make its vapor pressure match the lower external pressure, resulting in a lower boiling point.
How does heat of vaporization correlate with intermolecular forces?
The enthalpy of vaporization (ΔHvap) represents the energy required to overcome intermolecular attractions in the liquid phase. Substances with strong intermolecular forces (like water with hydrogen bonds) require more energy to vaporize, resulting in high ΔHvap values and lower vapor pressures at room temperature.
What is the difference between evaporation and boiling?
Evaporation is a surface phenomenon that occurs at any temperature; only molecules at the liquid-gas interface with sufficient kinetic energy escape into the vapor phase. Boiling occurs throughout the entire bulk of the liquid when vapor bubbles form inside the liquid, which happens only at the boiling point temperature.