Boiling Point Calculator
Calculates boiling point at a different pressure using the Clausius-Clapeyron equation.
Calculation Result:
The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. At this temperature, the liquid converts into a vapor.
Factors Affecting Boiling Point:
- Intermolecular Forces (IMFs): Stronger IMFs (like hydrogen bonding, dipole-dipole interactions) require more energy (and thus higher temperature) to overcome, leading to higher boiling points.
- Molecular Weight/Size: For similar types of compounds, larger molecules with greater surface area tend to have stronger London dispersion forces and thus higher boiling points.
- External Pressure: This is the primary factor this calculator considers.
- If external pressure increases, the boiling point increases (more pressure pushing down on the liquid, requires higher vapor pressure, thus higher temperature to boil).
- If external pressure decreases (e.g., at high altitudes), the boiling point decreases. This is why water boils at a lower temperature on a mountain.
- Purity of the Substance: Impurities can alter the boiling point (e.g., boiling point elevation in solutions). This calculator assumes a pure substance.
The normal boiling point is the boiling point at a standard external pressure of 1 atmosphere (atm).
The Clausius-Clapeyron equation describes the relationship between the vapor pressure of a liquid and its temperature. It's particularly useful for estimating the boiling point at one pressure if you know it at another pressure, along with the enthalpy of vaporization (ΔHvap).
A common two-point form of the equation is:
ln(P₂ / P₁) = -ΔHvap / R * (1/T₂ - 1/T₁)
Rearranging to solve for the new boiling point (T₂):
1/T₂ = 1/T₁ - (R * ln(P₂ / P₁)) / ΔHvap
So, T₂ = 1 / (1/T₁ - (R * ln(P₂ / P₁)) / ΔHvap)
P₁
= Initial pressure (e.g., 1 atm)T₁
= Boiling point at P₁ (normal boiling point, in Kelvin)P₂
= New pressureT₂
= Boiling point at P₂ (what we want to find, in Kelvin)ΔHvap
= Enthalpy of vaporization of the substance (in J/mol)R
= Ideal gas constant (8.314 J/(mol·K))
Assumptions for Clausius-Clapeyron:
- ΔHvap is constant over the temperature range (reasonable for small ranges).
- The vapor behaves as an ideal gas.
- The volume of the liquid is negligible compared to the volume of the vapor.
1. Enter Known Values:
- Normal Boiling Point (T₁): Enter the boiling point of the substance at a known pressure (P₁). This is typically the boiling point at 1 atm. Select its unit (°C, K, or °F).
- Pressure at Normal Boiling Point (P₁): Enter the pressure at which T₁ was measured. This is often 1 atm but can be adjusted. Select its unit.
- Enthalpy of Vaporization (ΔHvap): Enter the substance's enthalpy of vaporization. This is the amount of energy needed to vaporize one mole of the liquid at its boiling point. Select its unit (kJ/mol or J/mol).
- New Pressure (P₂): Enter the new external pressure for which you want to calculate the boiling point. Select its unit.
2. Select Desired Output Unit:
- Choose the unit in which you want the new boiling point (T₂) to be displayed (°C, K, or °F).
3. Calculate:
- Click the "Calculate New Boiling Point" button.
4. View Results:
- The calculated boiling point (T₂) at the new pressure (P₂) will be displayed.
- A detailed step-by-step solution will show how the result was obtained, including unit conversions and application of the Clausius-Clapeyron equation.
- If there are errors in your input (e.g., non-numeric values, zero ΔHvap, non-positive temperatures in Kelvin), an error message will appear.
5. Reset:
- Click the "Reset" button to clear all input fields and results.
6. Explore Information:
- Click on the "Theory," "Clausius-Clapeyron Equation," or "Instructions" bars to expand or collapse detailed information.