Heat of Vaporization

Heat of Vaporization Definition

Also known as enthalpy of vaporization, the heat of vaporization (∆Hvap) is defined by the amount of enthalpy (heat energy) that is required to transform a liquid substance into a gas or vapor. It is measured in Joules per mole (J/mol), or sometimes in Calories (C). Because enthalpy is always added to a system in order to vaporize a liquid, ∆Hvap always has a positive value.

The required increase in internal energy can be described as the energy needed to break the intermolecular interactions in the liquid. The weaker the bond between atoms, the less energy is needed to break those bonds.

The amount of energy required is a function of the pressure at which the transformation takes place, and is temperature dependent. The hotter the liquid already is, the less energy is required. At higher pressures, more energy is required. There is a critical temperature at which the heat of vaporization vanishes (Tr=1). Past this critical temperature, the substance is distinguishable neither as a liquid nor a vapor. Instead, it becomes known as a supercritical fluid.

In a solution containing both the liquid and gaseous states, the kinetic energy of the vapor is higher that that of the liquid because the particles of a vapor are able to flow more easily. The increased movement in gas particles compared to liquid particles creates heat and pressure.

Heat of Vaporization Equation

A very basic equation to calculate the heat of vaporization is:

ΔHvap = Hvapor – Hliquid

This calculates the difference in internal energy or heat of the vapor phase compared to the liquid phase.

However, this equation does not take into consideration the additional energy needed for the gas particles to push back against atmospheric pressure to allow for the increase in volume when a liquid boils.

Hence, a more complete equation to calculate the heat of vaporization is:

ΔHvap = ΔUvap + pΔV

Where ΔUvap is the difference in internal energy between the vapor phase and the liquid phase (ΔUvap = Hvapor – Hliquid), and pΔV is the word done against the ambient pressure.

Quiz

1. The heat of vaporization always has what kind of value?
A. Neutral
B. Positive
C. Negative

Answer to Question #1
B is correct. The heat of vaporization measures the difference in internal enthalpies between the gaseous state and the liquid state (Hvapor – Hliquid). Because the particles in a gas move around far more than the particles in a liquid, the gas state has a higher value of internal enthalpy than the liquid state (remember: more kinetic energy or movement creates more heat.

2. If a liquid uses 40 Joules of heat to vaporize two moles of liquid, what is the heat of vaporization?
A. 10 J/mol
B. 20 J/mol
C. 30 J/mol
D. 40 J/mol

Answer to Question #2
B is correct. The answer is 20 J/mol because the heat of vaporization is measured per mole, meaning each mole of liquid used 20 Joules of heat for a total of 40 Joules.

3. Heat of vaporization calculates ______________.
A. The amount of heat needed to be released from a substance into the environment.
B. The amount of heat needed to be added to a substance from the environment.
C. The difference in volume between a substance’s liquid state and gaseous state.
D. The difference in kinetic energy between a substance’s liquid state and gaseous state.

Answer to Question #3
B is correct. The heat of vaporization calculates how much heat is needed to turn a liquid substance into a gas. Enthalpy of condensation calculates how much heat needs to be released from a substance to transform it from a gas into a liquid.

References

  • Enthalpy of vaporization. (2017, April 29). Retrieved May 30, 2017, from https://en.wikipedia.org/wiki/Enthalpy_of_vaporization