Consider 1kg of water heated at a constant pressure. One can measure the volume of the resulting liquid/vapour mixture. It will be noticed that with the initial increase in temperature, the volume does not change significantly. This continues until the water starts to boil. Given that unit mass of water is being heated, the volume being measured is termed the specific volume.

If the results are plotted on a temperature-volume graph, three stages can be seen. Between points A and B - A corresponding to the initial conditions - the temperature rises until the graph reaches point B, where the temperature is denoted by T_{sat} and the water boils. At B, the water is a saturated liquid with a specific volume V_{f}.

Between points B and C, the water converts to a vapour - commonly known as steam - at a constant temperature T_{sat}. During this stage, the water exists as a liquid/vapour mixture, termed a wet vapour. At C, all the water has converted to a vapour; this is known as a saturated vapour, and its specific volume is V_{g}.

From C to D, the temperature and specific volume both increase and the vapour is a superheated vapour.

As the pressure increases, so does the boiling point. Also, points B and C move closer together. Their loci together form what is termed the vapour dome. Eventually, the two points coincide at what is termed the critical point.

It will be remembered that δQ=dU+δW. Given that δW=pdV, δQ=dU+pdV. This gives rise to the equation h=U+pV, where h=enthalpy. For constant pressure, d(pV)=pdV, in which case dh=dU+pdV=δQ. The enthalpy of saturated liquid, h_{f}=U_{f}+pV_{f}. Remember that as we are dealing with unit mass of material, we are dealing with "specific" quantities.

The Specific Latent Heat of Vaporisation (L_{V}) is the amount of heat energy needed to change 1kg of a substance from liquid to gas (or vice versa). This is the amount of energy needed to move from point B to point C on the graph, so L_{V}=Q_{C}-Q_{B}. Because we are concerned with a constant pressure situation, dh=δQ. Therefore, L_{V}=h_{C}-h_{B}. This can also be written as L_{V}=h_{g}-h_{f}.

For a wet vapour, the dryness fraction (or quality; x) is the fraction of wet vapour present as saturated vapour. The enthalpy of wet vapour, to take an example, can be given by h=h_{f}+x(h_{g}-h_{f}), or h=h_{f}+xh_{fg}. Multiplying out the brackets gives h=h_{f}-xh_{f}+xh_{g}, and factorising gives h=(1-x)h_{f}+xh_{g}. Similarly, we can derive V=(1-x)V_{f}+xV_{g} and U=(1-x)U_{f}+xU_{g}.