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102    ENERGY AND THE FIRST LAW OF THERMODYNAMICS


                       How much energy do we require during a distillation?

                      The effect of work on  U: introducing enthalpy H

                      Performing a simple distillation experiment is every chemist’s delight. We gently
                      warm a mixture of liquids, allowing each component to boil off at its own charac-
                      teristic temperature (the ‘boiling temperature’ T (boil) ). Each gaseous component cools
                      and condenses to allow collection. Purification and separation are thereby effected.
                        Although we have looked already at boiling and condensation, until now we have
                      always assumed that no work was done. We now see how invalid this assumption
                      was. A heater located within the distillation apparatus, such as an isomantle, supplies
                      heat energy q to molecules of the liquid. Heating the flask increases the internal
                      energy U of the liquids sufficiently for it to vaporize and thence become a gas.
                        But not all of the heater’s energy q goes into raising U. We need some of it to
                      perform pressure–volume work, since the vapour formed on boiling works to push
                      back the external atmosphere. The difference between the internal energy U and the
                      available energy (the enthalpy) is given by

                                                   H =  U + p V                            (3.13)

                        H is a state function since p, V and U are each state functions. As a state function,
                      the enthalpy is convenient for dealing with systems in which the pressure is constant
                      but the volume is free to change. This way, an enthalpy can be equated with the
                      energy supplied as heat, so q =  H.


                      Worked Example 3.8 A mole of water vaporizes. What is the change in enthalpy,  H?
                      Take pressure as p .
                                      O
                      We have already seen in the previous section that  U =+40.7 kJ per mole of water,
                                                                               3
                      and from SAQ 3.5 the volume of 1 mol of water vapour is 0.031 m per mole of water.
                        Inserting values into Equation (3.13):

                                                       (     5               3    −1  )
                               ∆H =   40700 J mol    +     10  Pa  ×   0.031 m  mol
                                                −1

                                        ∆U                p              ∆V

                      so
                                                                         −1
                                           H = 40 700 J mol −1  + (3100 J mol )
                      and
                                                   H = 45.8kJ mol −1

                        In this example, the difference between  U and  H is about 11 per cent.
                      The magnitude of the difference will increase as the values of  H and  U get
                      smaller.
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