23. Freeze drying is a process to remove water from a material by sublimation. It is used to maintain the structure of biological materials for analysis and preserve food by preventing bacterial growth. Use the following information for your calculations in parts (c) and (d) : fisH(H2O)=6.01kJmol1vapH(H2O)=40.68kJmol1Cp(H2O,s)=[0.4564+(1.377101)T]JK1mol1Cp(H2O,g)=[33.61(3.975104)T]JK1mol1 where T is absolute temperature in kelvin. The reference temperature is 298.15K. For all questions involving calculations, write out any equations used in the solution. (a) Sketch the pressure versus temperature phase diagram for pure (4 marks) water. Include: axis labels, phases, the critical point and the triple point. Numbers are not required. (b) Draw and label an arrow on the phase diagram conditions (1 marks) illustrating a sublimation process. (c) Calculate the standard enthalpy change for sublimation (subH) (2 marks) of water at 298.15K. (d) At 260.0K the equilibrium pressure between solid water and water vapour is 123Pa. (i) Calculate the enthalpy change for fusion (fiscH) under (3 marks) these conditions. (ii) Calculate the enthalpy change for vaporisation ( vap H) (3 marks) under these conditions. (iii) Calculate the enthalpy change for sublimation ( sabH) (2 marks) under these conditions. (iv) Calculate the entropy change for sublimation ( wbblS ) under (2 marks) these conditions. (v) Describe how sutaH will change if the temperature and (1 mark) pressure are lower. (vi) Explain why the specific heat capacity of a solid increases (2 marks) with temperature