Читать книгу Geochemistry - William M. White - Страница 48
Example 2.1 Entropy in reversible and irreversible reactions
ОглавлениеAir conditioners work by allowing coolant contained in a closed system of pipes to evaporate in the presence of the air to be cooled, then recondensing the coolant (by compressing it) on the warm or exhaust side of the system. Let us define our “system” as only the coolant in the pipes. The system is closed since it can exchange heat and do work but not exchange mass. Suppose our system is contained in an air conditioner maintaining a room at 20°C or 293 K and exhausting to outside air at 303 K. Let's assume the heat of evaporation of the coolant (the energy required to transform it from liquid to gas) is 1000 joules. During evaporation, the heat absorbed by the coolant, dQ, will be 1000 J. During condensation, –1000 J will be given up by the system. For each cycle, the minimum entropy change during these transformations is easy to calculate from eqn. 2.51:
The minimum net entropy change in this cycle is the sum of the two, or 3.413 – 3.300 = 0.113 J/K. This is a “real” process and irreversible, so the entropy change will be greater than this.
If we performed the evaporation and condensation isothermally at the equilibrium condensation temperature (i.e., reversibly), then this result gives the exact entropy change in each case. In this imaginary reversible reaction, where equilibrium is always maintained, there would be no net entropy change over the cycle. But of course no cooling would be achieved either, so it would be pointless from a practical viewpoint. It is nevertheless useful to assume this sort of reversible reaction for the purposes of thermodynamic calculations, because exact solutions are obtained.