2.57 is telling us that although the heat gained or lost in the transformation from state 1 to state 2 will depend on the path taken, for a reversible reaction the ratio of heat gained or lost to temperature will always be the same, regardless of path.
If we return to our example of the combustion of gasoline above, the second law also formalizes our experience that we cannot build a 100% efficient engine: the transformation from state 1 to state 2 cannot be made in such a way that all energy is extracted as work; some heat must be given up as well. In this sense, the second law necessitates the automobile radiator.
Where P–V work is the only work of interest, we can combine the first and second laws as:
The implication of this equation is that if equilibrium is approached at prescribed S and V, the energy of the system is minimized. For the specific situation of a reversible reaction
where dS = dQ/T, this becomes
(2.58)
This expresses energy in terms of its natural or characteristic variables, S and V. The characteristic variables of a function are those that give the simplest form of the exact differential. Since neither T nor P may have negative values, we can see from this equation that energy will always increase with increasing entropy (at constant volume) and that energy will decrease with increasing volume (at constant entropy). This equation also relates all the primary state variables of thermodynamics, U, S, T, P, and V. For this reason, it is sometimes called the fundamental equation of thermodynamics. We will introduce several other state variables derived from these five, but these will be simply a convenience.
By definition, an adiabatic system is one where dQ = 0. Since dQrev/T = dSrev (eqn. 2.52), it follows that for a reversible process, an adiabatic change is one carried out at constant entropy, or in other words, an isoentropic change. For adiabatic expansion or compression, therefore, dU = –PdV.
Example 2.1 Entropy in reversible and irreversible reactions
Air 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.
2.7 ENTHALPY
We have now introduced all the fundamental variables of thermodynamics, T, S, U, P and V. Everything else can be developed and derived from these functions. Thermodynamicists have found it convenient to define several other state functions, the first of which is called enthalpy. Enthalpy is a composite function and is the sum of the internal energy plus the product PV:
(2.59)
As is the case for most thermodynamic functions, it is enthalpy changes rather than absolute enthalpy that are most often of interest. For a system going from state 1 to state 2, the enthalpy change is:
(2.60)
The first law states:
so:
If pressure is constant, then:
(2.61)
(we use the subscript P in ΔQP to remind us that pressure is constant). If the change takes place at constant pressure and P–V work is the only work done by the system, then the last two terms cancel and enthalpy is simply equal to the heat gained or lost by the system:
or in differential form:
(2.62)
H is a state function because it is defined in terms of state functions U, P, and V. Because enthalpy is a state function, dQ must also be a state function under the conditions of constant pressure and the only work done being P–V work.
More generally, the enthalpy change of a system may be expressed as:
or at constant pressure as:
(2.63)
In terms of its characteristic variables, it may also be expressed as:
(2.64)
From this it can be shown that H will be at a minimum at equilibrium when S and P are prescribed:
(2.65)
