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Pressure. Ingredients : A thermodynamic model of a system, Ω (U,V), which describes the number of microstates of the system when isolated at particular values of U and V, (i.e. in a particular macrostate).
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Pressure Ingredients: A thermodynamic model of a system, Ω(U,V), which describes the number of microstates of the system when isolated at particular values of U and V, (i.e. in a particular macrostate). A reservoir of energy is also a system in thermal contact with our system of interest, which is much ‘larger’ – dU/dTR >> dU/dT The weight and piston are a system that is in mechanical contact with our system – it describes the work done (f *dx) by our system in exchanging space with a reservoir of volume at constant pressure. Conservation of Energy: UT=U + pV + UR UR = UT -U - pV Fundamental Postulate When isolated at a particular UT for a sufficiently long time, the most probable values of U and V are those that maximize the function ΩC(U,V). Weight Piston System Reservoir of Energy at TR An isolated composite system ΩC(UT)
Again, for convenience we can define entropy S= log Ω When total U is fixed, the change in the entropy of the composite is given by: (The weight contributes nothing to the change in entropy of the composite) Recalling conservation of energy:
We now have the entropy of the composite as a function of two independent internal variables, U and V: We can therefore apply the fundamental postulate: after a long time the composite will be in the macrostate with the largest number of microstates i.e. highest entropy. We can extract two equilibrium conditions required to satisfy this statement
Recognizing We have Which defines
We now know how to count the states of a system capable of independent variations of energy and volume: • Measure T, P hold V constant and add δU (slowly) • Measure T, P hold U constant and add δV (slowly)
Weight Piston What happens when we begin to consider not just U,V, but also N? If we fix U,N and allow V to vary, at equilibrium we recover the definition of pressure “Entropic elasticity” Just a consequence of N things exploring a volume V – but, the system must be dilute for each molecule to be independent!
‘Osmotic’ pressure Consider a semipermeable rigid membrane with volume V immersed in fluid. Again, we ask about the number of states contributed by N molecules of solute exploring a volume: Here the tendency for water to permeate the membrane is balanced by the pressure exerted by the rigid walls of the membrane on the fluid Osmosis is just greek for ‘a pushing’, so osmotic pressure is the ‘pushing pressure.’ I suspect we hid our ignorance about the phenomena by giving it a greek name! Perhaps ‘entropic’ pressure would be better.
Atmosphere as a reservoir of volume at constant p Two conservation statements: Conservation of Energy: UT=U + UR Conservation of Volume: VT=V+VR Atmosphere at pR Ω(VR) System Reservoir of Energy at TR Equilibrium condition:
Vapor Wine Liquid Cheese We now ask…. How do we study what happens when two systems can exchange matter in addition to space and energy? We consider that the number of states of each system is now a function of three variables: Ω(U,V,N) Again, it is convenient to define the entropy of a system as S = log Ω(U,V,N) An isolated composite system composed of two systems: wine (in two phases, vapor and liquid) and cheese. Consider the cheese wrapped in an impermeable membrane such that only water can move in or out. We can label the two systems A` and A`` for the wine and cheese respectively. For a given partition of energy, volume and water between the wine and cheese, the number of microstates in such a macrostate of the composite is the product of the number of microstates in the corresponding macrostates of the wine and of the cheese….
Vapor Wine Liquid Cheese Therefore the entropy of the composite is just the sum: The variation in the entropy of the composite is given by the sum When the composite is isolated, the quantities of U,V, N in the composite are fixed, and the variations in the cheese are equal and opposite those of the wine:
We have identified three independent variables for the composite We can now apply the fundamental postulate: Any change in the internal variables of the isolated system must increase the entropy of the isolated system. That is, for any combination of we require
When the entropy of the composite is maximized, and the wine and cheese are in their equilibrium configurations and their partial derivatives are equal. We already know some of these: What about the last term? What this says is that when the gain in entropy of the wine for adding water is greater than the loss in entropy of the cheese for losing water, water will flow spontaneously from the cheese to the wine. We can give this tendency a label:
Where we give μthe name ‘chemical potential.’ Referring back to our inequality we see what we really mean – that when pressure and temperature have equilibrated, if water still flows from the cheese to the wine we say the chemical potential of the water in the cheese is higher!
We can now approach the question of how to determine S(U,V,N) experimentally (remember, we need this function because for all our other physics we take it as given!) For small changes in the independent variables, the entropy of the composite changes as: We have discussed how to measure temperature and pressure, and small flows of energy and changes in volume; counting N does not seem to problematic, but how do we measure μ? We will here outline an approach based a nice property of pure substances in a single phase.
The nice property is just this: for a system composed of a pure substance in a single phase, the energy, entropy and volume of the system are all proportional to the number of molecules of the substance. We define: We can then prescribe a re-scaled thermodynamic function s(u,v), the entropy per molecule. For a small variation in u and v the entropy per molecule varies as How can we interpret these partial derivatives? Holding N constant it is easy to see
And we therefore recover our familiar definitions of the partial derivatives We can then obtain the function s(u,v) according to Furthermore, if we know N, we now have our thermodynamic function in hand
We may now derive an expression for the chemical potential based on experimental variables – things we can actually measure! Recall Changing variables and the chain rule unpacks a lot of terms out of that one partial derivative: But this simplifies to just Evidently, for a pure substance in a single phase, chemical potential is the Gibbs free energy per molecule
recalling yields Evidently, for a pure substance of a single phase, the chemical potential is a function of the temperature of and the pressure acting on the system: That’s nice, but the world is dirty, and we need to measure it!
Weight Piston In the vapor phase, if we go back to counting states, immediately we can say a few things: (Ideal part) Ideal gas law For a single species of N, we can define the volume per molecule as and the gas law becomes Recall at constant temperature we find 0 We can define po as the partial pressure in equilibrium with a liquid phase, and define μ(T, po) as zero.
Weight When the vapor phase is in equilibrium with a vapor phase, the pressure in vapor phase is by definition the saturated vapor pressure. Piston We now show that this pressure is a function of the temperature. For a small change in temperature and pressure we find: We may then regard the pressure as the function of the temperature (but we could equally pose the inverse relationship!) Clapeyron equation
We can obtain the form of this functional relationship with by noting that: And define the enthalpy of vaporization ~ the energy in the molecular bonds of the liquid We are almost ready to do an experiment!
In general however, we need to talk about water as one component of a gas composed of k species. If the probability of any one molecule being in a location is again independent of the location of any other molecule, the number of states of such a system is given by : The point is that in a multi-component system, the pressure that the chemical potential of a species depends on is the partial pressure of that species, and not the total system pressure. To recap, if we can find pw, we can measure T and calculate pw* for air in equilibium with a piece of cheese, we can determine the chemical potential of the water in the cheese. How do we find pw ?
For the liquid phase, takes a different form. Tm T The ‘chilled mirror’ chemo-meter We place a mirror backed with peltier plate connected to a current source, and monitor the resulting temperature with a thermocouple. We then lower the temperature of the mirror until it fogs. Saturated vapor pressure is a known function of temperature, so once the temperature at which condensation occurs is known, we know the vapor pressure and, with Ta, can calculate the chemical potential in the cheese. Ta What about the liquid phase?
We cannot count the states contributed per molecule, as the location of each molecule is no longer independent. However, the density variations of water over the changes in pressure typically encountered are very small, and so we can approximate v as independent of the pressure (although it is still a function of temperature). At constant temperature, for pure water Pl Mpa Equilibrating an undersaturated vapor with a liquid results in huge potential tensions p/psat
Res. of V at pR System = water Res. of N at μR Res. of U at TR An explicit view of incompressibility Notice, what matters is not that water transferred from the reservoir to the system is incompressible, but that the system is incompressible!
Using the reservoir chemical potential and pressure as our reference gives By raising the pressure in the fluid we can balance a negative chemical potential and bring it to equilibrium with pure water in the reference state
μ / v p
