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Thermodynamics: Lecture 3 Chris Glosser February 13, 2001 1 OUTLINE I. Work: Two Types (A) Configuration Work (B) Dissipative Work (C) Work done in an adiabatic process II. Heat (A) Units and theMechanical Equivalence of Heat (B) Heat Capacity (C) Latent Heat III. The First Law IV. Mayer’s Equation V. Enthalpy: Definition VI. More on Enthalpy 2 Work: Two Types There are basically two types of work that you can do on a system. Either the system can return the energy, in which you have a reversable process, or it cannot, in which case we call the process irreverseable. Here are some examples of reversable and irreversable processes: Reverseable: 1. Compressing a piston adiabatically. 1 System Gas Battery Magnet Intensive P (pressure) E(electromotive force) B(magnetic field) − extensive dW V (volume) P dV V (charge) Edq M (Magnetization) BdM 2. Sending current through an (ideal) inductor in a circuit. Irreverseable: 1. Stiring your coffee with a straw. 2. Sending current through a resistor. 2.1 Configuration Work If the work done on a system is reverseable, we call it configuration work. This is because in almost all cases, reverseable processes do something to change the macroscopic configuration of the system. This in general can be undone. Consider the compression of an ideal gas by a piston. Of the system is adiabatically isolated, then by “easing up” on the piston, I allow the gas to return the energy to the piston. We usually think of the configuration work done on an ideal gas, but it could take a variety of forms (see table ??). The most common use is the work done on magnets. We shall examine this in excruciating detail toward the end of the semister. For the time being, we will focus on an ideal gas. Let’s calculate the configuration work for an isothermal process, and an isobaric process (isochoric implies no work is done): Isothermal Configuration Work - Ideal Gas W = nRT Z Vf Vi Vf dV = nrT ln V Vt (1) Isobaric Configuration Work - Ideal Gas W =P Z Vf Vi dV = P (Vf − Vi ) (2) I will ask you to calculate the work done in the case of an inductor in a circuit as a Homework exercise. 2 2.2 Dissipative Work Irrecoverable energy is said to be dissipated. Therefore, an Irreverseable process does dissipative work on a system. An example is work done by stirring a fluid. The work done on such a system by an external torgue τ is: − dW = −τ dθ, (3) Where θ is the angle that the rod moves. It is quite clear that this work is always negative, and therefore the work done by stirring cannot be recovered. The second example is that of a resistor in a circuit. As you all know, the battery “works” on the charge in order to produce a current of charge (q). Recall the expression for the energy dissipation of a resistor over time; − dW = −I 2 R dt. (4) We may re-express this in terms of our state variables, − dW = −IR I dt = −E dq (5) Again, if you reverse the direction of charge flow, you do not recover any energy. You just dissipate more energy. As we shall soon see, all processes have an irreversable component. However, we shall assume that there are processes for which the configuration work is much, much greater than the dissipative work, and that our arguments will hold for these as well. 2.3 Work done in an adiabatic process An adiabatic process is a process in which no heat is added to the system. In such a casr, the first law of thermodynamics tells us that the total work done must be exact, since it is equivalent to the change in internal energy. Wad = Z b a − dW ad (6) In such a case, it makes sense to write down a formula for the work done for, say an Ideal Gas. This is one way to compare our models for the internal energies of the system. 3 3 Heat What if the system is not adiabatic? Well, the work is no longer exact, that’s for sure. Recalling the first law, and noting that the “adiabatic work” is equivalent to the (negative) internal energy of the system, we may write down, for any process; − − − dQ = dW − dW (7) ad. That is: The heat flow into a system is the work done minus the adiabatic work done. This is, of course, equivalent to the first law, which states: The heat supplied is the work done by the system, plus the internal energy of the system This is, of course, just the conservation of energy. 3.1 Units and the Mechanical Equivalence of Heat From the first law, it is evident that heat is a form of energy. Therefore, the fundamental unit of heat, the calorie, should be related to the fundamental unit of (mechanical) energy, the Joule. This is know as the Mechanical Equivalence of Heat 3.2 Heat Capacity Now, let’s turn our discussion to some of the more familiar concepts of heat transport. The heat capacity is defined to be the heat added over a change in temperature. That is; − dQ C= (8) dT clearly, if this expression is to be concrete we must specify something else about the system. Namely, we will “fix” one of the state variables. If we fix P , then refer to the heat capacity at constant pressure CP . If we fix V , then this is the heat capacity at constant volume CV . The heat capacity is extensive, and we can make it intensive by dividing by n and calling the resultant quantity the sppecific heat capacity c: c= − 1 dQ n dT as usual, “specific” quantities are denoted with lower case letters. 4 (9) 3.3 Latent Heat During a phase transition, when the temperature doesn’t change when heat is added or taken away, The definition of the specific heat doesn’t really make any sense. In such cases, we define the latent heat of transformation, L: Q (10) L= m We generally insist that no work is done on the system while the transformation is taking place (hence; dU = Q = Lm, etc.). Hopefully, you remember how to use this from elementary physics classes. When we discuss enthalpy later, I will come up with a more precise definition of the latent heat. 4 The First Law Now, If you haven’t already figured it out, The first law of thermodynamics is eqivalent to the following statements: • Heat is a form of energy. • Energy is conserved. The equation form of this statement is: − − dU = dQ + dW (11) − Where I think of dW as the work done by the system. It is obvious that this is true for reversable processes. For irreversable processes, we will have to wait until I introduce entropy later in teh semister. 4.1 Mayer’s Equation Anyway, why don’t we combine our ideas of heat capacity with the first law, and see where we get. We shall assume that the process in question is reversable. Recall the (intensive) equation of state for an ideal gas: P v = RT (12) Now, assuming that the number of particles is constant, we may write the first law as; − du = dq − P dv (13) 5 . Since the equation of state allows to eliminate one of the variables, we may also write the energy differential as: du = ∂u ∂v dv + T ∂u ∂T dT. (14) v We may eliminate the energy differential and solve for the specific heat: − dq = ∂u ∂v + P dv + T ∂u ∂T dT. (15) v If we divide this by dT , we get the specific heat capacity; c= ∂u ∂v +P T dv + dT ∂u ∂T . (16) v We therefore obtain an alternate expression for the specific heat capacity at constant V ; ∂u cv = . (17) ∂T v In the specific case that we have an Ideal Gas, you can differentiate the equation of state to get; P dv + v dP = R dT. (18) Now we can write Equation 16 in terms of the specific heat capacity at constant volume, and use the previous equation for the ideal gas to eliminate dv in favor of dP ; − dq = (cv + R)dT − vdP (19) at constant pressure, then, we must have that the coefficient dT is the heat capacity at constant pressure: cP = cv + R (20) This is Meyer’s Equation. Be careful, as this relationship only applies for an ideal gas. 4.2 Enthalpy: Definition Enthalpy is basically the energy required to create a system out of nothing in a constant pressure environment. Therefore, it is to be defined as: Enthalpy (H): H = P V + U . 6 Now let’s examine how enthalpy behaves in an environment of constant pressure: − dH = V dP + P dV + dU = dQ + V dP (21) Of course, for constant pressure, we just have that the change of enthalpy equals the heat flow. The enthalpy is related to the heat capacity at constant pressure by the relation; ∂H (22) CP = ∂T P The derivation of this proceeds along the same lines as the derivation of Meyer’s equation, and is left as an exercise. 4.3 More on Enthalpy With our definition, we have a way of expressing the Latent heat in a more concrete manner. It is just the difference in enthalpy between the two states. That is, L = HGas − HLiquid (23) Is the the latent heat of evaporation. This idea is very natural if we consider what enthalpy is. We completely destroy a system, then set up another system in it’s place, with the same state variables, P, V, T . The difference in Enthalpy gives tells us how much energy we must expend to do this, and is therefore the logical equivalent to the latent heat. 7