Unformatted text preview: The 1st Law of thermodynamics δq + δW = dU U is a function of state (while q or W are not). U is called energy (or internal energy) of the system. If δq = 0 this is the standard energy conservation law from the mechanics textbooks. If both δq and δW are 0, we get dU = 0 – the energy is conserved. For a process involving a transition from 1 to 2 we can also write this as: q + W = U(2) – U(1) = ΔU Both q and W would depend on the path of the process while their sum only depends on the initial and final states. Example: in a cyclic process q =
W (since ΔU=0) Constant volume process: the relationships between U(T,V), CV, and q. U = U(V, T). Consider a process at a constant volume: δW =
P dV = 0 so that δq = dU. Therefore Cv = δq/dT = ( ∂U/∂T )V. Here the subscript T indicates that the volume is kept constant when the partial derivative is taken. Generally, we can write: ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ dV + ⎜ dT ⎝ ∂V ⎟ T ⎠ ⎝ ∂T ⎟ V ⎠
(this is just an identity for any function U=U(V,T)). For an ideal gas, U is a function of T only: ( ∂U/∂V )T =0 and dU = CV dT, U = If CV does not depend on temperature (which is generally not true but in some cases it is a reasonable approximation in a limited range of temperatures) then our expression further reduces to U = CV T + constant Constant pressure process: Relationships between enthalpy, heat, and Cp. q = U(2) – U(1) –W = U(2)
U(1) + P(V(2) – V(1)) = H(2) – H(1) where H = U + PV is called enthalpy . Note that we have used W = − P ΔV , which is only true if the pressure is constant. So in a process with a constant P the heat given to the system is equal to the change of its enthalpy H. It is convenient to view the enthalpy as a function of P and T (rather than V and T, which we used for the internal energy – remember than any two out of the three variables P,V, and T will describe the state). Then we can write dH = ( )T dP + ( )P dT )P If pressure is constant, we find dP=0, dH=δq= CP dT , so that Cp = ( Connection between Cv and Cp for an ideal gas. Consider an isobaric process, P=constant. We can write: δq = Cp dT = dU – δW = Cv dT + P dV = Cv dT + d(P V) The last equality follows from the fact that dP = 0 in our process. Using PV = nRT we get Cp dT = Cv dT + nR dT or Cp = Cv + Rn or This equation is valid ONLY for ideal gases. What the 1st law cannot tell us: reversible and irreversible processes Some processes proceed in one direction only. Examples are: 1. A cup of coffee that gets cold. We have q = ΔU <0. Why can’t we just change sign in this equation to get a reverse process? 2. chalk falling on a desk: mgh = ΔU. 3. If a bottle of perfume is opened in the middle of the classroom, the molecules will spread throughout the room. They don’t spontaneously gather inside a small volume. 4. wood + O2 → CO2 + H2O (burning). Why can’t we mix CO2 + H2O to get wood? Yet plants do just that. ...
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This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas.
 Spring '11
 Makarov

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