lecture 5 - The 1st Law of thermodynamics δq ...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

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.

Ask a homework question - tutors are online