PHY1039 Properties of Matter Entropy Changes in a Thermodynamic Universe and the Maxwell Equations May 14, 2012 Lectures 21

Thermodynamic Universe: Thermally Isolated System and Reservoirs System Res. 1 Q1Q1 Res. 2 Q2Q2 Res. 3 Q3Q3 Adiabatic Wall Not the same as THE Universe

Entropy Change,  S, in a Thermodynamic Universe  S univ =  S syst +  S res1 +  S res2 + …. Entropy can decrease or increase within the various parts of the thermodynamic universe. For an irreversible process within the universe,  S univ  0. For a reversible process within the universe,  S univ =0. Thus, S goes to a maximum within a thermodynamic universe (i.e., a thermally-isolated system). Implication: In a thermodynamic universe, a higher entropy state must follow a lower entropy state. Entropy gives us the “arrow of time”!

Which Carnot Engine Will Do More Work? T1T1 T2T2 C Q1Q1 Q2Q2 W T1T1 T2T2 C Q1Q1 Q2Q2 W* Q1Q1 T1*T1* T 1 * < T 1 Irreversible heat flow

How Much Less Work is Being Done by the Modified Carnot Engine? As the second (modified) Carnot engine is operating from a reservoir at a lower T, it is less efficient. With the same amount of heat input, it is doing less work. How much less? Simplify: Factor out T 2 : As T 1 * 0. We do more work without the irreversible heat flow.

Irreversible heat flow T1T1 T2T2 C Q1Q1 Q2Q2 W* Q1Q1 T1*T1* T 1 * < T 1  S = 0 Reversible heat flow Entropy Change in the Modified Carnot Engine Entropy change in hot reservoir: Entropy change in auxiliary reservoir: The modified Carnot engine has some additional entropy change because of the auxiliary reservoir.

What is the Difference in  S for the Two Carnot Engines?  S for the original Carnot engine is simply 0, because it only has reversible heat flows.  S for the modified Carnot engine is given as: This equation represents the extra entropy associated with the irreversible process of heat flow from the hot reservoir to the auxiliary reservoir. What can we conclude?

Helmholtz Free Energy, F For any irreversible process in a thermodynamic universe, the energy that is “unavailable” for work is T o  S universe. The potential for work in the universe decreases by T o  S for every irreversible process. The total energy is still conserved, however, as required by the First Law of Thermodynamics. We can define a state function to provide the energy that is “available” (or free) to do work: F = U – TS  F =  U – T  S – S  T F is the Helmholtz free energy.

Maxwell Relations -S P V T -S P V T -S P V T -S P V T