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# Illustration 21.4: Engines and Entropy

Please wait for the animation to completely load.

There is a time delay-since the system must be in equilibrium-before the initial change of state occurs.

You must go in order.

**In this animation N = nR** (i.e., k_{B} = 1). This, then, gives the ideal gas law as PV = NT. Restart.

This engine cycle, the Carnot cycle, is considered to be a reversible process because it can run forward or backward. For a reversible process, we define a change in entropy as dS = dQ/T or ΔS = ΔQ/T. (Note that if T changes, some calculus is involved so that ΔS = ∫dQ/T.)

For the Carnot cycle, we can calculate the change in entropy over a cycle by finding the change in entropy for each step. What is the heat input or output divided by the temperature in each step? Note that for the two adiabatic steps, although the temperature changes, the heat input is zero and the process is reversible, so you do not need to use calculus, and ΔQ = 0. Adding the two nonzero terms, you should find that the entropy change is zero for this cycle. The second law of thermodynamics says that to keep the entropy change at zero is the best you can do (the entropy of a cyclic process either increases or remains zero).

Entropy is important for calculations regarding engines because it tells you the best efficiency you can hope for. The efficiency of any engine is the (work out)/(heat in) = |W|/|Q_{H}| where Q_{H} is the heat input from the high temperature reservoir (from whatever heats the gas: burning gasoline, propane, boiling water, etc.). To calculate the efficiency of this engine, take the work done by the gas (698 total for all steps) and divide by the heat absorbed by the engine (2079 during step 1) and we get 0.33.

For an ideal engine (no frictional losses, reversible processes), |W| = |Q_{H}| -|Q_{L}|, and the efficiency is |W|/|Q_{H}| = 1 - |Q_{H}|/|Q_{L}|, where Q_{L} is the heat exhausted to the low temperature reservoir. Since the entropy change of this cycle is zero, this means that Q_{H}/T_{H} + Q_{L}/T_{L} = 0. Thus, for an engine operating between two temperature reservoirs, the maximum efficiency is 1 - |T_{H}|/|T_{L}|.

Entropy, S, is a state variable (does not depend on the process) just like pressure, volume, and temperature unlike work and heat, which do depend on the process. We can thus describe a thermodynamic process by including entropy on a graph. We often describe a process on a TS diagram, because the area under a curve in a TS diagram is the heat. Click here to change the graph from a PV diagram to a TS diagram. (Note that the initial entropy is arbitrary—we are simply interested in the change in entropy.)

Illustration authored by Anne J. Cox.

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