![]() ![]() We can put one molecule into the left side of the box into the box in M ways. Let's first calculate the entropy of the initial state. ![]() (Small compared to the size of the box, but large compared to the size of an atom, so we don't have to worry about atoms "filling up one of the small volume".) One way to do this is to imagine breaking up the left side of the box into M small volumes. When the partition separating the two halves of the box is removed and the system reaches equilibrium again, how does the new entropy of the gas compare to the energy of the original system? An ideal gas occupies half of the container and the other half is empty. Presenting a sample problemĬonsider an example of an isolated box of volume $2V$ divided into two equal compartments. We choose a model that is simple enough that we can see all the details. Entropy change calculator free#(This example will be of particular interest when we start thinking about free energy.) Now, let's look at a toy model example that lets us see how the microstate counting form of entropy, $S = k_B \ln W$, can also give us insights into what happens spontaneously. Which is zero as expected because a Carnot refrigerator operates reversibly and the second law of thermodynamics states that for any reversible process, the entropy of the universe remains constant.In our analysis of the entropy change in heat flow, we analyzed a toy model to see how the macroscopic form of entropy, $S = Q/T$, gave us insight into what would happen spontaneously. The entropy change of the universe is the sum of the entropy change of the system (the refrigerator in this case) and of its surroundings (the thermal reservoirs): In this page you can see in detail how the entropy change of a thermal reservoir is calculated. The entropy change of the thermal reservoirs is given by:Īs you can see in the previous expressions, Q 1 is positive from the point of view of the hot thermal reservoir (because Q 1 is absorbed by the hot reservoir) and Q 2 is negative from the point of view of the cold thermal reservoir (as Q 2 is removed from it). Entropy is a state function, so its change depends only on the initial and final states of the system, and in a cycle they are the same. The entropy change of the system (the working fluid of the refrigerator) is zero. The first law of Thermodynamics must apply so the energy entering the system has to be equal to the energy leaving it (see upper figure), So: In order to calculate the entropy change of the thermal reservoirs we need to know the amount of heat discarded (Q 1) to the hot thermal reservoir. We will use the expression for the coefficient of performance of a refrigerator to determine the heat removed (Q 2) by the refrigerator from the cold thermal reservoir: Its coefficient of performance is (both temperatures in kelvin):Īfter substituting temperatures T 1 and T 2 we get:Īs you can see, the coefficient of performance of a refrigerator is greater than 1. The most efficient refrigerator operating between two thermal reservoirs is the Carnot refrigerator.
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