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Making use of the results of Problems 17.8 and 18.7 try to find the connection between the entropy and the thermodynamical probability. |
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Answer» `Q_T = W_T = m/M RT ln (V)/(V_0)` Therefore, the change in the entropy is `DeltaS = S - S_0 = (Q_T)/(T) = m/M R ln (V)/(V_0) = Nk ln V/(V_0)` (1) Here N is the number of molecules. Since the compression of the gas took place at a CONSTANT temperature, the energy of molecular motion does not change, and the change in the entropy is due solely to the change in the volume. Estimate the probabilities of the initial and final states. The mathematical probability of a gas occupying the entire volume `V_0` is UNITY because this is a certainty, hence `w_0 = 1`. The mathematical probability of the gas occupying volume `V lt V_0` was found in Problem 18.7, it is W = `(V//V_0)^N`. The thermodynamic probabilities of macroscopic states are proportional to their mathematical probabilities: `W/(W_0) = w/(w_0) = ((V)/(V_0))^(N)` Taking logarithms, we obtain `ln W/(W_0) = N ln V/(V_0)` (2) COMPARING equalities (1) and (2), we obtain `S- S_0 = k ln W - k ln W_0` whence the relation between the entropy and the thermodynamic probability. |
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