SLec13 - 1) A gas described by the heat capacity C P = n c...

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Lecture 13: Entropy When we have discussed the first law, we found the expressions for the coefficients We can also calculate them from the equation of state and the entropy differentials: Since entropy is a state variable, the terms in dS should fulfill the Euler condition: We can find: - proof of the Joule’s law
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The entropy of perfect crystals approaches zero as T approaches 0 K. We can formulate the change of entropy from T = 0 to 298.15 K as follows: S 0 for 1 mole of an element defines its standard entropy. Example: The Third Law of Thermodynamics
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From the third law, we can calculate entropies associated with reactions. Consider the reaction at P = 1 bar and T = 298.15 K. The change in entropy is (we use standard entropies S 0 ) For a general reaction: we obtain, analogously as for enthalpy: We can also find the change Δ S at different temperatures from the T-derivative Entropy changes in chemical reactions
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Unformatted text preview: 1) A gas described by the heat capacity C P = n c n T n is heated at constant pressure from T 1 to T 2 (try constant volume). Find the change in entropy of the gas. 2) We can find the entropy change when the gas is isothermally expanded (dT=0) 3) For a process with a general path, we can also integrate along the actual path (assume that V(T) is given, P(T) is analogous try) 4) Even if the process is irreversible, you must calculate S of the system along a reversible path (see problem 4.10 p. 180, book). Examples 5) Consider a van der Waals gas that is reversibly and isothermally expanded. new possibility 6) Consider an isothermal reaction (298.15 K): We can see that the reaction is an irreversible and spontaneous process. We also can find S at T=400 K. See briefly also the solution of Exam 1....
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SLec13 - 1) A gas described by the heat capacity C P = n c...

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