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S is a state variable, so we can evaluate the integral over any convenient path (ie., an isothermal path even if the process was not isothermal).
The Helmholtz free energy F ≡ E - T S, giving
Finally, the Gibbs free energy G ≡ E - T S + P V yields
By holding one variable constant, we can obtain a whole host of relations, such as
Since entropy is associated with the transfer of heat, it is useful to think of an increase in entropy as the distribution of thermal energy into
a larger number of degrees of freedom.
Entropy
dS = dQ / T.
where T is the temperature at which dQ is transferred.
S differs from other state variables in that for any real process, ΔS > 0. This is one statement of the
second law of thermodynamics.
dE = T dS - P dV.
Defining the enthalpy as H ≡ E + P V, it takes the form
dH = T dS + V dP.
Since chemical processes commonly take place at constant pressure, the enthalpy in a chemical reaction indicates whether the reaction is
exothermic (releases heat) or
endothermic (absorbs heat).
dF = -S dT - P dV.
In an isothermal process, the Helmholtz free energy is the maximum amount of energy available to do work.
dG = -S dT + V dP.
The Gibbs free energy can also be written as H - T S or F + P V. The Gibbs free energy is constant during phase transitions
in which both the temperature and pressure are constant.
S = k ln Ω.
A useful analogy: if order is not an observable, Ω would correspond to the number of permutations, and S would correspond to the number of combinations.
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©2011, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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