The entropy change is negative which is what we expect when going from a liquid to solid state. b) What is the entropy of the gaseous chloroform at this temperature if the standard entropy of chloroform is 295.6 J/(mol K)? a) Given that the heat of vaporization of chloroform is 29.24 kJ/mol, calculate the entropy change, S of the system when 1.00 mol of CHCl 3 evaporates at its boiling point (61.2 ☌). It is the configuration corresponding to the maximum of entropy at equilibrium. Boltzmann's entropy describes the system when all the accessible microstates are equally likely. The entropy change can be calculated using this equation as long as the temperature remains constant.įor example, Chloroform, CHCl 3, is a common organic solvent once used as an anesthetic. Since is a natural number (1,2,3.), entropy is either zero or positive ( ln 1 0, ln 0 ). Δ S is the entropy change of the system measured in J/K, and T – the temperature in Kelvin Rearranging the equation, we get an expression for Δ S: Remember, also that when the system is at equilibrium the Gibbs free energy is equal to zero: To derive an equation for the entropy change associated with phase transitions, we need to remember that these changes occur when the system is at equilibrium. Remember this pattern, and the corresponding terms for each pair of opposite processes: melting/fusion vs freezing, evaporation/vaporization vs condensation.Ĭalculating the Entropy Change of State Changes These are the points of phase transitions where, for example, the liquid turns into a gas even at the same temperature. 2Freeexpansion Anexamplethathelpselucidatethedi erentde nitionsofentropyisthefreeexpansionofagas fromavolumeV 1toavolumeV 2. The hot body loses entropy Q/TH, the cold one gains entropy Q/TC, so the net entropy increase as a result is Q/TCQ/TH, a positive quantity. Notice how the entropy is still increasing in the regions where the temperature is not changing. The analysis of the processes proceeding in an isolated system. The phase transition graph showing the entropy vs temperature is very useful to visualize this concept: Sine dQsys 0 and p dVsys 0, it is clear from Eq. This behavior is explained by the increasing freedom of motion when molecules go from the most ordered solid state to liquid, and then a gas state where the degree of randomness is the highest. On the basis of determining the entropy change associated with phase transitions is the third law of thermodynamics: the entropy of a perfect crystalline substance is zero at the absolute zero temperature.Īn implication of this is that the entropy is the lowest in solids, and it keeps increasing in the order of going to liquid and gas states: solid < liquid < gas.
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