Thermodynamics_III--Gibbs (1)

Thermodynamics_III--Gibbs (1) - ENTROPY AND SPONTANEOUS...

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ENTROPY AND SPONTANEOUS PROCESSES Zumdahl
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FOR A PROCESS TO BE SPONTANEOUS, S universe > 0 SECOND LAW OF THERMODYNAMICS—ENTROPY OF THE UNIVERSE IS ALWAYS INCREASING S universe = S system + S surroundings The positional entropy of a gas is given by S = nR ln V EXAMPLE 5 : Show that the expansion of a gas of (say 3 moles of gas from V 1 to V 2 = 2V 1 ) is possible without any influence by the surroundings ( S sur = 0). S sys =nRlnV 2 –nRlnV 1 =nRln(V 2 /V 1 ) = (3 mol ){8.31 J/(mol ּ K)} ln2 = 17J/K Units: J/K S uni = S sys + S sur =17 + 0 = 17 J/K > 0
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FOR A PROCESS TO BE SPONTANEOUS, S universe > 0 If S uni = 0, then the system is at equilibrium and any change is reversible. Example: ice and water mixture at 0ºC and atmospheric pressure. What does it mean if S uni = 0? Ice (10ºC) Æ water (10ºC) S uni = ? (sign) > 0 Zumdahl
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T h T c EXAMPLE 6 : The second law of thermodynamics implies that heat flows spontaneously from hot to cold and not cold to hot. Where T h > T c Let δ q be a small positive quantity of reversible heat flow. For heat flow from hot to cold: S h = - δ q /T h S c = δ q /T c S uni = S h + S c = - δ q /T h + δ q /T c > 0 For heat flow from cold to hot: S h = δ q /T h S c = - δ q /T c S uni = S h + S c = δ q /T h - δ q /T c < 0
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WE WANT TO BE ABLE TO DETERMINE IF A PROCESS IS SPONTANEOUS BY ONLY ANALYZING SYSTEM PROPERTIES ( DIRECTLY CALCULATIING S uni MAY BE DIFFICULT ) Done reversibly and at constant T Gibbs Energy or Free Energy = G = H -TS H is negative for an exothermic process G > 0 spontaneous in reverse direction New state function for system T in K G = H - T S at constant T and P - T S
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Thermodynamics_III--Gibbs (1) - ENTROPY AND SPONTANEOUS...

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