Entropy and the 3rd Law

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Entropy and the 3rd Law

• 9/24/03

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From 1st and 2nd law
What is the T-dependence of S? If we know Cv(T)
we can get DS in an isothermal process
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A similar derivation starting from dH
If we know CP(T) we can get DS in a process at
constant P
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Integrating at constant V or P,we need S(T0)

• T0 0K
• Nernst postulated that DS(rxn) -gt0 as
• T-gt0.
• Planck extended Nernst postulate suggesting that
  S(0K)0. (for a pure substance).
• 3rd Law Every substance has a positive entropy,
  but at 0K, S may become zero, if the substance
  is a perfect crystal.

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From what we learnt last class

• S kB ln W
• W of ways the total energy may be distributed
  among various energy states.
• At 0K, ordered state, W 1, S0.

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Therefore,
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Cp for solids

• For 0 lt T lt 15 K
• Cp (solid)-gt T3 (non metallic systems)
• Cp (solid) -gt a T B T3 as T-gt0
• Integrating we can find S(T) at low T for solids.

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Fig. 7.1 and 7.2
Benzene melting and vaporization
Nitrogen s-s, melting, and vaporization
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Entropies of gases
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In the limit T-gt0

• Since E (excited state) gt E (ground state)

If n Avogadros number, s7.56x10-22 J/(mol K)
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See Table 2

• Sgases gt Sliquids gt Ssolids

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See table 3

• Relate entropy to Cp (vibrational and
  translational degrees of freedom)
• Case of isomers (different degrees of freedom)
  acetone (CH3)2CO vs. trimethylene oxide (CH2)3O
  (cyclic)

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Calculation of So through a thermodynamic cycle
Calculation of So (gas) at 298.15 K
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Heating liquid Br2 to its boiling point
Vaporize
Summing all contributions, s0 (Br2, gas) 245.5
J/(mol K) Alternatively, from Q for the
diatomic gas s0 (Br2, gas) 245.5 J/(mol K)
Cool to 298.15 K
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See Table 4

• Residual entropy (Sexp may not be exactly zero at
  0K)