STRUTTURA DELLA MATERIA 1 - 21 Sept 2022 - solutions
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3 components; 9.574 mm
- 128.8 pm; the 186.8 μm line (L=4)
- Define x=ε/(kB T).
(a) CV = N kB x²exp(x)/(1+exp(x))²;
(b) CV ≈ (kB/N) x²exp(x)/(1+exp(x)/N)².
Good agreement for exp(x) >> N,
alias T << ε/(kB log(N)).
For the problem data x=23.209,
(a) CV = 6.192×10-25 J/K;
CV = 6.192×10-21 J/K
(b) CV = 6.191×10-25 J/K;
CV = 1.844×10-21 J/K;
- 742669 Pa; 3.369 mJ/(mol K)
