The heat capacity, CP, of liquid carbon disulfide is a relatively constant 78 J/

Question

The heat capacity, CP, of liquid carbon disulfide is a relatively constant 78 J/(mol·K). However, the heat capacity of solid carbon disulfide varies greatly with temperature. From 89 K to its melting point at 161 K, the heat capacity of solid carbon disulfide increases linearly from 42 J/(mol·K) to 57 J/(mol·K). The enthalpy of fusion of carbon disulfide is Hfus = 4390 J/mol. The absolute entropy of liquid carbon disulfide at 298 K is S = 151 J/(mol·K). Estimate the absolute entropy of carbon disulfide at 89 K.

Explanation / Answer

The Third Law of Thermodynamics tells us that as T 0 K, S 0. Therefore, for any substance, the absolute entropy at any temperature can be obtained by summing up all the entropy changes from 0 K to to that temperature. For CS,

S(89 K) = S(0 89 K, s)
and
S°(298, ) = S(0 89 K, s) + S(89 K 161 K, s) + Sfus(s ) + S(161 K 298 K, )
or, solving for S(0 89 K, s),
S(0 89 K, s) = S°(298, ) – S(89 K 161 K, s) – Sfus(s ) – S(161 K 298 K, )

S(89 K 161 K, s) = (CpdT)/T (89 K 161 K)
The information given about Cp in this temperature range can be expressed (in J/mol•K) as
Cp = 42 + 14•[(T – 89)/72] = 23.358 + (14/72)T (check that increases linearly from 42 to 57)
Then, S(89 K 161 K, s) = (CpdT)/T = 23.358 • ln(161/89) + (14/72)(72) = 13.90 J/mol•K

Sfus = Hfus/Tm,
Sfus(s ) =(4390 J/mol)/(161 K) = 27.27 J/mol•K

S(161 K 298 K, ) = (CpdT)/T, but Cp is constant so
S(161 K 298 K, ) = (78 J/mol·K) • ln(298/161) = 48.02 J/mol·K

Adding up all the terms, S(0 94 K, s) = 298 – 101.86 = 49 J/mol·K

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