(The answer is not 49 or 74.51 J/(mol K) The heat capacity, C_p, of liquid carbo
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(The answer is not 49 or 74.51 J/(mol K)
The heat capacity, C_p, of liquid carbon disulfide is a relatively constant 78 J/(mol middot K). However, the heat capacity of solid carbon disulfide varies greatly with temperature. From 85 K to its melting point at 161 K, the heat capacity of solid carbon disulfide increases linearly from 41 J/(mol middot K) to 57 J/(mol middot K). The enthalpy of fusion of carbon disulfide is Delta H_fus = 4390 J/mol. The absolute entropy of liquid carbon disulfide at 298 K is S = 151 J/(mol middot K). Estimate the absolute entropy of carbon disulfide at 85 K.Explanation / Answer
Just add all entropy values from 0K to 85 K , also the room temperature value must also be included (298.15K)
S161 - S85 = Cpo ln (T2 / T1) - bT ln (T2 / T1) + b (T2 - T1)
T1 = 85 K ; T2 = 161 K ; T = 85 K ; Cpo = 41 J/mol K
b= (Cp2 - Cp1) / (T2 - T1) = [( 57 - 41) J/mol K] / [161 - 85] K = 0.21052 J/mol K2
Sfus = Hfus / Tm = 4390 J/mol / 161 K = 27.26708 J/mol K
therefore,
S85 = S298 + Cp ln (T2 / 298.15) + Cpo ln (T2 / T1) - bT ln (T2 / T1) + b (T2 - T1) - Sfus
S85 = 151 + 78 ln (161 / 298.15) + 41 ln(161 / 85) - 0.21052 ln (161 / 85) + 0.21052 (161-85) - 27.26708
S85 = [151 - 48.063 + 26.18887 - 0.13447 +15.99952 -27.26708 ] J/mol K
S85 = 117.7238 J/mol K
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