The heat capacity, C_P, of liquid carbon disulfide is a relatively constant 78 J
ID: 948744 • Letter: T
Question
The heat capacity, C_P, of liquid carbon disulfide is a relatively constant 78 J/(mol.K). However, the heat opacity of solid carbon disulfide varies greatly with temperature. From 75 K to its melting point at 161 K the heat capacity of solid carbon disulfide increases linearly from 39 J/(mol.K) to 57 J/(mol.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 K). Estimate the absolute entropy of carbon disulfide at 75 K.Explanation / Answer
According to Third Law of Thermodynamics 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 any temperature. For CS,
S(75K) = S(0 75K, s)
and
S°(298, ) = S(0 75 K, s) + S(75 K 161 K, s) + Sfus(s ) + S(161 K 298 K, ) Equation 1
on rearranging the above equation solving for S(0 75 K, s),
S(0 75 K, s) = S°(298, ) – S(75 K 161 K, s) – Sfus(s ) – S(161 K 298 K, )
S(75 K 161 K, s) = (CpdT)/T (75K 161 K)
The information given about Cp in this temperature range can be expressed (in J/mol•K) as
Cp = 39 + 18•[(T – 75)/86] = 23.30+ (18/86)T (check that increases linearly from 39K to 57K)
Then, S(75K 161 K, s) = (CpdT)/T = 23.30 • ln(161/75) + (18/86)(86) = 19.55 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 in the equation 1,we get S(0 75K, s) = 151 – 94.84 = 56.16J/mol·K
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