Map a Sapling Learning macmillan learning The heat capacity, CP, of liquid carbo
ID: 484466 • Letter: M
Question
Map a Sapling Learning macmillan learning The heat capacity, CP, of liquid carbon disulfide is a relatively constant 78 Jl(mol. 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. K) to 57 J/(mol. K). The enthalpy of fusion of carbon disulfide is AH 4390 J/mol. fus The absolute entropy of liquid carbon di at 298 K is SE 151 J/(mol. K). Estimate the absolute entropy of carbon disulfide at 85 K. Number J maol K A Previous 3 Give Up & View Solution Check Answer Next ExitExplanation / Answer
we should firstly know that for a system that does not undergo any reactions (e.g., phase transitions):
(S/T)_p = Cp/T
where S is the molar entropy and Cp is the molar constant-pressure heat capacity.
At constant pressure, dS = Cp/T dT
If Cp is independent of temperature, then this is easily integrated to obtain:
S(T) - S(T) = Cpln(T/T)
We can use this equation to calculate the entropy change associated with cooling the material from 298 K to the freezing point at 161 K.
In this case, we are told that Cp is a linear function of T.
Cp(T) = Cp° + b*(T - T°)
In this case,b = (57-41)/(161-85) J/(mol*K^2) = 0.2105 J/(mol*K), Cp° = 41 J/(mol*K) and T° = 85 K
Again,using differential equation,
S = (Cp° + b*(T - T°))/T dT
dS = (Cp°/T - b*(T°/T - 1)) dT
S(T) - S(T) = Cp°*ln(T/T) - b*T°*ln(T/T) + b*(T - T)
We can use this equation to calculate the entropy change associated with cooling the solid carbon disulfide from it's freezing point at 161 K to the temperature of interest, 85 K.
At the melting point, the solid and liquid phases are in equilibrium so G = 0 = Hfusion - T_melting*Sfusion. In this case, we are told that Hfusion = 4,390 J/mol and Tmelting is 161 K, so Sfusion = 27.27 J/(mol*K) .
Now by putting all this together, the entropy of the solid at 85 K is given by: -
S(85 K) = [151 + 78*ln(161/298) - 27.27 + 41*ln(85/161) - 0.2105*(85)*ln(85/161) + 0.2105*(85 - 151)] J/(mol*K)
S(85 K) = 47 J/(mol*K)