At low temperatures, the heat capacity of solid KCl may be expressed as 0.006 T 3 cal /(deg mol). Calculate the entropy of KCl at 15K, using the third law.
(b)
At 234.3 K, the enthalpy of fusion of mercury is 548.6 cal/mol. Calculate the entropy change for the fusion process.
Solution
(a)
S o for KCl may be taken to be 0 at 0 K.
ds = dq rev = Cp dT at constant pressure\ S = oTCP / T dT
S = S 15 –S 0 = o15 CP / T dT = o15 0.006 T 3 /T dT
= 6.75 cal/(deg.mol)
(b)
Hg(s) = Hg (l)
S = S l – S S = q rev / T =H/ T = 548.6/234.3 = 2.34 cal /(deg mol)
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