Question

Please show all steps and explain each step, thanks!

The heat capacity of solid lead oxide is given by

Cp,m = (44.35 + 1.47 x 10^-3 T) in units of J K^-1 mol^-1

Calculate the entropy change that occurs when 1 mole of PbO(s) is cooled from 500 to 300 K.

Answer 1

Given that Cp,m = (44.35 + 1.47 x 10^-3 T)

1 mole of PbO(s) is cooled from 500 K to 300 K .

So, T1 = 500 K

T2 = 300 K

Entropy change = Integral of (nCp/T) dT

= n  Integral of [ 44.35 + 1.47 x 10^-3 T] / TdT

= n  Integral of { [ 44.35/T] +  [1.47 x 10^-3] dT

= n { 44.35 [InT]³⁰⁰₅₀₀ + [1.47 x 10^-3] [T]³⁰⁰₅₀₀

= 1 mol { 44.35 [ In300- In500] + [1.47 x 10^-3] [ 300 -500]

= - 22.94 J/K/mol

Therefore,

entropy change that occurs when 1 mole of PbO(s) is cooled from 500 to 300 K = - 22.94 J/K/mol