For a solid at T KST − S0 = ΔS = ∫TOCpdTT= CP ln T = 2.303 CP log T
True
Yes this statement is true
According to the third law of Thermodynamics, "At absolute zero, the entropy of a perfectly crystalline substance is taken as zero”
(i) For a solid at TK
ST − S0 = ΔS =∫TOCpdTT Cp In T = 2.303 Cp log T
Where ST and S0 are the entropies at TK and 0K respectively and S0 = 0 (according to third law of thermodynamics).
The value of integral can be calculated from the graph of CPT against temperature (T). The area under the curve between 0K and TK gives the value of integral and, thus, the value of S at temperature T.
(ii) For liquids and gases, the absolute entropy at a given temperature T is given by the expression
S=∫TfOCp(s)dTT+ΔHfTf+∫TbTfCp(1)dTT+ΔHvapTb+TTbCp(g)dTT