The entropy versus temperature plot for phases α and β at 1 bar pressure is given. S T and S 0 are entropies of the phases at temperatures T and 0 K , respectively. The transition temperature for α to β phase change is 600 K and C p , β - C p , α = 1 J mol - 1 K - 1 . Assume C p , β - C p , α is independent of temperature in the range of 200 to 700 K . C p , α and C p , &#

Step-by-step solution

S = S 0 + ∫ C p d T T S α = S 0 + ∫ C p α d T T S β = S 0 + ∫ C p β d T T Now, S β - S α = S 0 + ∫ C p β - C p α d T T Given C p β - C p α = 1 S β - S α = lnT + C at any temperature of T S β - S α T 2 - S β - S α T 1 = lnT 2 - lnT 1 T 2 = 600 K   ,   T 1 = 300   K   and   S β - S α = 1   at   600   K 1 - S β - S α 300 = ln 600   - ln 300 ⇒ S &