The entropy versus temperature plot for phases α and β at 1 bar...

$S = {S}_{0} + ∫{C}_{p}\dfrac{d T}{T}$

${S}_{\alpha} = {S}_{0} + {∫}_{}{\left[{C}_{p}\right]}_{\alpha}\dfrac{d T}{T}$

${S}_{\beta} = {S}_{0} + {∫}_{}{\left[{C}_{p}\right]}_{\beta}\dfrac{d T}{T}$

Now,

${S}_{\beta} - {S}_{\alpha} = {S}_{0} + ∫ \left[{\left[{C}_{p}\right]}_{\beta} - {\left[{C}_{p}\right]}_{\alpha}\right]\dfrac{d T}{T}$

Given ${\left[{C}_{p}\right]}_{\beta} - {\left[{C}_{p}\right]}_{\alpha} = 1$

${S}_{\beta} - {S}_{\alpha} = \ln T + C$ at any temperature of $T$

${\left[{S}_{\beta} - {S}_{\alpha}\right]}_{{T}_{2}} - {\left[{S}_{\beta} - {S}_{\alpha}\right]}_{{T}_{1}} = {\ln T}_{2} - {\ln T}_{1}$

${T}_{2} = 600K, {T}_{1} = 300Kand{S}_{\beta} - {S}_{\alpha} = 1at600K$

$1 - {\left[{S}_{\beta} - {S}_{\alpha}\right]}_{300} = \ln 600 - \ln 300 ⇒ {S}_{\beta} - {S}_{\alpha} = 1 - 0 . 69 = 0 . 31$