JEE Advanced 2023 · Paper 2 · numerical answer

JEE Advanced 2023 Chemistry question on Chemical Thermodynamics

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The entropy versus temperature plot for phases α and β at 1 bar pressure is given. STS_{\text{T}} and S0S_{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 Cp,β−Cp,α=1C_{\text{p},\beta} - C_{\text{p},\alpha} = 1 J mol−1^{-1} K−1^{-1}. Assume (Cp,β−Cp,α)(C_{\text{p},\beta} - C_{\text{p},\alpha}) is independent of temperature in the range of 200 to 700 K. Cp,αC_{\text{p},\alpha} and Cp,βC_{\text{p},\beta} are heat capacities of α and β phases, respectively.

The value of entropy change, Sβ−SαS_{\beta} - S_{\alpha} (in J mol−1^{-1} K−1^{-1}), at 300 K is ___.

[Use: ln 2 = 0.69
Given: Sβ−Sα=0S_{\beta} - S_{\alpha} = 0 at 0 K]
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Answer: 0.31

[JEE Advanced 2023 · Paper 2]

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