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JEE Advanced 2023 Paper 2 Solutions

Paragraph for next two questions

The entropy versus temperature plot for phases

\alpha

and

\beta

at 1 bar pressure is given.

S_{T}

and

S_0

are entropies of the phases at temperatures

T

and

0\ \mathrm{K}

, respectively.

The transition temperature for

\alpha

\beta

phase change is

600\ \mathrm{K}

and

C_{p,\beta}-C_{p,\alpha}=1\ \mathrm{J\,mol^{-1}\,K^{-1}}

. Assume

(C_{p,\beta}-C_{p,\alpha})

is independent of temperature in the range of 200 to

700\ \mathrm{K} .\ C_{p,\alpha}

and

C_{p,\beta}

are heat capacities of

\alpha

and

\beta

phases, respectively.

Section: Chemistry

Question : 31 of 51

Marks: +1, -0

S_{\beta} - S_{\alpha}

\mathrm{J} \ \mathrm{mol}^{-1} \ \mathrm{K}^{-1}

300\ \mathrm{K}

\ln 2 = 0.69

S_{\beta} - S_{\alpha} = 0

0\ \mathrm{K}]

Solution:

\Delta S_{600} - \Delta S_{300} = \int\limits_{300}^{600} \; \frac{1 \times (C_{p,\beta} - C_{p,\alpha})\, dT}{T}

= 1 \times 1 \times \left(\ln \; \frac{T_2}{T_1}\right) \;\; \left\{ \begin{array}{l} T_2 = 600\ \mathrm{K} \\ T_1 = 300\ \mathrm{K} \end{array} \right.

1 - \Delta S_{300} = 1 \times 1 \times \ln 2

\Delta S_{300} = 1 - 0.69

\Delta S_{300} = 0.31\ \mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}

$ $

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