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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 : 32 of 51

Marks: +1, -0

H_{\beta} - H_{\alpha}

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

300\,\mathrm{K}

Solution:

\Delta H_{600} - \Delta H_{300} = 1 \times (C_{p,\beta} - C_{p,\alpha}) (600-300)

Now, at transition temperature,

\Delta H_{600}\; = T \Delta S_{600}

\;= 600 \times (6-5)

\;= 600\,\mathrm{J}\,\mathrm{mol}^{-1}

600 - \; \Delta H_{300} = 1 \times 1 \times 300

\Delta H_{300}\; = 600 - 300

\;= 300\,\mathrm{J}\,\mathrm{mol}^{-1}

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