The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapours are $4.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$ and $2.0 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$, heat of liquid fusion and vapourisation of water are $334 \mathrm{~kJ} \mathrm{~kg}^{-1}$ and $2491 \mathrm{~kJ} \mathrm{~kg}^{-1}$, respectively) ( $\log 273=2.436, \log 373=2.572, \log 383=2.583$ )

(D) $9.26 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$

(A) $7.90 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$

(B) $2.64 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$

(C) $8.49 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Chemical Thermodynamics

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapours are

4.2 kJ K^{- 1} kg^{- 1}

and

2.0 kJ K^{- 1} kg^{- 1}

, heat of liquid fusion and vapourisation of water are

334 kJ kg^{- 1}

and

2491 kJ kg^{- 1}

, respectively) (

lo g 273 = 2.436, lo g 373 = 2.572, lo g 383 = 2.583

)

(A) $7.90 kJ K^{- 1} kg^{- 1}$
(B) $2.64 kJ K^{- 1} kg^{- 1}$
(C) $8.49 kJ K^{- 1} kg^{- 1}$
(D) $9.26 kJ K^{- 1} kg^{- 1}$ ✓ Correct

Answer: (D)

Solution

Build the entropy change step by step as

1 kg

of ice at

273 K

becomes vapour at

383 K

. Melting at

273 K

gives

\frac{334}{273} \approx 1.22

. Warming the liquid from

273

373 K

gives

4.2 ln \frac{373}{273} \approx 1.32

. Boiling at

373 K

gives

\frac{2491}{373} \approx 6.68

. Heating the vapour from

373

383 K

gives

2.0 ln \frac{383}{373} \approx 0.05

. Adding the four pieces,

Δ S \approx 1.22 + 1.32 + 6.68 + 0.05 \approx 9.26 kJ K^{- 1} kg^{- 1}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Chemical Thermodynamics

Solution

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