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JEE MAIN 2019
09-01-19 S2
Question
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 vapour 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). $(\log 273=2.436, \log 373=2.572, \log 383=2.583)$
Select the correct option:
A
$8.49 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
B
$7.90 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
C
$9.26 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
D
$2.64 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & \Delta \mathrm{S}_{\text {tus }}=\frac{\Delta \mathrm{H}_{\text {tus }}}{273}=\frac{334}{273}=1.22 \\ & \Delta \mathrm{~S}_{\text {vap }}=\frac{\Delta \mathrm{H}_{\text {vap }}}{373}=\frac{2491}{373}=6.67 \\ & \Delta \mathrm{~S}_{\text {water }}=\frac{\mathrm{mCdT}}{\mathrm{~T}}=\mathrm{mCln}\left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right) \\ & =4.2 \times \ln \left(\frac{273}{273}\right)=1.31 \\ & \Delta \mathrm{~S}_{\text {vap }}=\mathrm{mCln}\left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right) \\ & =2 \times \ln \left(\frac{383}{373}\right)=0.05 \end{aligned} $$ Total entropy change $$ \Delta \mathrm{S}=9.26 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $$
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