Thermodynamics and Thermochemistry - Result Question 18
18. 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 $4.2$ $kJ$ $\mathrm{K}^{-1} $ $\mathrm{~kg}^{-1}$ and $2.0$ $ \mathrm{kJ}$ $ {K}^{-1}$ $ \mathrm{~kg}^{-1}$; heat of liquid fusion and vapourisation of water are $334 $ $\mathrm{~kJ} $ $\mathrm{~kg}^{-1}$ and $2491$ $ \mathrm{kJ}$ $ kg^{-1}$ respectively). $\quad(\log 273=2.436, \quad \log 373=2.572$, $\log 383=2.583$ )
(2019 Main, 9 Jan II)
(a) $9.26 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
(b) $8.49 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
(c) $7.90 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
(d) $2.64 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
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Answer:
Correct Answer: 18. $(a)$
Solution:
The conversion of $1 $ $kg$ of ice at $273 K$ into water vapours at $383$ $K$ takes place as follows:
$\Delta S _1 =\dfrac{\Delta H _{\text {Fusion }}}{\Delta T _{\text {Fusion }}}=\dfrac{334 kJ kg^{-1}}{273 K}=1.22 $ $kJ $ $kg^{-1}$ $ K^{-1} $
$\Delta S _2 =C \ln \dfrac{T _2}{T _1}=4.2 $ $kJ $ $K^{-1} $ $kg^{-1} \ln \left(\dfrac{373 K}{273 K}\right) $
$=4.2 \times 2.303(\log 373-\log 273) $ $kJ $ $K^{-1} $ $kg^{-1} $
$=4.2 \times 2.303(2.572-2.436)=1.31$ $ kJ$ $ K^{-1}$ $ kg^{-1}$
$\Delta S _3 =\dfrac{\Delta H _{\text {vap. }}}{\Delta T _{\text {vap. }}}=\dfrac{2491 kJ kg^{-1}}{373 K} $
$=6.67$ $ kJ $ $kg^{-1} $ $K^{-1}$
$\Delta S _4=C \ln \dfrac{T _2}{T _1}=2 $ $kJ $ $K^{-1}$ $ kg^{-1} \ln \left(\dfrac{383 K}{373 K}\right) $
$=2 \times 2.303(\log 383-\log 373)$ $ kJ$ $ K^{-1} $ $kg^{-1} $
$=2 \times 2.303(2.583-2.572) $ $kJ $ $K^{-1} $ $kg^{-1} $
$=0.05 $ $kJ $ $K^{-1}$ $ kg^{-1} $
$\Delta S _{\text {Total }} =\Delta S _1+\Delta S _2+\Delta S _3+\Delta S _4 $
$=1.22+1.31+6.67+0.05 $
$ =9.26$ $ kJ $ $kg^{-1} $ $K^{-1}$
BETA
