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For the following thermodynamic process Dry ice $\longrightarrow CO_2$ (g)

(a) Both $\Delta H$ and $\Delta S$ are positive
(b) $\Delta H$ is -ve while $\Delta S$ is +ve
(c) $\Delta H$ is +ve while $\Delta S$ is -ve
(d) Both $\Delta H$ and $\Delta S$ are negative

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Answer: Both $\Delta H$ and $\Delta S$ are positive
Dry ice is solid $CO_2$.
$\Delta H$ = +ve, because heat is absorbed during sublimation.
$\Delta S$ = +ve, because entropy of gaseous state is more than solids.
answered Feb 24, 2014 by mosymeow_1

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