If the enthalpy change for the transition of liquid water to steam is $30 kJ mol^{-1} \ at \ 27^{o}C$, the entropy change for the process would be Option 1) 10 $J mol^{-1}K^{-1}$ Option 2) 1.0 $J mol^{-1}K^{-1}$ Option 3) 0.1 $J mol^{-1}K^{-1}$ Option 4) 100 $J mol^{-1}K^{-1}$

Answers (1)
P Prateek Shrivastava

Entropy for phase transition at constant pressure -

$\Delta S= \frac{\Delta H_{Transition}}{T}$

- wherein

Transition $\Rightarrow$Fusion, Vaporisition, Sublimation

$\Delta H\Rightarrow$Enthalpy

$\Delta E\Rightarrow$Internal Energy

$T\Rightarrow$Transitional temperature

Entropy change = $\frac{\Delta H}{T}$

$= \frac{30\times 10^{3} J/mdi-k}{300}$

= 100 $J mol^{-1}K^{-1}$

Option 1)

10 $J mol^{-1}K^{-1}$

Incorrect

Option 2)

1.0 $J mol^{-1}K^{-1}$

Incorrect

Option 3)

0.1 $J mol^{-1}K^{-1}$

Incorrect

Option 4)

100 $J mol^{-1}K^{-1}$

Correct

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