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The latent heat of vaporisation of water at 100 $^\circ C$ is 540 $cal \cdot {g^{ - 1}}$ . Calculate the entropy increase when 1 mole of water at 100 $^\circ C$ is evaporated.
A. 26 $cal.{K^{ - 1}}mo{l^{ - 1}}$
B. 1.45 $cal.{K^{ - 1}}mo{l^{ - 1}}$
C. 367 $cal.{K^{ - 1}}mo{l^{ - 1}}$
D. 1.82 $cal.{K^{ - 1}}mo{l^{ - 1}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: During the phase change it is seen that the temperature remains constant though there is an increase in the amount of heat that is being provided to change the phase. It has been inferred that such heat goes into the breaking of the bonds that are essential for the change of matter from one phase to another.

Complete step by step answer:
All naturally occurring processes tend to proceed in a single direction spontaneously. Here spontaneity means the potential to proceed without the assistance of an external agent.
S is called the entropy of the reaction and is the measure of the disorder. The entropy is usually defined for an isolated system where there is always a tendency for the system to become more disordered. The greater the disorder in an isolated system the higher the entropy. Neither decrease in enthalpy nor the increase in entropy can determine the spontaneity of the system alone.
The entropy of the reaction can be written as $\Delta S = \dfrac{{\Delta H}}{T}$
Where the entropy that is provided to the process is given as $\Delta H = 540cal/g$
The molecular mass of water is known as 18 $g$ from its chemical formula.
So, $\Delta {H_{(per{\text{ mole}})}} = \Delta {H_{(per{\text{ gram}})}} \times molar{\text{ }}weight$
$\Delta {H_{({\text{per mole}})}} = 540 \times 18 = 9720cal/mol$
As we have stated that $\Delta S = \dfrac{{\Delta H}}{T}$
And the temperature is given as 100 $^\circ C$ which upon conversion to kelvin becomes 373 K
We can calculate the entropy by the following $\Delta S = \dfrac{{9720}}{{373}} = 26cal/(mol.K)$

So, the correct answer is Option A.

Note: For a spontaneous process, the change in entropy should be greater than 0 in other words
$\Delta S > 0$
In the process of reduction, the change in enthalpy of the system is less than 0 since it is generally an endothermic process.
Using the above points we can infer that for a reduction reaction $\Delta H < 0$ and $\Delta S > 0$ Which upon putting in the Gibbs equation makes $\Delta G < 0$