Question

Calculate the entropy increase in the evaporation of 1 mole of a liquid when it boils at ${100^ \circ }C$ having heat of evaporation at ${100^ \circ }C$ as $540cal/gm$.

Answer 1

Hint: In the question we are given Heat of evaporation, but the formula requires the use of molal heat of evaporation and hence we need to multiply it by the molecular weight of ${H_2}O$.
Then, we use the formula given by the second law of thermodynamics which relates entropy change with molal heat of evaporation to obtain an answer.
Formula used:
$\Delta S = \dfrac{q}{T}$
$\Delta S = entropy\,change,q = molal\,heat\,of\,vaporisation,T = Temperature$

Complete step by step answer:
Let us understand the terms which have been used in the given question.
Entropy is a measure of randomness or disorder in the system, it can be explained by considering that a gaseous system has a higher degree of disorder as compared to a liquid system.
The freer the molecules are, higher will be the degree of disorder.
Entropy change can also tell us about the feasibility of a reaction, if the entropy change is positive, which means the reaction is proceeding towards a more disordered stage then the reaction will be spontaneous.
However, according to the second law of thermodynamics, it states that the entropy of the universe will never change. This statement is in accordance with the law of conservation of energy, Energy can never be created nor be destroyed.
In the question, we are given heat of evaporation which is the amount of heat required to convert a given quantity of liquid to gas. Since the process is a conversion of liquid to gas, we can say that there will be an increase in the degree of disorder in the system and hence the entropy change should be positive.
Now, the formulae we need to use is :
$\Delta S = \dfrac{q}{T}$
where; $\Delta S = entropy\,change,q = molal\,heat\,of\,vaporisation,T = Temperature$
From the question we know, $heat\,of\,vaporisation = 540cal/gm,T = {100^ \circ }C \Rightarrow 373K$
The liquid used is water and hence the molecular mass of water can be calculated as
$mass\,of\,H = 1g,mass\,of\,oxygen = 16g$
hence mass of water= $1 \times 2 + 16$
$\therefore $ the mass of water is $18g$
$Molal\,heat\,of\,vaporisation = Heat\,of\,vaporisation \times Molecular\,mass$
Substituting the value of molecular mass, we get
$Molal\,heat\,of\,vaporisation = 540cal/gm \times 18g$
Substituting this value in final equation=$\Delta S = \dfrac{{540 \times 18}}{{373}}$
$\Delta S = 26.06cal/gm$
Hence the correct answer to this question is option “D”.

Note: The value of entropy change is positive this means that the reaction is moving towards a state of higher degree of disorder. Heat of condensation is equal in magnitude to the heat of evaporation but opposite in magnitude. This is because the reaction is proceeding towards a state with less degree of disorder and hence external force is required for the reaction to proceed.