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Thermodynamics question

2024 · Q119
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Thermodynamics question

2024 · Q119

NEETChemistryThermodynamicsMCQ+4 / −1

In which of the following processes entropy increases?

A. A liquid evaporates to vapour.

B. Temperature of a crystalline solid lowered from 130 K130 \mathrm{~K}130 K to 0 K0 \mathrm{~K}0 K.

C. 2NaHCO3( s)→Na2CO3( s)+CO2( g)+H2O(g)2 \mathrm{NaHCO}_{3(\mathrm{~s})} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(\mathrm{~s})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}2NaHCO3( s)​→Na2​CO3( s)​+CO2( g)​+H2​O(g)​

D. Cl2(g)→2Cl(g)\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{Cl}_{(g)}Cl2(g)​→2Cl(g)​

Choose the correct answer from the options given below:

  1. A
    A and C
  2. B
    A, B and D
  3. C
    A, C and D
  4. D
    C and D
View written solutionFree

Correct answer: C

The concept of entropy in thermodynamics refers to the degree of randomness or disorder in a system. An increase in entropy is generally associated with processes in which disorder increases. Let's analyze each option listed:

A. A liquid evaporates to vapour.

During the evaporation of a liquid to form a vapour, the molecules of the substance move from a relatively ordered state (liquid) to a more disordered state (vapour). In a vapour, the molecules have more freedom of motion and are less confined than in a liquid. This transition from liquid to vapour increases the randomness or disorder of the system, hence, the entropy increases.

B. Temperature of a crystalline solid lowered from $130 \mathrm{~K}$ to $0 \mathrm{~K}$.

Reducing the temperature of a crystalline solid generally decreases the entropy of the system. As the temperature decreases, the molecular motion within the solid becomes more restricted, leading to a decrease in randomness. At absolute zero ($0 \mathrm{~K}$), the entropy is at its lowest possible value (ideally zero for a perfect crystal), as the molecular motion is minimized to only quantum mechanical vibrations.

C. $$2 \mathrm{NaHCO}_{3(s)} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(s)}+\mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(g)}$$

In this chemical reaction, solid sodium bicarbonate decomposes to form solid sodium carbonate, carbon dioxide gas, and water vapor. The formation of gases from a solid significantly increases the entropy of the system because gases have much higher randomness due to their free molecular motion compared to solids.

D. $$\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{Cl}_{(g)}$$

The dissociation of chlorine gas ($\mathrm{Cl}_2$) into atomic chlorine ($\mathrm{Cl}$) represents a transition from a diatomic molecule to two separate atoms. This increases the number of particles in the gas phase, leading to increased randomness and disorder, hence, increasing the entropy.

Conclusion: Based on our analysis, processes A, C, and D involve an increase in entropy, as they each lead to greater disorder or randomness in the system. Option B is incorrect as it wrongly includes lowering the temperature of a solid as increasing entropy. Therefore, the correct answer is:

Option C :

A, C and D

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