Consider the entropy changes in a system undergoing transformation, as depicted in the diagram, below
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The correct statement among the following is

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CSIR-UGC (NET) Chemical Science: Held on (15 Dec 2019)
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  1. ΔS1 = ΔS2 and ΔS ≠ ΔS1 + ΔS2
  2. ΔS1 > ΔS2 and ΔS ≠ ΔS+ ΔS2
  3. ΔS1 = ΔSand ΔS = ΔS1 + ΔS2
  4. ΔS1 > ΔS2 and ΔS = ΔS1 + ΔS2

Answer (Detailed Solution Below)

Option 4 : ΔS1 > ΔS2 and ΔS = ΔS1 + ΔS2
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Concept:

→ In thermodynamics, ΔS represents the change in entropy of a system, which is a measure of the system's disorder or randomness.  

→ ΔS can be calculated as the difference between the entropy of the final state and the entropy of the initial state of a process. It is path independent.

Explanation:

→ The entropy change of a system due to a temperature change is given by:

ΔS = \(\frac{q}{T}\)

This equation implies that as the temperature of a system increases, the amount of entropy change that occurs for a given amount of heat transfer decreases.

→ In summary, the entropy of a system is directly proportional to the temperature of the system.

→ Thus, the change in entropy for the first process is greater than the change in entropy for the second process. This could occur, for example, if the first process involves a larger increase in disorder or randomness than the second process.

Therefore, ΔS1 > ΔS2

→ If  ΔS1 and  ΔS2 represent the changes in entropy for two separate processes, and  ΔS represents the total change in entropy for both processes combined, then we can say that:

 ΔS =  ΔS1 +  ΔS2

→ This equation simply states that the total change in entropy for the combined processes is equal to the sum of the changes in entropy for each individual process.

Conclusion:
The equation satisfies ΔS1 > ΔS2 and ΔS = ΔS1 + ΔS2, i.e. option 4.

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