Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

The direct conversion of \(\mathrm{A}\) to \(\mathrm{B}\) is difficult, hence it is carried out by the following path: Given \(\Delta \mathrm{S}(\mathrm{A} \longrightarrow \mathrm{C})=50 \mathrm{e.u}\) \(\Delta \mathrm{S}(\mathrm{C} \longrightarrow \mathrm{D})=30 \mathrm{e.u}\) \(\Delta \mathrm{S}(\mathrm{B} \longrightarrow \mathrm{D})=20 \mathrm{e.u}\) where e.u. is entropy unit then \(\Delta \mathrm{S}(\mathrm{A} \longrightarrow \mathrm{B})\) is (a) \(+100\) e.u. (b) \(+60\) e.u. (c) \(-100\) e.u. (d) \(-60\) e.u.

Step-by-Step Solution

Answer

\(\Delta S(\mathrm{A} \rightarrow \mathrm{B}) = +60\) e.u. (option b)

Step 1: Understanding the Problem

We are given the entropies for different paths and asked to find the entropy change from A to B. This involves utilizing the concept of entropy change being a state function, implying it is independent of the path.

Step 2: Identifying Given Data

We note the given entropy changes are: \(\Delta S(\mathrm{A} \to \mathrm{C}) = 50\) e.u., \(\Delta S(\mathrm{C} \to \mathrm{D}) = 30\) e.u., \(\Delta S(\mathrm{B} \to \mathrm{D}) = 20\) e.u.

Step 3: Using Entropy as a State Function

Since entropy is a state function, the total \(\Delta S\) from \(\mathrm{A} \rightarrow \mathrm{B}\) can be found by rearranging given routes. The sum of entropies from \(\mathrm{A} \to \mathrm{B}\) via \(\mathrm{C} \to \mathrm{D}\) should equal \(-\Delta S(\mathrm{B} \to \mathrm{D})\).

Step 4: Setting up the Equation

According to the path independence of state functions, \(\Delta S(\mathrm{A} \rightarrow \mathrm{B}) = \Delta S(\mathrm{A} \rightarrow \mathrm{C}) + \Delta S(\mathrm{C} \rightarrow \mathrm{D}) - \Delta S(\mathrm{B} \rightarrow \mathrm{D})\).

Step 5: Calculating \(\Delta S(\mathrm{A} \rightarrow \mathrm{B})\)

Plugging in the values: \(\Delta S(\mathrm{A} \rightarrow \mathrm{B}) = 50 + 30 - 20 = 60\) e.u. Hence, \(\Delta S(\mathrm{A} \rightarrow \mathrm{B}) = +60\) e.u.

Step 6: Conclusion and Answer Selection

The calculated \(\Delta S(\mathrm{A} \rightarrow \mathrm{B})\) matches with option (b), which is \(+60\) e.u.

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the study of energy, energy transfer, and the changes in matter. The key aspect of thermodynamics important to our exercise is the concept of entropy. Entropy is a measure of randomness or disorder within a system. In thermodynamics, we're often concerned with entropy changes, denoted as \(\Delta S\), which describe how much the entropy of a system has increased or decreased during a process.
Entropic changes provide insight into the spontaneity of processes. A positive \(\Delta S\) typically indicates a spontaneous process, meaning the energy distribution has become more disordered. Entropy and thermodynamics concepts are crucial for understanding why certain processes occur over others and grasping how they can be manipulated in different applications.

• Understanding the relationship between heat transfer and work done on or by a system.
• Grasping how these factors cause changes in a system's entropy.

Grasping thermodynamics helps in understanding the natural direction of energy transformations that increase the overall disorder of the universe.

State Functions

State functions are properties of a system that depend solely on the current state of the system, not on the path the system took to reach that state. This concept is key in thermodynamics because it means that no matter how a system changes state, the change in a state function depends only on the initial and final states.
Entropy (\(S\)), as discussed in the previous section, is a perfect example of a state function. Whether a substance goes through several intermediate transformations, or one direct transformation, the change in entropy when going from one state to another will remain the same. This can be visualized as:

• The initial state (A) and final state (B) define the \(\Delta S\), regardless of the intermediate states (like C and D in our example).
• Mathematically, \(\Delta S\) from A to B is the same irrespective of the number or type of processes between them.

In our problem, the calculation of \(\Delta S\) relies on this state function property. No matter the path, the entropy change between any two states depends only on those two states.

Path Independence

Path independence in thermodynamics is a fascinating property that stems from the concept of state functions. Essentially, because state functions depend only on the initial and final states of a process, the path taken does not influence the overall change in the state function.

• For entropy, this means that even if we take a long, complex path from point A to B, the total change will be the same as if we took a direct route, provided the starting and ending points are the same.
• This is what allows us to confidently use thermodynamic equations without needing to map every possible state change along the way.

In the original exercise, path independence simplifies our calculations: by knowing the state function property of entropy, we can calculate \(\Delta S(\mathrm{A} \rightarrow \mathrm{B})\) using whatever path data connects these points without needing to verify every possible journey. This unique feature saves time and reduces complexity when tracking changes in systems.