Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

The change in entropy, in the conversion of one mole of water at \(373 \mathrm{~K}\) to vapour at the same temperature is (Latent heat of vaporization of water = \(\left.2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\right)\) (a) \(99 \mathrm{JK}^{-1}\) (b) \(129 \mathrm{JK}^{-1}\) (c) \(89 \mathrm{JK}^{-1}\) (d) \(109 \mathrm{JK}^{-1}\)

Step-by-Step Solution

Answer

The change in entropy is approximately \(109 \mathrm{JK}^{-1}\), so the answer is (d).

Step 1: Convert Latent Heat from g to moles

First, convert the latent heat of vaporization from per gram to per mole. The latent heat given is \(2.257 \mathrm{kJ} \mathrm{~g}^{-1}\). 1 mole of water is \(18 \mathrm{~g}\), so the latent heat per mole is \(2.257 \mathrm{kJ} \mathrm{~g}^{-1} \times 18 \mathrm{~g} = 40.626 \mathrm{kJ} \mathrm{~mol}^{-1}\).

Step 2: Convert Latent Heat to Joules

Convert \(40.626 \mathrm{kJ} \mathrm{~mol}^{-1}\) to joules: \(40.626 \mathrm{kJ} \mathrm{~mol}^{-1} = 40626 \mathrm{~J} \mathrm{~mol}^{-1}\).

Step 3: Use Entropy Change Formula

The entropy change \(\Delta S\) in the conversion is calculated by \(\Delta S = \frac{q}{T}\), where \(q\) is the heat absorbed (latent heat) and \(T\) is the temperature in Kelvin. Here, \(q = 40626 \mathrm{~J} \mathrm{~mol}^{-1}\) and \(T = 373 \mathrm{~K}\).

Step 4: Calculate Entropy Change

Plug in the values into the formula: \(\Delta S = \frac{40626}{373} = 108.96 \mathrm{JK}^{-1}\).

Step 5: Choose the Closest Answer

The calculated entropy change is approximately \(109 \mathrm{JK}^{-1}\), which correlates with option (d).

Key Concepts

Latent Heat of Vaporization

Latent heat of vaporization is the amount of heat required to transform a liquid into a gas at a constant temperature and pressure.
This heat is necessary because it overcomes the intermolecular attractions in the liquid.

• For water, the latent heat of vaporization is often given in kJ per gram, such as 2.257 kJ/g, indicating the energy needed for each gram to vaporize.
• To apply this concept in calculations, like the conversion to joules per mole, multiply the given value by the molar mass of the liquid. For water, this means converting using 18 g/mol (since 1 mole of water weighs 18 g).
• In our exercise, converting the latent heat from per gram to per mole involves multiplying by 18, resulting in 40.626 kJ/mol, which is then converted into joules for precise calculations.

Understanding this step is crucial for accurately calculating changes in state properties, such as entropy.

Entropy Calculation

Entropy is a measure of disorder or randomness in a system, often related to the dispersal of energy.
During a phase transition, such as vaporization, entropy specifically measures how energy distribution changes.
Calculating entropy change involves understanding that energy (heat) absorbed during a process is distributed across the system at a specific temperature. Here's a simple breakdown of the process:

• The formula used for calculating the entropy change (6): \[ 6 = \frac{q}{T} \] where \( q \) is the heat involved in the transition, and \( T \) is the absolute temperature.
• In scenarios like heating or phase changes, it's essential to convert energy into joules to ensure consistency in units (6 often uses J/K as the unit).
• In our exercise, converting 40.626 kJ/mol into 40626 J/mol allows us to use this value with the temperature of 373 K to find the entropy change. 6 is calculated as \( \frac{40626 \, J}{373 \, K} = 108.96 \, JK^{-1} \).

This formula highlights the relationship between heat, temperature, and entropy in thermodynamics.

Thermodynamics

Thermodynamics deals with the principles of energy transfer and entropy in physical systems. It's the science of how energy moves and changes form.
This field covers everything from engines to the vaporization of water in our exercise.
Let's break down some core concepts:

• First Law of Thermodynamics: This is the law of energy conservation, stating that energy cannot be created or destroyed, only transformed. In phase changes, energy input typically equals the energy expressed as latent heat.
• Second Law of Thermodynamics: This principle states that entropy will either increase or remain the same in an isolated system, driving the natural direction of processes. During vaporization, entropy increases because energy dispersal becomes more random in a gaseous state than a liquid one.
• Understanding these laws helps us predict and describe the behavior of systems during phase changes, making it easier to comprehend why energy such as latent heat is necessary for changes like vaporization.

By grasping these fundamental concepts, you can easily follow and solve problems involving energy changes in physical systems, such as our initial entropy exercise.