Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

Determine \(\Delta \mathrm{H}\) and \(\Delta \mathrm{E}\) for reversible isothermal evaporation of \(90 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\). Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal \(\mathrm{g}^{-1}\left(\mathrm{R}=2.0 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) 48600 cal, 44870 cal (b) \(43670 \mathrm{cal}, 47700 \mathrm{cal}\) (c) 47700 cal, 43670 cal (d) \(44870 \mathrm{cal}, 48670 \mathrm{cal}\)

Step-by-Step Solution

Answer

The values are (a) 48600 cal for \(\Delta H\) and 44870 cal for \(\Delta E\).

Step 1: Understanding the Significance of Given Data

We are provided with the mass of water (90 g) that is being evaporated, the temperature (100°C), heat of evaporation (540 cal/g), and the gas constant \(R\). The task is to determine \(\Delta H\) (change in enthalpy) and \(\Delta E\) (change in internal energy) for the process.

Step 2: Calculation of Moles of Water

First, calculate the moles of water evaporated. The molar mass of water is 18 g/mol. So, the moles of water \(n\) is: \[ n = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol} \]

Step 3: Calculation of Change in Enthalpy (\(\Delta H\))

The change in enthalpy of the process, \(\Delta H\), is given by: \[ \Delta H = \text{Heat of evaporation} \times \text{Mass of water} \] \[ \Delta H = 540 \text{ cal/g} \times 90 \text{ g} = 48600 \text{ cal} \] This represents the total heat required for the phase change.

Step 4: Calculation of Work Done (W) in Reversible Isothermal Process

Since the vapor behaves like an ideal gas, the work done \(W\) can be calculated using the formula: \[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \] For phase change at constant temperature, this simplifies to \(W = \Delta nRT\), where \(\Delta n = 1\). Thus, \(W = 5 \text{ mol} \times 2 \text{ cal/mol·K} \times 373 \text{ K} \) \[ W = 3730 \text{ cal} \]

Step 5: Calculation of Change in Internal Energy (\(\Delta E\))

Using the formula \(\Delta E = \Delta H - W \), we find \(\Delta E\) as follows: \[ \Delta E = 48600 \text{ cal} - 3730 \text{ cal} = 44870 \text{ cal} \]

Step 6: Verifying the Correct Option

The calculated values are \(\Delta H = 48600 \text{ cal}\) and \(\Delta E = 44870 \text{ cal}\), which corresponds to option (a).

Key Concepts

Enthalpy

Enthalpy is a measure of the total energy of a thermodynamic system, often viewed as the heat content of a system at constant pressure. In simple terms, it's the amount of heat absorbed or released during a constant pressure process.
When you boil water, for instance, you are adding heat, which increases its enthalpy. This is why the process of evaporation, as in the exercise, involves calculating the change in enthalpy, \( \Delta H \).
For the reversible isothermal evaporation of water in the problem: - The heat of evaporation per gram is given as 540 cal/g.
- With a mass of 90 g of water, the total change in enthalpy \( \Delta H \) is \( 540 \, \text{cal/g} \times 90 \, \text{g} = 48600 \, \text{cal} \).
Understanding these computations helps you see how energy transformations occur during phase changes in a controlled setting.

Internal Energy

Internal energy, denoted as \( U \), is the total energy contained within a system, encompassing both kinetic and potential energy of the molecules. It's a core concept to grasp in thermodynamics because it reflects how energy balance is maintained.
In the context of the exercise, the change in internal energy \( \Delta E \) is calculated using the relationship between work, enthalpy, and internal energy. During an isothermal process, any heat added to the system becomes internal energy plus the work done by the system:

• Using the equation: \[ \Delta E = \Delta H - W \] where \( W \) is the work done.
• Here, \( W \) is calculated based on the ideal gas laws since water vapor is assumed to behave as an ideal gas. In this exercise, the work was calculated to be 3730 cal.
• Therefore, \( \Delta E = 48600 \, \text{cal} - 3730 \, \text{cal} = 44870 \, \text{cal} \).

Recognizing how energy is conserved through heat transfer and work is a fundamental aspect of solving such problems.

Ideal Gas

The concept of an ideal gas is a significant simplification that allows one to predict how a gas behaves under different conditions. An ideal gas is one in which the gas particles:

• Do not experience any intermolecular forces.
• Occupy no volume, essentially being "point particles."
• Undergo perfectly elastic collisions with each other and the walls of their container.

Using the ideal gas law \( PV = nRT \), one can compute various properties of gases under ideal conditions. Here, during the water vapor phase of evaporation:

• Work done by the gas is calculated using \( W = nRT\), considering temperature at 373 K.
• The assumption of ideal gas behavior simplifies the computations substantially, eliminating the need to consider intermolecular forces or the finite size of molecules.

This simplification allows for the calculation of work done by the system during the isothermal expansion, which is critical in finding the change in internal energy from enthalpy in this scenario.