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 \(540 \mathrm{cal} \mathrm{g}^{-1}\left(\mathrm{R}=2.0 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(48600 \mathrm{cal}, 44870 \mathrm{cal}\) (b) \(43670 \mathrm{cal}, 47700 \mathrm{cal}\) (c) 47700 cal, \(43670 \mathrm{cal}\) (d) \(44870 \mathrm{cal}, 48670 \mathrm{cal}\)

Step-by-Step Solution

Answer

Option (a): ΔH = 48600 cal, ΔE = 44870 cal

Step 1: Calculate Moles of Water

First, we need to find the number of moles of water in 90 grams. The molar mass of water (H₂O) is approximately 18 g/mol. Therefore, the number of moles, \( n \), can be calculated using the formula: \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ moles} \]

Step 2: Calculate Heat of Evaporation (ΔH)

The heat of evaporation, \( \Delta H \), is given as 540 cal/g for water. To find the total heat required for 90 g, we multiply this value by the mass of water:\[ \Delta H = 540 \text{ cal/g} \times 90 \text{ g} = 48600 \text{ cal} \]

Step 3: Calculate Work Done by the System

In the reversible isothermal process, the work done on or by the system is given by:\[ W = nRT \cdot \ln\left(\frac{V_2}{V_1}\right) \] Since the final and initial molar volumes of water vapour and liquid water differ greatly, and since water vapour behaves as an ideal gas:\[ V_2 \gg V_1 \text{, thus } \ln\left(\frac{V_2}{V_1}\right) \approx RT \] However, for the evaporation of liquid to gas, work done \(W\) simplifies to \(nRT\):\[ W = 5 \text{ mol} \times 2 \text{ cal/(mol K)} \times 373 \text{ K} = 3730 \text{ cal} \]

Step 4: Calculate Change in Internal Energy (ΔE)

The change in internal energy, \( \Delta E \), is given by the equation:\[ \Delta E = \Delta H - W \] Plugging in the values, we have:\[ \Delta E = 48600 \text{ cal} - 3730 \text{ cal} = 44870 \text{ cal} \]

Key Concepts

Isothermal Processes

In thermodynamics, an isothermal process is a change in a system where the temperature remains constant. For the case of water vaporizing at 100°C, the temperature does not change throughout the evaporation. This stable temperature is critical because it ensures that all the added energy is used for phase change rather than raising temperature.

In an isothermal process, the heat added to the system is equal to the work done by the system, making calculations slightly different from other processes. Here, since we are dealing with an ideal gas situation (as assumed for water vapor), we utilize the Ideal Gas Law and properties in simplifications. When vapors behave ideally, it implies that they follow the equation:

• For isothermal expansion: \[ \Delta T = 0 \]
• The work done (W) can be calculated using: \[ W = nRT \cdot \ln\left(\frac{V_2}{V_1}\right) \]

This shows how the system exchanges heat and work without altering internal energy from a temperature perspective.

Enthalpy Change

Enthalpy, denoted as \( \Delta H \), is a measure of the total energy of a thermodynamic system. In the context of evaporation, \( \Delta H \) represents the heat absorbed to convert liquid water into vapor at constant pressure. For our example, this is calculated as follows:

• Heating water to its boiling point at a constant temperature of 100°C requires a specific amount of thermal energy, termed as the heat of evaporation.
• Given as \( 540 \text{ cal/g} \) for water, the total \( \Delta H \) for 90g becomes:\[ \Delta H = 540 \text{ cal/g} \times 90 \text{ g} = 48600 \text{ cal} \]

This energy amount signifies the enthalpy change needed for the phase transition of water from liquid to gas. It's a key factor in understanding energy requirements in isothermal processes.

Internal Energy Change

Internal energy, represented by \( \Delta E \), is the total of all kinetic and potential energy within a system. In thermodynamics, internal energy change is a crucial concept to understand how energy transitions between states.For isothermal processes, although the temperature remains constant, the system's internal energy might still change if there's work done on or by the system.

• The relationship linking \( \Delta E \) to \( \Delta H \) is expressed via: \[ \Delta E = \Delta H - W \]
• In our scenario, we calculated work using the formula: \[ W = nRT \] Where \( n \) is moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.
• Plugging in the values for the example: \[ W = 5 \times 2 \times 373 \] resulting in \( W = 3730 \text{ cal} \).

Thus, the internal energy change is derived as:\[ \Delta E = 48600 \text{ cal} - 3730 \text{ cal} = 44870 \text{ cal} \]This highlights how energy distribution occurs within the system during phase changes, even when external temperature remains stable.