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

The values are \( \Delta H = 48600 \, \text{cal} \) and \( \Delta E = 44870 \, \text{cal} \). So, option (a) is correct.

Step 1: Calculate Moles of Water

First, determine the number of moles of water evaporating. The molecular weight of water is approximately 18 g/mol. The given mass of water is 90 g. Use the formula \( n = \frac{\text{mass}}{\text{molar mass}} \) to calculate the number of moles, \( n = \frac{90 \, \text{g}}{18 \, \text{g/mol}} = 5 \, \text{mol} \).

Step 2: Determine \( \Delta H \) (Enthalpy Change)

The heat of evaporation is given as 540 cal/g. For 90 g of water, the enthalpy change \( \Delta H \) is calculated using \( \Delta H = \text{mass} \times \text{heat of evaporation} = 90 \, \text{g} \times 540 \, \text{cal/g} = 48600 \, \text{cal} \).

Step 3: Determine \Delta E (Internal Energy Change)

Use the relation \( \Delta H = \Delta E + \Delta nRT \). \( \Delta H \) has been calculated as 48600 cal. For the reversible isothermal process, \( \Delta n = 5 - 0 = 5 \) (the moles of the gas formed), and \( T = 100^{\circ} \mathrm{C} = 373 \, \text{K} \). Substituting into the equation: \( 48600 = \Delta E + (5)(2.0)(373) \). Calculate \( 5 \times 2.0 \times 373 = 3730 \, \text{cal} \).Rearrange to find \( \Delta E \): \( \Delta E = 48600 - 3730 = 44870 \, \text{cal} \).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is an important concept in thermodynamics. It represents the total heat change in a system during a constant pressure process. For the reversible isothermal evaporation of water, this change reflects the energy required to convert liquid water to vapor at a specific temperature. Since the process is isothermal, the temperature remains constant, and the system absorbs energy as heat from the surroundings.

In our example, the heat of evaporation is 540 cal/g for water. When 90 grams of water evaporates, the enthalpy change can be calculated by multiplying the mass of the water by the heat of evaporation, resulting in \( \Delta H = 48600 \text{ cal} \). This calculation shows how much energy in the form of heat is needed to transition the water from liquid to vapor under these conditions.

Internal Energy

Internal energy, symbolized as \( \Delta E \), reflects the total energy change within a system due to internal factors such as molecular motion. For reversible isothermal processes, calculating internal energy involves considering not only the energy transfer due to temperature but also the work done due to volume changes.

In this case, we use the equation \( \Delta H = \Delta E + \Delta nRT \) to determine internal energy. We already know \( \Delta H \) is 48600 cal. \( \Delta n \) is the change in moles, which is 5, and \( R \) is the gas constant at 2.0 cal/mol K, with the temperature \( T \) being 373 K. Plugging these into the formula yields \( 48600 = \Delta E + 3730 \). Solving, we find \( \Delta E = 44870 \text{ cal} \). This highlights how internal energy changes even in a process where temperature remains constant.

Reversible Isothermal Process

A reversible process is a theoretical concept where the system and surroundings can be returned to their initial states without any net change in the universe. In an isothermal process, the temperature is held constant. This is crucial in thermodynamics as it ensures that any heat transfer results in work being done by the system.

During the reversible isothermal evaporation of water, the system, initially at 100°C, stays at the same temperature. The enthalpy change accounts for the energy needed to overcome intermolecular forces and change the state of water from liquid to vapor. Simultaneously, additional energy absorbed manages the expansion against atmospheric pressure, converting some internal energy into mechanical work.

Ideal Gas

The ideal gas concept simplifies understanding gas behavior under various conditions. It assumes that the gas particles do not attract or repel each other and occupy negligible space. This model effectively explains most gases' behavior under normal conditions but is particularly handy in calculations involving temperature, pressure, and volume relationships.

In our exercise, water vapor is assumed to behave like an ideal gas. This approximation allows us to utilize the equation \( \Delta nRT \) while calculating changes in internal energy. Here, \( \Delta n \) represents the change in the number of moles transitioning into the gaseous state during evaporation. These simplifications make it easier to compute property changes without delving into more complex real gas equations.