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) \(47700 \mathrm{cal}, 43670 \mathrm{cal}\) (d) \(44870 \mathrm{cal}, 48670 \mathrm{cal}\)

Short answer

\( \Delta H = 48600 \, \text{cal} \), \( \Delta E = 44870 \, \text{cal} \) (option a).

Step-by-step solution

Calculate total heat of evaporation

The heat of evaporation per gram of water is given as 540 cal/g. We have 90 g of water. Therefore, the total heat required for evaporation, which is equal to \( \Delta H \), is calculated as follows:\[ \Delta H = 540 \, \text{cal/g} \times 90 \, \text{g} = 48600 \, \text{cal} \]

Calculate moles of water

First, we need to calculate the number of moles of water. We know the molar mass of water is approximately 18 g/mol. So, the moles of water are calculated as:\[ \text{moles of water} = \frac{90 \, \text{g}}{18 \, \text{g/mol}} = 5 \, \text{mol} \]

Calculate work done by gas

Since water vapor behaves as an ideal gas and evaporation occurs isothermally, the work done during evaporation is given by:\[ W = nRT \Delta V \]where \( R = 2.0 \, \text{cal/mol K} \) and \( T = 373 \, \text{K} \). However, \( \Delta V \) for complete evaporation from liquid to gas can use the ideal gas work equation:\[ W = nRT \ln \frac{V_f}{V_i} \approx nRT \].Given that isothermally, \( W = nRT \), we have\[ W = 5 \, \text{mol} \times 2.0 \, \text{cal/mol K} \times 373 \, \text{K} = 3730 \, \text{cal} \]

Calculate change in internal energy \( \Delta E \)

The change in internal energy is given by the relation:\[ \Delta E = \Delta H - W \].Substitute \( \Delta H = 48600 \, \text{cal} \) and \( W = 3730 \, \text{cal} \):\[ \Delta E = 48600 \, \text{cal} - 3730 \, \text{cal} = 44870 \, \text{cal} \]

Key concepts

Enthalpy change

The enthalpy change, denoted as \( \Delta H \), is a key concept in thermodynamics, especially when discussing phase transitions like evaporation. - Enthalpy is the total heat content of a system, representing the energy required to transform a substance from one state to another, at constant pressure.- When water evaporates, it absorbs heat. This absorbed heat is the enthalpy change needed for the phase transition.For our exercise, the given heat of evaporation of water is \( 540 \, \text{cal/g} \), and we determined \( \Delta H \) for \( 90 \, \text{g} \) of water. This calculation shows how much energy is needed for this process, which was found to be \( 48600 \, \text{cal} \). This value represents how much heat energy must be supplied for water's reversible isothermal evaporation.

Internal energy

Internal energy, denoted by \( \Delta E \), is another crucial concept. It reflects all the energy within a system, including kinetic and potential energy of the particles.- It's important to remember that internal energy does not include external energy like pressure or thermal energy exchanged with the surroundings.- The relationship between enthalpy change and internal energy change is given by the equation: \( \Delta E = \Delta H - W \).In our problem:1. We calculated \( \Delta H = 48600 \, \text{cal} \).2. The work done during the expansion or contraction of a gas, calculated as \( W = 3730 \, \text{cal} \).3. Therefore, \( \Delta E = 48600 \, \text{cal} - 3730 \, \text{cal} = 44870 \, \text{cal} \).This result means that the change in internal energy accounts for the energy change once the work done by the system is considered.

Ideal gas behavior

Ideal gas behavior is a simplifying assumption that helps us understand gas properties using mathematical models.- Under this assumption, gas molecules are considered point particles with no forces acting between them, except during elastic collisions.- The ideal gas law expressed as \( PV = nRT \) connects pressure \( P \), volume \( V \), and temperature \( T \) with the number of moles \( n \) and the gas constant \( R \).In our exercise, we assume water vapor to behave ideally during evaporation. Using ideal gas behavior helps in simplifying calculations like the work done \( W \), knowing that the relationship holds as described without complex intermolecular influences.

Isothermal processes

Isothermal processes occur when a system changes state while the temperature remains constant.- Since temperature does not change, the energy transfer only involves work and heat.- For an ideal gas, we use equations like \( W = nRT \ln \frac{V_f}{V_i} \) to calculate work done, under isothermal conditions.In the provided exercise, evaporation is isothermal, taking place at \( 100^{\circ}\text{C} \), equivalent to \( 373 \, \text{K} \).- Here, we calculated \( W = 3730 \, \text{cal} \) using \( W = nRT \), where \( n \) is moles of water, \( R \) the gas constant, and \( T \) the absolute temperature.Isothermality simplifies calculations since temperature is held constant, affecting how energy and work are distributed in the process.