Problem 164 - A Complete Resource Book in Chemistry for JEE Main

Question

Asuming that water vapour is an ideal gas, the internal energy change \((\Delta U)\) when \(1 \mathrm{~mol}\) of water is vapourized at 1 bar pressure and \(100^{\circ} \mathrm{C}\), (Given: Molar enthalpy of vaporization of water at 1 bar and \(373 \mathrm{~K}\) \(=41 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\left.\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) will be \(\quad\) [2007] (a) \(3.7904 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(37.904 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(41.00 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(4.100 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

\(41 \text{ kJ/mol}\) (option c) is the internal energy change.

Step 1: Understand the Relationship

For an ideal gas, the relationship between the change in internal energy \( \Delta U \) and the change in enthalpy \( \Delta H \) is given by:\[\Delta H = \Delta U + \Delta (PV)\] where \( P \) is pressure and \( V \) is volume.

Step 2: Calculate \( \Delta (PV) \) using Ideal Gas Law

Since water vapour is assumed to be an ideal gas, we use the ideal gas equation \( PV = nRT \) to calculate \( \Delta (PV) \). For 1 mole, \( \Delta (PV) = R \Delta T \). Here the temperature change \( \Delta T = 0 \) since vaporization is at constant temperature (isothermal), thus \( \Delta (PV) = 0 \).

Step 3: Apply the Formula for Internal Energy Change

From Step 1, since \( \Delta (PV) = 0 \), the formula simplifies to \( \Delta U = \Delta H - R \Delta T = \Delta H - 0 = \Delta H \). Given that the molar enthalpy of vaporization \( \Delta H = 41 \text{ kJ/mol} \), \( \Delta U \) is also \( 41 \text{ kJ/mol} \).

Step 4: Verify the Calculation

Check the calculation: Since \( \Delta (PV) = 0 \) and we have a straightforward subtraction, our conclusion that \( \Delta U = 41 \text{ kJ/mol} \) is correct given that the conditions are at constant temperature and pressure.

Key Concepts

Internal Energy

Internal energy refers to the total energy contained within a thermodynamic system. This includes the energy due to the movement and interaction of molecules within the system. Internal energy, denoted as \( U \), is a crucial concept in thermodynamics as it symbolizes a system's ability to do work or transfer heat. It is dependent on the state of the system and can change through interactions with surroundings.

When examining isothermal processes, the internal energy changes are typically considered. However, for an ideal gas, internal energy changes are specifically linked to temperature changes. Since vaporization in this problem occurs isothermally (at constant temperature), there's no change in internal energy contributed by temperature, leading us to focus on the enthalpy changes.

In the context of the given exercise, the calculation of internal energy change relies heavily on understanding the relationship between enthalpy change and internal energy. This concept is pivotal when considering phase changes like vaporization.

Enthalpy of Vaporization

The enthalpy of vaporization is the heat required to convert a unit mole of a substance from liquid to vapor at constant temperature and pressure. It's a measure of the energy needed to overcome intermolecular forces in a liquid, allowing it to transform into a gas.

For water at 1 bar pressure and 100°C, the molar enthalpy of vaporization is given as \(41\ \text{kJ/mol}\). This serves as a direct input in the calculation of internal energy changes during the vaporization process.

Understanding how to use the enthalpy of vaporization is crucial when calculating energy changes within a system undergoing a phase change. It also helps in explaining why vaporization requires energy input, highlighting the breakage of molecular attractions within liquid water.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle described by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. This law provides a simple description of the behavior of gases under various conditions.

In the exercise, when assuming water vapor behaves as an ideal gas, this law is used to calculate changes in the pressure-volume product, \( \Delta(PV) \). Since the process is isothermal (constant temperature), \( \Delta T = 0 \) and thus \( \Delta(PV) = R \Delta T = 0 \). This simplifies the equation related to changes in internal energy and helps streamline the calculations.

Understanding the application of the Ideal Gas Law is key in thermodynamics, as it helps predict gas behavior accurately under specified conditions, as long as the gas behaves ideally.

Isothermal Processes

Isothermal processes are thermodynamic processes in which the temperature remains constant. This results in the energy exchanges being primarily in the form of work done on or by the system due to a consistent temperature. Isothermal conditions are ideal for analytically manageable calculations of energy transformations.

In the original exercise, vaporization occurs isothermally, meaning there is no change in temperature \( \Delta T = 0 \). This implies that the ideal gas calculations for changes in pressure-volume work \( \Delta(PV) \) account for no additional work done when calculated.

Recognizing isothermal processes simplifies many thermodynamic problems, as it reduces the number of variables impacting a system, allowing more straightforward conclusions regarding energy and enthalpy changes to be drawn.