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 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

The internal energy change \( (\Delta U) \) is approximately \( 37.904 \text{ kJ} \text{ mol}^{-1} \).

Step 1: Understanding the relationship

The change in internal energy of a system, \( \Delta U \), is related to the change in enthalpy, \( \Delta H \), by the relation: \[ \Delta U = \Delta H - P \Delta V \]. Since we are dealing with an ideal gas, we can use the ideal gas law.

Step 2: Apply ideal gas law

For 1 mole of an ideal gas, the change in volume, \( \Delta V \), can be determined using the ideal gas law: \[ P \Delta V = nRT \]. Here, \( n = 1 \) mol, \( R = 8.3 J \cdot mol^{-1} \cdot K^{-1} \), \( T = 373 K \). Thus, \[ P \Delta V = (1)(8.3)(373) = 3086.9 \text{ J} \text{ mol}^{-1} = 3.0869 \text{ kJ} \text{ mol}^{-1} \].

Step 3: Calculate internal energy change

Now, using \( \Delta U = \Delta H - P \Delta V \): \[ \Delta U = 41 \text{ kJ} \text{ mol}^{-1} - 3.0869 \text{ kJ} \text{ mol}^{-1} = 37.9131 \text{ kJ} \text{ mol}^{-1} \].

Step 4: Choose the closest option

The value \( 37.9131 \text{ kJ} \text{ mol}^{-1} \) is very close to the option (b) \( 37.904 \text{ kJ} \text{ mol}^{-1} \). Therefore, option (b) is the best choice.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics that describes how gases behave under different conditions of temperature, pressure, and volume. It is frequently represented by the equation: \[ PV = nRT \]- **P** stands for pressure
- **V** denotes volume
- **n** is the number of moles
- **R** is the ideal gas constant
- **T** represents temperature This equation helps us understand the relationship between these variables for a perfect gas, assuming no interactions between gas molecules. When applied in calculations, like determining changes in volume or predicting behavior during expansion or compression, the Ideal Gas Law simplifies the complexity of real-world gases to a more manageable ideal case. This is particularly useful in scenarios where gases behave nearly ideally, such as high temperature and low pressure conditions, which closely align with the situation of water vapor in this context.

Internal Energy Change

Internal energy change \((\Delta U)\) is a crucial thermodynamic concept that reflects the energy difference within a system between two states. It accounts for all forms of energy within the system except external forces like work or heat flow. The expression for internal energy change in relation to enthalpy change is defined as:\[ \Delta U = \Delta H - P \Delta V \]This formula links the energy change in a system to its enthalpy change \(\Delta H\) and the pressure-volume work \(P \Delta V\). Enthalpy is often easier to measure, so this relation provides a practical pathway to determine internal energy. In our exercise involving water vapor, once we computed \(P \Delta V\) using the ideal gas law, the internal energy change was calculated. This highlights how energy is transferred and transformed within a system during vaporization.

Enthalpy of Vaporization

Enthalpy of vaporization is the heat needed to convert a substance from a liquid to a gas at constant pressure and temperature. For our case, the enthalpy of vaporization for water is given as 41 kJ/mol. This quantity indicates the energy required to break intermolecular forces in the liquid and allow molecules to escape into the gas phase. It is a specific form of enthalpy change associated with phase changes, reflecting the energy cost or benefit of transforming states without changing temperature. The enthalpy of vaporization is crucial for processes like boiling and evaporation. During vaporization, this energy is absorbed, leading to increased molecular freedom and spacing. Understanding this concept aids in comprehending energy transfers involved in phase transformations, and it connects directly with the internal energy calculations by illustrating how much energy is taken up solely for the phase change.