Question

Molar enthalpy change for vaporization of 1 mol of water at $100^{\circ}C$ is $41kJ~mol^{-1}$. Calculate the internal energy change ($kJ~mol^{-1}$) assuming ideal gas behavior. ________.

• A. 43.1
• B. 37.9
• C. -43.1
• D. -37.9
• E. 41.0

Hint:

$\Delta U$ is usually less than $\Delta H$ for vaporization.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem explores the relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta U\)) for a physical process, specifically the vaporization of water. Enthalpy includes both the internal energy and the work done on or by the system due to volume changes at constant pressure.
Step 2: Key Formula or Approach
The fundamental relationship between enthalpy and internal energy is \(\Delta H = \Delta U + P\Delta V\). For processes involving gases that can be treated as ideal, this can be expressed as: \[ \Delta H = \Delta U + \Delta n_g RT \] where: - \(\Delta n_g\) is the change in the number of moles of gas during the process. - R is the ideal gas constant. - T is the absolute temperature. We can rearrange this formula to solve for \(\Delta U\).
Step 3: Detailed Explanation
1. Analyze the process and find \(\Delta n_g\).
The process is the vaporization of 1 mole of water: \[ H_2O(l) \to H_2O(g) \] The change in the number of moles of gas (\(\Delta n_g\)) is: \[ \Delta n_g = (\text{moles of gas, final}) - (\text{moles of gas, initial}) = 1 - 0 = 1 \] 2. Identify the given values and ensure consistent units.
- Molar enthalpy change, \(\Delta H = 41 \text{ kJ mol}^{-1}\). - Temperature, \(T = 100^\circ\text{C} = 100 + 273.15 \approx 373 \text{ K}\). - Gas constant, \(R = 8.3 \text{ J K}^{-1}\text{mol}^{-1}\). Since \(\Delta H\) is in kJ, we must convert R. \(R = 8.3 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1}\). 3. Calculate \(\Delta U\).
Rearrange the formula: \(\Delta U = \Delta H - \Delta n_g RT\). Substitute the known values: \[ \Delta U = 41 \text{ kJ mol}^{-1} - (1 \text{ mol}) \times (8.3 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1}) \times (373 \text{ K}) \] \[ \Delta U = 41 - (0.0083 \times 373) \] \[ \Delta U = 41 - 3.0959 \] \[ \Delta U \approx 37.9 \text{ kJ mol}^{-1} \] Since vaporization is an endothermic process where the system absorbs heat, both \(\Delta H\) and \(\Delta U\) are positive.
Step 4: Final Answer
The internal energy change for the vaporization of 1 mole of water is 37.9 kJ mol\(^{-1}\).