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Step-by-Step Solution
Step 1: Understand the given data
We are given the following information:
• Enthalpy of vaporization of water, \Delta_{\mathrm{vap}} H = 41 \text{ kJ mol}^{-1} at 373 K and 1 bar.
• The gas constant, R = 8.3 \text{ J mol}^{-1} \text{K}^{-1} .
• We assume that water vapor behaves ideally and the volume of vapor is much larger than that of liquid water.
Step 2: Recall the relationship between \Delta H and \Delta U
When a substance changes from liquid to gas under constant pressure, the change in enthalpy ( \Delta H ) is related to the change in internal energy ( \Delta U ) by the equation:
\Delta H = \Delta U + \Delta n_g \, R T
Here, \Delta n_g is the change in the number of moles of gas during the process, R is the gas constant, and T is the temperature.
Step 3: Determine \Delta n_g
The process is the vaporization of 1 mole of liquid water to gas:
\text{H}_2\text{O (l)} \rightarrow \text{H}_2\text{O (g)}
Initially, there are 0 moles of gas (liquid water does not count toward \Delta n_g ), and finally there is 1 mole of water vapor. Thus,
\Delta n_g = 1 - 0 = 1
Step 4: Calculate the term \Delta n_g \, R T
Substitute \Delta n_g = 1 , R = 8.3 \text{ J mol}^{-1} \text{K}^{-1} , and T = 373 \text{ K } :
\Delta n_g \, R T = 1 \times 8.3 \text{ J mol}^{-1} \text{K}^{-1} \times 373 \text{ K}
First convert Joules to kilojoules by dividing by 1000:
\Delta n_g \, R T = \frac{8.3 \times 373}{1000} \text{ kJ mol}^{-1}
\Delta n_g \, R T \approx 3.09 \text{ kJ mol}^{-1}
Step 5: Use the enthalpy-internal energy relation
We know:
\Delta H = 41 \text{ kJ mol}^{-1}
Plug the values into the equation:
41 \text{ kJ mol}^{-1} = \Delta U + 3.09 \text{ kJ mol}^{-1}
Therefore,
\Delta U = 41 - 3.09 \text{ kJ mol}^{-1}
\Delta U \approx 37.91 \text{ kJ mol}^{-1} \approx 38 \text{ kJ mol}^{-1}
Final Answer
\Delta U \approx 38 \text{ kJ mol}^{-1}
