The molar enthalpy of vaporization of water at 1 bar and $100^\circ$C is 41 kJ mol$^{-1}$. What is the internal energy change when 1 mol of water is vaporized at 1 bar pressure and $100^\circ$C? Assume water vapor as a perfect gas. (R = 8.3 J K$^{-1}$ mol$^{-1}$)
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- $37.9$ kJ mol$^{-1}$
- $44.1$ kJ mol$^{-1}$
- $34.7$ kJ mol$^{-1}$
- $47.9$ kJ mol$^{-1}$
- $34.9$ kJ mol$^{-1}$
The Correct Option is A
Solution and Explanation
Step 1: The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is given by: \[ \Delta H = \Delta U + P\Delta V \] For 1 mole of an ideal gas, the volume change at constant pressure can be calculated as: \[ P\Delta V = R T \] Step 2: Given: \[ \Delta H = 41 { kJ mol}^{-1}, \quad R = 8.3 { J K}^{-1} { mol}^{-1}, \quad T = 373 { K} \] \[ P\Delta V = (8.3 \times 373) \times 10^{-3} { kJ} \] Step 3: Computing the expansion work: \[ P\Delta V = 3.1 { kJ} \] Step 4: Substituting in the equation: \[ \Delta U = 41 - 3.1 = 37.9 { kJ mol}^{-1} \] Step 5: Therefore, the correct answer is (A).
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