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Chemistry · The first law of thermodynamics: Concept of work, heat, internal energy and enthalpy, heat capacity, molar heat capacity

The molar enthalpy change for \rightleftharpoons \boldsymbol{H}_{2} \boldsymbol{

The molar enthalpy change for \( \boldsymbol{H}_{2} \boldsymbol{O}(l) \rightleftharpoons \boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{g}) \) at \( \boldsymbol{3} \boldsymbol{7} \boldsymbol{3} K \) and \( \boldsymbol{1} \) at \( \boldsymbol{m} \) is \( 41 k J / \) mol. Assuming ideal behavior the internal energy change for vaporization of 1 mol of water at \( 373 K \) and 1 atm in \( \mathrm{kJ} \mathrm{mol}^{-1} \) is:

  • A. 30.2
  • B. 41.0
  • C. 48.1
  • D. 37.9

Step-by-step solution

Using the relation ΔH = ΔU + Δ(PV). For vaporization at constant pressure, assuming ideal gas behavior and neglecting liquid volume, Δ(PV) = RT. Thus ΔU = ΔH - RT = 41 kJ/mol - (8.314 J/mol·K × 373 K)/1000 = 41 - 3.10 = 37.9 kJ/mol.
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