Question

Standard enthalpy of vapourisation $\Delta_{\text {vap }} H ^{\Theta}$ for water at $100^{\circ} C$ is $40.66\, kJ\, mol ^{-1}$. The internal energy of vaporisation of water at $100^{\circ} C$ (in $kJmol ^{-1}$) is : (Assume water vapour to behave like an ideal gas)

• A. 43.76
• B. 40.66
• C. 37.56
• D. -43.76

Answer 1

The Correct Option is C

  To determine the internal energy of vaporization of water at \(100^{\circ} C\), we need to relate the enthalpy change and internal energy change during the vaporization process. This can be done using the first law of thermodynamics. For phase changes involving gases at constant pressure, the relationship between the enthalpy change \((\Delta H)\) and the internal energy change \((\Delta U)\) is given by:

\(\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \Delta V\)

Where:

• \(\Delta_{\text{vap}} H\) is the enthalpy of vaporization.
• \(\Delta_{\text{vap}} U\) is the internal energy of vaporization.
• \(P \Delta V\) is the work done on the system, which can be expressed using the ideal gas law as \(nRT\) at constant temperature.

For water at \(100^{\circ} C\), the pressure is 1 atm, and assuming 1 mole of water, the gas constant \(R = 8.314 \, J \, K^{-1} \, mol^{-1}\) and temperature \(T = 373 \, K\). Calculate \(P\Delta V\) using:

\(P \Delta V = RT = (8.314 \, J \, K^{-1} \, mol^{-1})(373 \, K)\)

Converting to \(kJ\), we have:

\(P \Delta V = \frac{8.314 \times 373}{1000} = 3.10 \, kJ \, mol^{-1}\)

Now substitute into the equation to find the internal energy of vaporization:

\(40.66 = \Delta_{\text{vap}} U + 3.10\)

Solving for \(\Delta_{\text{vap}} U\) gives:

\(\Delta_{\text{vap}} U = 40.66 - 3.10 = 37.56 \, kJ \, mol^{-1}\)

Therefore, the internal energy of vaporization of water at \(100^{\circ} C\) is 37.56 kJ mol^-1. This supports the answer being option:

37.56

.