Question

What is the work done (in J $mol^{-1}$) to vaporise 1 mole of $H_2O(l)$ to $H_2O(g)$ at 1 bar pressure and $100^{\circ}C$? ($\Delta_{vap}H = 41$ kJ $mol^{-1}$; $\Delta U = 37.9$ kJ $mol^{-1}$)

• A. 3.1
• B. 3100
• C. 44.1
• D. 44100

Hint:

Work done during phase change is the difference between enthalpy and internal energy changes.

Answer 1

The Correct Option is B

Step 1: Connect $\Delta H$, $\Delta U$ and work.
For a process at constant pressure, the first law gives \[ \Delta H = \Delta U + P\Delta V \] Here $P\Delta V$ is the expansion work done by the system.

Step 2: Pick out the work term.
Rearranging, the work done by the gas as the liquid turns to vapour is \[ W = P\Delta V = \Delta H - \Delta U \]
Step 3: Make the units match.
$\Delta_{vap}H = 41$ kJ/mol $= 41000$ J/mol and $\Delta U = 37.9$ kJ/mol $= 37900$ J/mol.

Step 4: Subtract.
\[ W = 41000 - 37900 = 3100 \text{ J mol}^{-1} \]
Step 5: Quick sanity check.
The liquid expands into a large vapour, so positive work is done against the $1$ bar pressure. A value of about $3100$ J fits $P\Delta V \approx RT$ at $373$ K.

Step 6: Conclusion.
So the work done is $3100$ J/mol. \[ \boxed{3100\ \text{J mol}^{-1}} \]