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 $mol^{-1}$)
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 $mol^{-1}$)
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Work done during phase change is the difference between enthalpy and internal energy changes.
Updated On: Jun 6, 2026
- 3.1
- 3100
- 44.1
- 44100
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Relationship between $\Delta H$ and $\Delta U$: $\Delta H = \Delta U + P\Delta V$.
Step 2: Meaning
Work done $W = P\Delta V = \Delta H - \Delta U$.
Step 3: Analysis
$\Delta H = 41000$ J/mol. $\Delta U = 37900$ J/mol. $W = 41000 - 37900 = 3100$ J/mol.
Step 4: Conclusion
The work done is 3100 J/mol.
Final Answer: (B)
Relationship between $\Delta H$ and $\Delta U$: $\Delta H = \Delta U + P\Delta V$.
Step 2: Meaning
Work done $W = P\Delta V = \Delta H - \Delta U$.
Step 3: Analysis
$\Delta H = 41000$ J/mol. $\Delta U = 37900$ J/mol. $W = 41000 - 37900 = 3100$ J/mol.
Step 4: Conclusion
The work done is 3100 J/mol.
Final Answer: (B)
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