Question

Molar enthalpy change for vaporization of 1 mol of water at $100^{\circ}C$ is $41kJ~mol^{-1}$. Calculate the internal energy change ($kJ~mol^{-1}$) assuming ideal gas behavior. ________.

• A. 43.1
• B. 37.9
• C. -43.1
• D. -37.9
• E. 41.0

Hint:

$\Delta U$ is usually less than $\Delta H$ for vaporization.

Answer 1

The Correct Option is B

Step 1: Concept
Use the relation $\Delta H = \Delta U + \Delta n_g RT$.

Step 2: Meaning
$H_2O(l) \rightarrow H_2O(g)$. $\Delta n_g = 1 - 0 = 1$. $T = 373 K$, $R = 8.3 \times 10^{-3} kJ/K/mol$.

Step 3: Analysis
$41 = \Delta U + (1 \times 8.3 \times 10^{-3} \times 373)$. $41 = \Delta U + 3.09$.

Step 4: Conclusion
$\Delta U = 41 - 3.09 = 37.91 kJ/mol$. Final Answer: (B)