Search any question & find its solution
Question:
Answered & Verified by Expert
The milar enthalpy change for $\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{H} 2 \mathrm{O}(\mathrm{g})$ at $373 \mathrm{~K}$ and $1 \mathrm{~atm}$ is $41 \mathrm{~kJ} / \mathrm{mol}$. Assuming ideal behavior, the internal energy change for vaporization of 1 mol of water at $373 \mathrm{~K}$ and 1 atm in $\mathrm{kJ} \mathrm{mil}^{-1}$ is
Options:
Solution:
2404 Upvotes
Verified Answer
The correct answer is:
$37.9$
$\quad \Delta \mathrm{E}=\Delta \mathrm{H}-\Delta(\mathrm{PV})=\Delta \mathrm{H}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.