Question

Calculate the work done when 1 mole of water at 373k vaporizes against 1 atmosphere. Assume ideal gas behavior.

• A. 3100 J
• B. -3100 J
• C. 3200 J
• D. -3200 J

Answer 1

The correct option is B

-3100 J

Volume of 1 mole of steam at 1 atm pressure. (As we know that $PV=nRT$)

$V=\frac{nRT}{P}=\frac{1\times 8.314\times 373}{101.3\times {10}^3}$

$\therefore V=30.6\times {10}^{-3}{m}^3$
$\Delta V=volumeofsteam-volumeofwater$

$\Delta V=30.6\times {10}^{-3}$ - negligible ( As the volume of water is negligible)

We know that the work done can be given by $w=-P\times \Delta V$

$\therefore w=-101.3\times {10}^3\times 30.6\times {10}^{-3}$

$\therefore w=-3100J$