Question

Calculate the work done when $2.5mol$ of $H_{2}O$ vaporizes at $1.0atm$ and $25^{\circ }C$. Assume the volume of liquid $H_{2}O$ is negligible compared to that of vapour. Given $1L$ atm $=101.3J$ and $R=0.082Latm{\text{mol}}^{-1}{K}^{-1}$.

• A. 6190 kJ
• B. 6.19 kJ
• C. 61.1 kJ
• D. 5.66 kJ

Answer 1

Answer: (B)

Volume of water vapour at $1.0atm$ and $298K$ is given by
$V=\frac{nRT}{P}$
$=\frac{2.5mol\times 0.082Latmmo{l}^{-1}{K}^{-1}\times 298K}{1atm}$
$=61.09L$
Now change volume
$\Delta V=V_{\text{final }}-V_{\text{initial }}$
$=61.09L-0L=61.09L$
So work done against constant pressure of $1atm=P\Delta V$
$=1atm\times 61.09L=61.09Latm$
$=61.09Latm\times \frac{101.3J}{1Latm}$
$=6.19kJ$