Question

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

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

Answer 1

Answer: (B)

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