Question

1 g of water (volume $1{cm}^3$) becomes $1671{cm}^3$ of steam when boiled at a pressure of 1 atm. The latent heat of vaporization is 540 cal/g, then the external work done is:
($1atm=1.013\times {10}^{5}N/m^2$)

• A. 499.7 J
• B. 40.3 J
• C. 169.2 J
• D. 128.57 J

Answer 1

Answer: (C)

169.2 J

Work done, $W=p\Delta V$
$=1.013\times {10}^5\times (1671-1)\times {10}^{-6}$
$=1.013\times {10}^5\times 1670\times {10}^{-6}$
$=169.2J$