Question

One mole of water at $100^{\circ} C$ is converted into steam at $100^{\circ} C$ at a constant pressure of $1\, atm .$ The change in entropy is [heat of vaporisation of water at $100{ }^{\circ} C =540\,cal / gm ]:$

• A. $ 8.74 $
• B. $ 18.76 $
• C. $ 24.06 $
• D. $ 26.06 $

Answer 1

Answer: (D)

The entropy change
$=\frac{\text { heat of vaporisation }}{\text { temperature }}$
Here, heat of vaporisation $=540 \,cal / gm$
$=540 \times 18\, cal\, mol ^{-1}$
Temperature of water $=100+273=373\, K$
$\therefore $ entropy change $=\frac{540 \times 18}{373}$
$=26.06\, cal\, mol ^{-1} K ^{-1}$