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Chemistry Question on Thermodynamics

One mole of water at 100∘C100^{\circ} C is converted into steam at 100∘C100^{\circ} C at a constant pressure of 1 atm.1\, atm . The change in entropy is [heat of vaporisation of water at 100∘C=540 cal/gm]:100{ }^{\circ} C =540\,cal / gm ]:

A

8.748.74

B

18.7618.76

C

24.0624.06

D

26.0626.06

Answer

26.0626.06

Explanation

Solution

The entropy change = heat of vaporisation  temperature =\frac{\text { heat of vaporisation }}{\text { temperature }} Here, heat of vaporisation =540 cal/gm=540 \,cal / gm =540×18 cal mol−1=540 \times 18\, cal\, mol ^{-1} Temperature of water =100+273=373 K=100+273=373\, K ∴\therefore entropy change =540×18373=\frac{540 \times 18}{373} =26.06 cal mol−1K−1=26.06\, cal\, mol ^{-1} K ^{-1}