One mole of water at 100° C is converted into steam at 100° C at a constant pressure of 1 atm . The change in entropy is [heat of vaporisation of water at 100 ° C =540 cal / gm ]:

Question

One mole of water at 100° C is converted into steam at 100° C at a constant pressure of 1 atm . The change in entropy is [heat of vaporisation of water at 100 ° C =540 cal / gm ]: (A)  8.74  (B)  18.76  (C)  24.06  (D)  26.06

Tardigrade Admin · Accepted Answer

The entropy change =( text heat of vaporisation / text temperature ) Here, heat of vaporisation =540 cal / gm =540 × 18 cal mol -1 Temperature of water =100+273=373 K ∴ entropy change =(540 × 18/373) =26.06 cal mol -1 K -1

Q. One mole of water at $10 0^{\circ} C$ is converted into steam at $10 0^{\circ} C$ at a constant pressure of $1 a t m .$ The change in entropy is [heat of vaporisation of water at $100^{\circ} C = 540 c a l / g m] :$

$8.74$

$18.76$

$24.06$

$26.06$

The entropy change
$= \frac{heat of vaporisation}{temperature}$
Here, heat of vaporisation $= 540 c a l / g m$
$= 540 \times 18 c a l m o l^{- 1}$
Temperature of water $= 100 + 273 = 373 K$
$∴$ entropy change $= \frac{540 \times 18}{373}$
$= 26.06 c a l m o l^{- 1} K^{- 1}$

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

8.74

18.76

24.06

26.06