Entropy of vaporisation of water at 100° C, if molar heat of vaporisation is 9710 cal mol -1, will be:

Question

Entropy of vaporisation of water at 100° C, if molar heat of vaporisation is 9710 cal mol -1, will be: (A) 20 cal mol -1 K-1 (B) 26 cal mol -1 K -1 (C) 24 cal mol -1 K-1 (D) 28 cal mol -1 K -1

Tardigrade Admin · Accepted Answer

Use the following formula to solve the problem

Δ S = \frac{Δ H}{T}

where

Δ S =

entropy

= ?

Δ H =

enthalpy

= 9710 c a l m o l^{- 1}

T =

temperature

= 100^{\circ} C

= 100 + 273 = 373 K

∴ Δ S = \frac{9710}{373}

= 26.032 c a l m o l^{- 1} K^{- 1}

\approx 26 c a l m o l^{- 1} K^{- 1}

Q. Entropy of vaporisation of water at $10 0^{\circ} C$ , if molar heat of vaporisation is $9710 c a l m o l^{- 1}$ , will be:

$20 c a l m o l^{- 1} K^{- 1}$

$26 c a l m o l^{- 1} K^{- 1}$

$24 c a l m o l^{- 1} K^{- 1}$

$28 c a l m o l^{- 1} K^{- 1}$

Q. Entropy of vaporisation of water at 100∘C, if molar heat of vaporisation is 9710calmol−1, will be:

20calmol−1K−1

26calmol−1K−1

24calmol−1K−1

28calmol−1K−1

Use the following formula to solve the problem ΔS=TΔH​ where ΔS= entropy =? ΔH= enthalpy =9710calmol−1 T= temperature =100∘C =100+273=373K ∴ΔS=3739710​ =26.032calmol−1K−1 ≈26calmol−1K−1

Q. Entropy of vaporisation of water at $10 0^{\circ} C$ , if molar heat of vaporisation is $9710 c a l m o l^{- 1}$ , will be:

$20 c a l m o l^{- 1} K^{- 1}$

$26 c a l m o l^{- 1} K^{- 1}$

$24 c a l m o l^{- 1} K^{- 1}$

$28 c a l m o l^{- 1} K^{- 1}$