The enthalpy of vaporization of substance is 840 J - mol -1 and its boiling point is -173° C. Its entropy of vaporization is:

Question

The enthalpy of vaporization of substance is 840 J - mol -1 and its boiling point is -173° C. Its entropy of vaporization is: (A) 42 J mol -1 K-1 (B) 21 J mol -1 K -1 (C) 84 J mol -1 K-1 (D) 8.4 J mol-1 K-1

Tardigrade Admin · Accepted Answer

Δ S=(Δ H/T) where Δ S= entropy of vaporisation. Δ H= enthalpyof vaporisation =840 J / mol T=-173° C=-173+273=100 K ∴ Δ S=(840/100)=8.4 J mol -1 K -1

Q. The enthalpy of vaporization of substance is $840 J - m o l^{- 1}$ and its boiling point is $- 17 3^{\circ} C$ . Its entropy of vaporization is:

$42 J m o l^{- 1} K^{- 1}$

$21 J m o l^{- 1} K^{- 1}$

$84 J m o l^{- 1} K^{- 1}$

$8.4 J m o l^{- 1} K^{- 1}$

$Δ S = \frac{Δ H}{T}$
where $Δ S =$ entropy of vaporisation.
$Δ H =$ enthalpyof vaporisation $= 840 J / m o l$
$T = - 17 3^{\circ} C = - 173 + 273 = 100 K$
$∴ Δ S = \frac{840}{100} = 8.4 J m o l^{- 1} K^{- 1}$

Q. The enthalpy of vaporization of substance is 840J−mol−1 and its boiling point is −173∘C. Its entropy of vaporization is:

42Jmol−1K−1

21Jmol−1K−1

84Jmol−1K−1

8.4Jmol−1K−1