The enthalpy of vaporisation of a substances is 840 J mol- 1 and its boiling point is -173° C . Its enthalpy of vaporisation is

Question

The enthalpy of vaporisation of a substances is 840 J mol- 1 and its boiling point is -173° C . Its enthalpy of vaporisation is (A) 8.4 Jmol- 1K- 1 (B) 49 mol- 1K- 1 (C) 21 mol- 1K- 1 (D) 12 mol- 1K- 1

Tardigrade Admin · Accepted Answer

Δ Sv a p=(qr e v/T)=(Δ Hv a p/T) =(840/(. - 173 + 273 .) K) =8.4Jmol- 1K- 1 .

Q. The enthalpy of vaporisation of a substances is 840 J $m o l^{- 1}$ and its boiling point is $- 17 3^{\circ} C$ . Its enthalpy of vaporisation is

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

$49$ $m o l^{- 1} K^{- 1}$

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

$12$ $m o l^{- 1} K^{- 1}$

$Δ S_{v a p} = \frac{q _{re v}}{T} = \frac{Δ H _{v a p}}{T}$

$= \frac{840}{( - 173 + 273 ) K}$

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

Q. The enthalpy of vaporisation of a substances is 840 J mol−1 and its boiling point is −173∘C . Its enthalpy of vaporisation is

8.4 Jmol−1K−1

49 mol−1K−1

21 mol−1K−1

12 mol−1K−1