Question

Molar heat of vaporisation of a liquid is $H_{2}O(l) \rightarrow H_{2}O(g);P = 1atm$. If the entropy change is $T = 373K$, the boiling point of the liquid is.

• A. $\Delta G = 0$
• B. $H_{2}O(l)$
• C. $H_{2}O(g)$
• D. $373K$

Answer 1

Answer: (B)

$H_{2}O(l)$

Answer 2

$227^{o}C$

$(R = 2.0cal.mol^{- 1}K^{- 1})$