Question

Assuming that water vapour is an ideal gas, the internal energy change $\left(\right.\Delta U\left.\right)$ when $1mole$ of water is vaporised at $1$ bar pressure and $100oC$ , (given : molar enthalpy of vaporization of water $41kJmol^{- 1}$ at $1$ bar and $373k$ and $R=8.3Jmol^{- 1}K^{- 1}$ ) will be :

• A. $4.100kJmol^{- 1}$
• B. $3.7904kJmol^{- 1}$
• C. $37.904kJmol^{- 1}$
• D. $41.00kJmol^{- 1}$

Answer 1

Answer: (C)

$H_{2}O_{\left(\right. \ell \left.\right)}\overset{vaporisation}{ \rightarrow }H_{2}O_{\left(\right. g \left.\right)}$
$Δn_{g}=1-0=1$
$\Delta H=\Delta U+\Delta n_{g}RT$
$\Delta U=\Delta H-\Delta n_{g}RT$
$=41-8.3\times 10^{- 3}\times 373$
$=37.9kJmol^{- 1}$