Question

Assuming that water vapour is an ideal gas, the internal energy change $\left(\Delta U\right)$ when 1 mol of water is vapourised at 1 bar pressure and 100${}^0$C will be:

[Given molar enthalpy of vapourisation of water at 1bar and 373 K=41 kJ mol${}^{-1}$]

• A. 41 kJ mol${}^{-1}$
• B. 4.1 kJ mol${}^{-1}$
• C. 3.7904 kJ mol${}^{-1}$
• D. 37.904 kJ mol${}^{-1}$

Answer 1

The correct option is D 37.904 kJ mol${}^{-1}$

$H_{2}O\left(l\right)\to H_{2}O\left(g\right)$

$\Delta H=\Delta U+\Delta n_{g}RT$

where

$\Delta H=$Enthalpy of vapourisation

$\Delta U=$ change in internal energy
$\Delta n_g=$ no of moles of water vapou$=1$

$R=$ Gas constant $=8.3$

$T=$ Temperature of veporisation$$ = 373 K$

$41kJ=\Delta U+RT$

$\Delta U=41-8.3\times 373\times {10}^{-3}$

$=41-3.095$

$=37.904$ kJ mol${}^{-1}$

Hence, the correct option is $D$