1. Tardigrade
  2. Question
  3. Chemistry
  4. Assuming that water vapour is an ideal gas, the internal energy change (Δ U ) when 1 mole of water is vaporised at 1 bar pressure and 100° C, (given: molar enthalpy of vaporization of water 41 kJ mol -1 at 1 bar and 373 K and R =8.3 J mol -1 K -1 ) will be :

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Q. Assuming that water vapour is an ideal gas, the internal energy change $(\Delta U )$ when $1$ mole of water is vaporised at $1$ bar pressure and $100^{\circ} C$, (given : molar enthalpy of vaporization of water $41\, kJ$ $mol ^{-1}$ at 1 bar and $373\, K$ and $R =8.3\, J \,mol ^{-1}\, K ^{-1}$ ) will be :

Solution:

$H _{2} O _{(\ell)} \xrightarrow{\text { vaporisation }} H _{2} O _{(g)}$
$\Delta n _{ g }=1-0=1$
$\Delta H =\Delta U +\Delta n _{9} RT$
$\Delta U =\Delta H -\Delta n _{g} RT$
$=41-8.3 \times 10^{-3} \times 373$
$=37.9\, kJ \,mol ^{-1}$