Assuming that water vapour is an ideal gas, the internal energy change (.Δ U.) 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- 1K- 1 ) will be :

Question

Assuming that water vapour is an ideal gas, the internal energy change (.Δ U.) 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- 1K- 1 ) will be : (A) 4.100kJmol- 1 (B) 3.7904kJmol- 1 (C) 37.904kJmol- 1 (D) 41.00kJmol- 1

Tardigrade Admin · Accepted Answer

H2O(. ℓ .) oversetvaporisation arrow H2O(. g .) Δng=1-0=1 Δ H=Δ U+Δ ngRT Δ U=Δ H-Δ ngRT =41-8.3× 10- 3× 373 =37.9kJmol- 1

Q. Assuming that water vapour is an ideal gas, the internal energy change $(Δ U)$ when $1 m o l e$ of water is vaporised at $1$ bar pressure and $100 o C$ , (given : molar enthalpy of vaporization of water $41 k J m o l^{- 1}$ at $1$ bar and $373 k$ and $R = 8.3 J m o l^{- 1} K^{- 1}$ ) will be :

$4.100 k J m o l^{- 1}$

$3.7904 k J m o l^{- 1}$

$37.904 k J m o l^{- 1}$

$41.00 k J m o l^{- 1}$

$H_{2} O_{(ℓ)} \to v a p or i s a t i o n H_{2} O_{(g)}$
$Δ n_{g} = 1 - 0 = 1$
$Δ H = Δ U + Δ n_{g} RT$
$Δ U = Δ H - Δ n_{g} RT$
$= 41 - 8.3 \times 1 0^{- 3} \times 373$
$= 37.9 k J m o l^{- 1}$

Q. Assuming that water vapour is an ideal gas, the internal energy change (ΔU) 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−1K−1 ) will be :

4.100kJmol−1

3.7904kJmol−1

37.904kJmol−1

41.00kJmol−1