for water is + 40.79 kJ at 1 bar and 100°C. Change in internal energy for this vapourisation under same condition is ________ kJ . (Integer answer)

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Solution

Q.
$Δ_{vap} H^{Θ}$ for water is + 40.79 kJ $m o l^{- 1}$ at 1 bar and 100°C. Change in internal energy for this vapourisation under same condition is ________ kJ $m o l^{- 1}$ . (Integer answer)

(Given R = 8.3 $J K^{- 1}$ $m o l^{- 1}$ ) [2024]

Ans.
(38)

$H_{2} O (l) ⇌ H_{2} O (g)$

$Δ n_{g} = Gaseous moles of products - Gaseous moles of reactants$

$= 1 - 0 = 1$

$Δ H = Δ U + Δ n_{g} R T$

$40.79 {kJmol}^{- 1} = Δ U + 1 \times 8.3 \frac{J}{molK} \times 373 K$

$40.79 {kJmol}^{- 1} = Δ U + 1 \times \frac{8.3}{1000} \frac{kJ}{molK} \times 373 K$

$Δ U = 37.69 {kJmol}^{- 1} \approx 38 {kJmol}^{- 1}$

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