The internal energy change (in J) When 90 g of
water undergoes complete evaporation at
100oC is ____________.
(Given : $\Delta$Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
Answer (integer)
189494
Solution
H<sub>2</sub>O(l) ⇌ H<sub>2</sub>O(g)
<br><br>90 gm of H<sub>2</sub>O = ${{90} \over {18}}$ moles of H<sub>2</sub>O = 5 moles of H<sub>2</sub>O
<br><br>$\Delta$H<sub>vap</sub> = $\Delta$U + $\Delta$n<sub>g</sub>RT
<br><br>$\Rightarrow$ $\Delta$U = $\Delta$H<sub>vap</sub> - $\Delta$n<sub>g</sub>RT
<br><br>= 41000 - 1$\times$8.314$\times$373
<br><br>= 37898.878
<br><br>For 5 moles, $\Delta$U = 37898.878 $\times$ 5 = 189494.39 Joule
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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