Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar internal energy change at boiling point (373.15K).

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4.26 ${kJmol}^{- 1}$

3.75 ${kJmol}^{- 1}$

42.6 ${kJmol}^{- 1}$

37.5 ${kJmol}^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} ΔH = work done = i \times V \times t \\ = 0 .50 A \times 12 V \times 300 s Molar enthalpy of vaporisation, \\ = 1800 J = + 1 .8 kJ \end{array}$
Molar internal energy change, $\begin{array}{l} {ΔE}_{m} = {ΔH}_{m} - RT \\ = 40 .6 - 8 .314 \times 10^{- 13} \times 373 .1 \\ = 37 .5 kJ {mol}^{- 1} \end{array}$
$\begin{array}{l} {ΔH}_{n} = \frac{ΔH}{moles of H_{2} O} = \frac{ΔH}{n_{H_{2} o}} \\ = \frac{1 .8 kJ}{\frac{0 .798}{18}} = 40 .6 {kJmolm}^{- 1} \\ {ΔH}_{n} = {ΔE}_{m} + {Δn}_{g} RT \\ {ΔH}_{n} = {ΔE}_{m} + RT \\ [{Δn}_{g} = 1 for H_{2} O (l) ⇌ H_{2} O (g)] \end{array}$