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Assuming that water vapour is an ideal gas, the internal energy change $Δ U$ () when 1 mol of water is vaporised at 1 bar pressure and 100° C, (Given : Molar enthalpy of vaporization of water at 1 bar and $373 K = 41 kJmol {mol}^{- 1}$ and $R = 8.3 {Jmol}^{- 1} K^{- 1})$ will be:

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$37.904 kJ {mol}^{- 1}$

$41.00 kJ {mol}^{- 1}$

$4.100 kJ {mol}^{- 1}$

$3.7904 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

Here ,the change in internal energy,

$\begin{array}{l} Δ U = Δ H - Δ n_{g} R T; \\ Δ U = 41 - 1 \times \frac{8.3 \times 373}{1000} \\ = 37.904 kJ / mol \end{array}$

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