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At 1 bar pressure and 373K, the enthalpy change for the vaporization of 1 mol of water is $41 kJ {mol}^{- 1} .$ The change in internal energy for the same change under the same conditions $(in kJ {mol}^{- 1})$ is
( $R = 8.3 {JK}^{- 1} {mol}^{- 1},$ Assume water vapor as an ideal gas)

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$- 37.9$

$+ 3.1$

$+ 37.9$

$+ 379$

answer is B.

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Detailed Solution

$∆ H = ∆ E + ∆ n_{g} RT$

$∆ E = ∆ H - ∆ n_{g} RT$

$∆ E = 41 - 8.3 \times 10^{- 3} \times 373 = 3.0959 kJ ≅ 3.1 kJ$

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