For water Δ vap   H = 41   kJ   mol - 1 at 373   K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation o

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For water &#916; vap &#160; H = 41 &#160; kJ &#160; mol - 1 at 373 &#160; K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ( kJmol - 1 ) : [ Use : R = 8 . 3 &#160; J &#160; mol - 1 &#160; K - 1

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Correct answer: 38

H 2 O ( l ) &#8652; H 2 O ( g ) ; &#160;&#160; &#916;H vap .&#160; = 41 &#160; kJ / mol &#916;H = &#916;U + &#916;n g RT 41 = &#916;U + ( 1 ) 8 . 3 1000 &#215; 373 &#916; U = 41 - 3 . 0959 &#916;U = 37 . 9 &#160; kJ / mol &#160; &#8771; &#160; 38 &#160; kJ / mol

For water Δ vap   H = 41   kJ   mol - 1 at 373   K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ( kJmol - 1 ) : [ Use : R = 8 . 3   J   mol - 1   K - 1

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