For water at 100 ° C and 1 bar, Δ v a p H - Δ v a p U = ¯ × 10 2 J mol - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31 J mol - 1 K - 1 [Assume volume of H 2 O ( l ) is much smaller than volume of H 2 O ( g

Question

For water at 100 &#176; C and 1 bar, &#916; v a p H - &#916; v a p U = &#175; &#215; 10 2 J mol - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31 J mol - 1 K - 1 [Assume volume of H 2 O ( l ) is much smaller than volume of H 2 O ( g ) . Assume H 2 O ( g ) treated as an ideal gas]

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H 2 O ( &#8467; ) &#8652; H 2 O ( v ) &#916;H = &#916;U + &#916;n g RT for 1 mole waters ; &#916;n g = 1 &#8756; &#160; &#916;n g RT = 1 &#160; mol &#215; 8 . 31 &#160; J / mol - k &#215; 373 &#160; K = 3099 . 63 &#160; J &#8773; 31 &#215; 10 2 &#160; J

For water at 100 ° C and 1 bar, Δ v a p H - Δ v a p U = ¯ × 10 2 J mol - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31 J mol - 1 K - 1 [Assume volume of H 2 O ( l ) is much smaller than volume of H 2 O ( g ) . Assume H 2 O ( g ) treated as an ideal gas]

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More from Thermodynamics (C)

For water at 100 &#176; C and 1 bar, &#916; v a p H - &#916; v a p U = &#175; &#215; 10 2 J mol - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31 J mol - 1 K - 1 [Assume volume of H 2 O ( l ) is much smaller than volume of H 2 O ( g ) . Assume H 2 O ( g ) treated as an ideal gas]

Correct answer

Step-by-step solution

More from Thermodynamics (C)

For water at 100 ° C and 1 bar, Δ v a p H - Δ v a p U = ¯ × 10 2 J mol - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31 J mol - 1 K - 1 [Assume volume of H 2 O ( l ) is much smaller than volume of H 2 O ( g ) . Assume H 2 O ( g ) treated as an ideal gas]