For water at 100° C and 1 bar, Δ text vap H-Δ text vap U=× 102 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]

Question

For water at 100° C and 1 bar, Δ text vap H-Δ text vap U=× 102 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] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H 2 O (l) leftharpoons H 2 O (v) Δ H =Δ U+Δ ng R T for 1 mole waters; Δ ng=1 ∴ Δ ng R T=1 mol × 8.31 J / mol -k × 373 K =3099.63 J ≅ 31 × 102 J

Q. For water at 100∘C and 1 bar, Δvap ​H−Δvap ​U=________×102Jmol−1. (Round off to the Nearest Integer) [Use: R=8.31Jmol−1K−1 ] [Assume volume of H2​O(l) is much smaller than volume of H2​O(g). Assume H2​O(g) treated as an ideal gas]

H2​O(l)​⇌H2​O(v)​ ΔH=ΔU+Δng​RT for 1 mole waters; Δng​=1 ∴Δng​RT=1mol×8.31J/mol−k×373K =3099.63J≅31×102J

$H_{2} O_{(l)} ⇌ H_{2} O_{(v)}$
$Δ H = Δ U + Δ n_{g} RT$
for 1 mole waters;
$Δ n_{g} = 1$
$∴ Δ n_{g} RT = 1 m o l \times 8.31 J / m o l - k \times 373 K$
$= 3099.63 J ≅ 31 \times 1 0^{2} J$