"For water at 100$^\circ$ C and 1 bar,$\Delta$vap H $-$ $\Delta$vap U = _____________ $\times$ 102 J mol$-$1. (Round off to the Nearest Integer)[Use : R = 8.31 J mol$-$1 K$-$1][Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]"

Question

Accepted Answer

"H₂O_(l) $\rightleftharpoons$ H₂O_(v)

$\Delta$H = $\Delta$U + $\Delta$n_gRT

For 1 mole waters;

$\Delta$n_g = 1

$\therefore$ $\Delta$n_gRT = 1 mol $\times$ 8.31 J/mol-k $\times$ 373 K

= 3099.63 J $\cong$ 31 $\times$ 10² J"

For water at 100$^\circ$ C and 1 bar,

$\Delta$_vap H $-$ $\Delta$_vap U = _____________ $\times$ 10² J mol^$-$1. (Round off to the Nearest Integer)

[Use : R = 8.31 J mol^$-$1 K^$-$1]

[Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) treated as an ideal gas]

Solution

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For water at 100$^\circ$ C and 1 bar,$\Delta$vap H $-$ $\Delta$vap U = _____________ $\times$ 102 J mol$-$1. (Round off to the Nearest Integer)[Use : R = 8.31 J mol$-$1 K$-$1][Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]

Solution

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More Thermodynamics questions

Drill 25 more like these. Every day. Free.

For water at 100$^\circ$ C and 1 bar,

$\Delta$_vap H $-$ $\Delta$_vap U = _____________ $\times$ 10² J mol^$-$1. (Round off to the Nearest Integer)

[Use : R = 8.31 J mol^$-$1 K^$-$1]

[Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) treated as an ideal gas]