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Question

For water at 100C and 1 bar,
ΔHvapΔUvapis x×102J mol1
The value of x (Round off to the Nearest Integer) is
[Use : R=8.31 mol1K1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) can be treated as an ideal gas]

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Solution

ΔHΔU=ΔngRT
Δng=10=1
ΔHΔU=1×8.31×373
ΔHΔU=30.99×102J mol1
ΔHΔU31×102J mol1
x=31

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