Calculate the boiling point of water at 24 torr pressure. The average $Δ H_{v a p .}$ over the temperature range is 10.12 kcal ${m o l}^{- 1}$ . Will all the water from gaseous state if placed in a closed container, if not then, how we can convert all the water into vapour state?

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Solution

From Tronton's law
$ln P = \frac{- Δ H_{v a p}}{R T} + ln K$
At, normal $B . p t$
$p = 1$ atm, & $T_{B . P} = 373 K$
$∴ ln 1 = \frac{- Δ H_{v a p}}{R \times 373} + ln K$
$\Rightarrow ln K = \frac{10.21 \times 10^{3}}{2 \times 373} = 13.56$
$∴ ln K = 13.56$
Now, at $p = 24 / 760 a t m$
$ln 24 / 760 = \frac{- 10.12 \times 10^{3}}{2 \times T_{B . P}} + 13.56$
$\Rightarrow \frac{10.12 \times 10^{3}}{2 \times T_{B . P}} = 17.01$
$\Rightarrow T_{B . P} = 297.38 K$

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Calculate the boiling point of water at 24 torr pressure. The average ΔHvap. over the temperature range is 10.12 kcal mol−1. Will all the water from gaseous state if placed in a closed container, if not then, how we can convert all the water into vapour state?

From Tronton's lawlnP=−ΔHvapRT+lnKAt, normal B.ptp=1 atm, & TB.P=373 K∴ ln1=−ΔHvapR×373+lnK⇒ lnK=10.21×1032×373=13.56∴ lnK=13.56Now, at p=24/760 atmln24/760=−10.12×1032×TB.P+13.56⇒ 10.12×1032×TB.P=17.01⇒ TB.P=297.38 K

Calculate the boiling point of water at 24 torr pressure. The average $Δ H_{v a p .}$ over the temperature range is 10.12 kcal ${m o l}^{- 1}$ . Will all the water from gaseous state if placed in a closed container, if not then, how we can convert all the water into vapour state?