Calculate the boiling point of water at $24 t o r r$ pressure. The average $△ H_{v a p .}$ over the temperature range is $10.12 k c a l m o l^{- 1}$ . Will all the water form 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

Pressure $= 24$ torr
$Δ H_{v a p} = 10.12$ kcal/mol
Using clausius-Clapeyron formula:
$l n (\frac{P_{2}}{P_{1}}) = \frac{- Δ H_{v a p}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
where; $P_{2}$ : vapour pressure at time $T_{1}$
$P_{1}$ : vapour pressure at time $T_{2}$
$Δ H_{v a p} =$ enthalpy
R: gas constant: $8.314$ J/kmol
Let us assume at $1$ atm pressure. vapour pressure of boiling point is $760$ torr
$l n (\frac{760}{24}) = \frac{- 10.12}{8.314} (\frac{1}{T_{2}} - \frac{1}{273})$
$3.45 = - 1.21 (\frac{273 - T_{2}}{273 T_{2}})$
$3.45 = \frac{- 330.33 + 1.21 T_{2}}{273 T_{2}}$
$79.13 T_{2} - 1.21 T_{2} = - 330.33$
$T_{2} = \frac{- 330.33}{77.92} = - 4.239$ .

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Calculate the boiling point of water at 24 torr pressure. The average △Hvap. over the temperature range is 10.12 kcal mol−1. Will all the water form 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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