The correct relationship between enthalpy (in cal mol $^{- 1}$ ) of vaporisation of liquid and its boiling points is?

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Solution

It is important to know that under normal conditions, everything boils at $1 a t m .$ A liquid boils when its vapor pressure equals the exterior pressure, and since we're at $1 a t m,$ that is what the vapor pressure for a substance at its boiling point will be,
The most useful form of the Clausius-Clapeyron is
$ln \frac{P_{2}}{P_{1}} = - \frac{Δ H v a p}{R} \frac{1}{T_{2}} - \frac{1}{T_{1}}$ ,where

$P_{1}$ - the vapor pressure of that substance at $T_{1}$ ;
$P_{2}$ - the vapor pressure of that substance at $T_{2}$ ;
$Δ H v a p$ - the enthalpy of vaporization.
$R$ - the gas constant - usually expressed as $8.314 \frac{J}{K} \times m o l$

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Similar questions

30 k J / m o l

75 J m o l^{- 1} K^{- 1}

1 a t m

26.0 k J m o l^{- 1}

{35.0}^{o} C .

△ S

{35.0}^{o} C

J K^{- 1} m o l^{- 1}

The enthalpy of vaporisation of liquid is 30KJ/mol and entropy of vaporisation is 75J/mol/k. The Boiling point of the liquid at 1 atm is

35^{\circ} C

24.64 k J {m o l}^{- 1}

{100.1}^{\circ} C

80

g^{- 1}

540

g^{- 1}

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