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Thermodynamic Processes

At 127∘ C ,a ...

Question

At $127^{\circ} C$ ,a compound that exists in a liquid state has a vapour pressure of $380 t o r r$ . The same compound has a vapour pressure of $1.5 a t m$ at $227^{\circ} C$ .
Calculate the enthalpy of vapourisation $(Δ H_{v a p})$ for the compound.

33.41 k J m o l^{- 1}

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18.27 k J m o l^{- 1}

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2.11 k J m o l^{- 1}

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67.19 k J m o l^{- 1}

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Solution

The correct option is B $18.27 k J m o l^{- 1}$
Effect of temperature on the vapour pressure of a liquid is given by Clausius – Clapeyron equation.

$ln (\frac{P_{2}}{P_{1}}) = \frac{Δ H}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

where,

$Δ H is the enthalpy of vapourisation P_{1} is the vapour pressure of the liquid at temperature T_{1} P_{2} is the vapour pressure of the liquid at temperature T_{2}$
Here,
$P_{1} = 380 t o r r T_{1} = 127^{\circ} C = 400 K P_{2} = 1.5 a t m = 1.5 \times 760 t o r r = 1140 t o r r T_{2} = 227^{\circ} C = 500 K$

Putting values in the equation:

$ln (\frac{1140}{380}) = \frac{Δ H_{v a p}}{8.314 J m o l^{- 1} K^{- 1}} (\frac{1}{400} - \frac{1}{500})$

$ln (3) \times 8.314 \times 2000 = Δ H_{v a p}$

$Δ H_{v a p} = 18267.72 J m o l^{- 1} Δ H_{v a p} = 18.27 k J m o l^{- 1}$

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