The normal boiling point of cyclopentane (enthalpy of vaporization: $28.7$ $k J m o l^{- 1}$ ) is ${49.25}^{\circ} C$ . What will be its vapour pressure at $25^{0} C$ ?

318.17 mm Hg

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112.4 mm Hg

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324.66 mm Hg

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120 mm Hg

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Solution

The correct option is A 318.17 mm Hg
Given $P_{1} = ?$ , $P_{2} = 760$ mm Hg;

$Δ H_{v a p} = 28700 J / m o l$

Applying Clausius Clapeyron's equation

log $\frac{P_{2}}{P_{1}} = \frac{Δ H_{v a p}}{2.303 \times R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$

or log $\frac{760}{P_{1}} = \frac{28700}{2.303 \times 8.314} (\frac{322.4 - 298.15}{322.4 \times 298.15})$

on solving, we get $P_{1} = 318.17 m m H g$

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The normal boiling point of cyclopentane (enthalpy of vaporization: 28.7 kJmol−1) is 49.25∘C. What will be its vapour pressure at 250C?

The correct option is A 318.17 mm HgGiven P1=?, P2=760 mm Hg;ΔHvap=28700J/molApplying Clausius Clapeyron's equation log P2P1=ΔHvap2.303×R(T2−T1T1T2)or log 760P1=287002.303×8.314(322.4−298.15322.4×298.15)on solving, we get P1=318.17 mmHg

The normal boiling point of cyclopentane (enthalpy of vaporization: $28.7$ $k J m o l^{- 1}$ ) is ${49.25}^{\circ} C$ . What will be its vapour pressure at $25^{0} C$ ?