Enthalpy of vaporisation of water is 40, 800 J/Mole at 1 atm. pressure. The mountain, where the pressure is 380 mm of H g, water will boil at

100° C

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81.6°C

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104° C

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50° C

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Solution

The correct option is B
81.6°C

$l o g \frac{P_{2}}{P_{1}} = \frac{Δ H_{v a p}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

or, $l o g \frac{380}{760} = \frac{40.8 \times 10^{3}}{2.303 R} [\frac{1}{373} - \frac{1}{T_{2}}]$

= $\frac{1}{373} - \frac{1}{T_{2}} = \frac{- 0.310 \times 2.303 \times 8.314}{40.8 \times 10^{3}}$

or, $T_{2} = 354.6 K$ or ${81.6}^{\circ} C$

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Enthalpy of vaporisation of water is 40, 800 J/Mole at 1 atm. pressure. The mountain, where the pressure is 380 mm of H g, water will boil at

The correct option is B 81.6°C logP2P1 = ΔHvap2.303R[1T1 − 1T2] or, log380760 = 40.8 × 1032.303 R[1373−1T2] = 1373 − 1T2 = −0.310 × 2.303 × 8.31440.8 × 103 or, T2 = 354.6K or 81.6∘C