Calculate the increase in vapour pressure of water for atmosphere rise in external pressure at 10 degree Celsius the vapour pressure of water at 10 degree Celsius and 1 atmosphere is equals to 92 millimetre

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Solution

Dear Priya Bhat,

$2.303 \log \frac{P 2}{P 1} = \frac{∆ H}{R} [\frac{T 2 - T 1}{T 1 T 2}] P 1 = 92 m m; P 2 = ? T 1 = 283 K; T 2 = 293 K ∆ H (h e a t o f v a p o r i z a t i o n o f w a t e r) = 540 c a l / g R = 2 c a l p u t t h e v a l u e s i n a b o v e e q u a t i o n 2.303 l o g \frac{P 2}{92} = \frac{540}{2} [\frac{293 - 283}{283 \times 293}] P 2 = 95 m m T h e V . P . w i l l i n c r e a s e b y 3 m m$
Regards

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