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The vapour pressure of water at 100^(@)C...

The vapour pressure of water at `100^(@)C` is `760mm`. What is the vapour pressure at `90^(@)C` if `triangle_(vap)H` of water is `41.25 kJ mol^(-1)`?

Text Solution

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`log""(P_2)/(P_1)=(triangle_(vap)H)/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2))) implies log""(P_2)/(769mm)= (41.25xx 10^(3)J mol^(-1))/(2.303 xx 8.314 JK^(-1)mol^(-1)) ((363K- 373 K)/((363K)(373K)))`
`log P_(2)-log 760 = -0.1591 implies logP_(2)= 2.8808 - 0.1591 implies (P_(2)= " antilog "2.7217)= 529.6 mm`.
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