1.

The boiling point of carbon tetrachloride is 77^(@)C and its heat of vaporisation is "31 kJ mol"^(-1). Calculate the vapour pressure of carbon tetrachloride in atmospheres at 25^(@)C.

Answer»

Solution :Applying Clausius - Clapeyron equation `log(P_(2))/(P_(1))=(Delta_("vap.")H)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`T_(1)=25^(@)C=25+273=298K,P_(1)=?`
`T_(2)=77^(@)C=77+273=350K, p_(2)="1 atm (PRESSURE at the b. pt.)"`
`Delta_("vap")H="31 kJ MOL"^(-1)="31000 J mol"^(-1),R=8.314"J K"^(-1)"mol"^(-1)`
`therefore""log.(1)/(P_(1))=(31000)/(2.303xx8.314)[(350-298)/(298xx350)]`
`"or"log.(1)/(P_(1))=0.8072"or"-logP_(1)=0.8072"or"logP_(1)=-0.8072=bar1.1928`
`therefore""P_(1)="Antilog "bar1.1928=0.1559 atm.`


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