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Normal boiling of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will be |
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Answer» 250 K `log.(P_(2))/(P_(1))=(Delta H_(V))/(2.303R)[(T_(2)-T_(1))/(T_(1)xxT_(2))]` `log.(760)/(23)=(40656)/(2.303xx8.314)[(373-T_(1))/(373T_(1))]` This gives `T_(1)=294.4 K`. |
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