For variations in Boiling suggest as a function of Vapor press use the Clausis-Clapeyron Equation ... However, because that variations in Vapor push as a duty of intrinsic physical and chemical properties Boiling Points room elevated for fluid phase substances with low Vapor push values when Boiling Points room low because that substances with relatively high Vapor push Values.


If a mentioned substance is subjected to variations in bordering atmospheric pressures, the boil Points will certainly decrease with decreasing atmospheric push values and also Increase with boosting atmospheric pressure values. If the boiling point values for a stated substance in ~ a specified BP temperature and pressure space given, one deserve to determine the Boiling point at various vapor press values utilizing the Clausis-Clapeyron Equation.

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The Clausis-Clapeyron Equation is derived from #VP_2 = VP_1e^(-DeltaH_v/(RT))# at two different temperatures. The classic type is ...

#ln((VP_2)/(VP_1))# = #(Delta H_v)/R##(1/T_1 - 1/T_2")#

Assume the following problems for water ... #BP_1 = T_1 = 373K# in ~ #VP_1 = 760mmHg#Determine #BP_2 = T_2# at #VP_2 = 700mmHg##R = 0.008314 (Kj)/("mole"*K)##Delta H_v = 40.66 (Kj)/"mole"#

#ln(700/760)# = #((40.66(Kj)/"mol")/(0.008314(Kj)/(mol*K)))(1/373-1/T_2)#

Solving this expression for #T_2 => 371K = BP_2#

However, if one is considering the boiling allude values of building material in regards to their chemical and also physical properties, the cook Points space inversely dependent upon the vapor pressure displayed by a liquid phase substance at a mentioned temperature. The problems here depends upon the an interpretation of boiling point of a fluid phase substance. The is, ...

by definition => Boiling Point the a provided substance is the temperature in ~ which the vapor pressure of a liquid substance is same to the bordering atmospheric pressure.

This method that in order to cook a liquid, enough heat need to be added to the body of the liquid to force enough molecules into the vapor phase such that its vapor pressure is equal to the surrounding atmospheric pressure. If, for example, the inter-molecular forces in between the liquid phase molecules space high, the liquid will certainly exhibit a short vapor press due to restricted evaporation. Consequently, a bigger quantity of warmth (higher temperature) have to be delivered into the fluid in bespeak to press its vapor press value high sufficient to same the surrounding atmospheric pressure and boil.

In contrast, a liquid substance v low inter-molecular pressures would show a high vapor pressure because the molecule at the liquid/atmosphere interface would evaporate much more easily and also thus achieve a vapor press equal come atmospheric pressure an ext easily. This means, the vapor pressure of the problem with lower inter-molecular pressures would with atmospheric vapor pressure with much less energy input and also boil in ~ a lower temperature.

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One can calculate the intrinsic boil boiling suggest of a liquid substance making use of Thermodynamic Properties and applying to the following expression to attain the "Thermodynamic boiling Point".

=> #DeltaH^o = TDeltaS^o# => #T_(bp)# = #(DeltaH^o)/(DeltaS^o) #

Thermodynamic Boiling allude of Water#(H_2O)#:(#DeltaH^o and also S^o# values space from traditional Thermodynamic nature Table. )

#DeltaH^o = - 285.8((Kj)/(mol))##DeltaH^o = - 241.8((Kj)/(mol))##H_2O(l) rightleftharpoons H_2O(g)#

#DeltaH^o(Rx) = DeltaH^o -DeltaH^o ##= -241.8((Kj)/(mol))(1 mol)+285.8((Kj)/(mol))(1mol#) #= 44Kj#

#DeltaS^o(Rx) = S^o(H_2O(g) - S^o(H_2O(l))##=188.7((j)/(mol*K)) - 69.95((j)/(mol*K))##= 118.75((j)/(mol*K)) = 0.11875(Kj)/(mol*K))#

#T_(bp)= ((44((Kj)/(mol))) /(0.11875((Kj)/(mol*K)))) = 370.53K##T_(bp)= (370.53 - 273)^oC = 97.5^oC#