Calculate the molar heat of vaporization of a fluid whose vapor pressure doubles when the temperature is raised from 75°C to 100°C ?

1 Answer
Truong-Son N.
May 3, 2018

#DeltabarH_(vap) ~~ "29.95 kJ/mol"#

The variation of vapor pressure #P_i# with the desired boiling temperature #T_i# is given by the Clausius-Clapeyron equation:

#ln (P_2/P_1) = -(DeltabarH_(vap))/R[1/T_2 - 1/T_1]#

where #DeltabarH_(vap)# is the molar enthalpy of vaporization and #R = "8.314472 J/mol"cdot"K"# is the universal gas constant.

Since we want to see the vapor pressure double due to raising the temperature from #75^@ "C"# and #100^@ "C"#, let:

  • #P_2 = 2P_1#
  • #T_1 = 75 + "273.15 K" = "348.15 K"#
  • #T_2 = 100 + "273.15 K" = "373.15 K"#

Therefore:

#ln(2) = -(DeltabarH_(vap))/("8.314 J/mol"cdotcancel"K")[1/(373.15 cancel"K") - 1/(348.15 cancel"K")]#

#= 2.315 xx 10^(-5)cdotDeltabarH_(vap) " ""mol"/"J"#

As a result:

#color(blue)(DeltabarH_(vap)) = (ln2)/(2.315 xx 10^(-5)) "J"/"mol"#

#=# #"29946 J/mol"#

#=# #color(blue)("29.95 kJ/mol")#

This kind of fluid would have moderate dipole-dipole forces, as water has #DeltabarH_(vap) = "40.67 kJ/mol"# at its boiling point.

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