He normal boiling point of liquid butanol is 391 k. assuming that its molar heat of vaporization is constant at 42.3 kj/mol, the boiling point of c4h9oh when the external pressure is 1.39 atm is

401 K
   The equation that expresses the vapor pressure of a liquid is the Clapeyron equation:
 Tb = (1/T0 - (R ln(P/P0))/Hvap)^(-1)
 where
 Tb = boiling point at pressure of interest
 R = Ideal gas constant
 P = vapor pressure of the liquid at the pressure of interest
 P0 = known pressure at T0
 Hvap = heat of vaporization
 T0 = boiling temperature
   So let's substitute the known values and calculate.
 Tb = (1/T0 - (R ln(P/P0))/Hvap)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol) ln(1.39atm/1atm))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol) ln(1.39))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol)*0.329303747)/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (2.73798276760652x10^-3 kJ/(K*mol))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (6.47277250025181x10^-5 1/K))^(-1)
 Tb = (2.49281703203073x10^-3 1/K))^(-1) Tb = 401.152586471767 K Rounding to 3 significant figures gives a boiling point of 401 K

I’m guessing you’re more alert to yourself or your surroundings if you’re more concentrated