What is the boiling point of methanol at 25 mmHg if its normal boiling point is 64.70 °C and #Δ_text(vap)H = "35.21 kJ·mol"^"-1"#?

What do I mean by #"normal boiling point"#? Namely that methanol has a vapour pressure of 1 atm precisely at this temperature - the liquid boils, and pushes back the atmospheric pressure, and bubbles of vapour form directly in the liquid. Now both #mm*Hg# and #"inches"*Hg# are units of length, but #760*mm*Hg# and #25" inches Hg"#, are the lengths of the columns of mercury that 1 atmosphere pressure will will support.

We can use the Clausius-Clapeyron equation to estimate the boiling point at another pressure.

#color(blue)(|bar(ul(color(white)(a/a) ln(P_2/P_1) = (Δ_"vap"H)/R( 1/T_1 -1/T_2) color(white)(a/a)|)))" "#

Your data are:

#P_1 = "25 mmHg";color(white)(ll) T_1 = "?"#
#P_2 = "760 mmHg"; T_2 = "(64.70 + 273.15) K" = "337.85 K"#

#Δ_"vap"H = "35.21 kJ·mol"^"-1"#; #R = "8.314 J·K"^"-1""mol"^"-1"#

# ln((760 color(red)(cancel(color(black)("mmHg"))))/(25 color(red)(cancel(color(black)("mmHg"))))) =("35 210" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))( 1/T_1 -1/"337.85 K")#

#3.414 = "4235 K"/T_1 - 12.535#

#15.949 = "4235 K"/T_1#

#T_1 = "4235 K"/15.949 = "265.5 K" = "-7.6 °C"#