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"?

2 Answers
Jun 2, 2016

The "normal boiling point of "CH_3OH is 64.7 ""^@C. You might have to restate this question.

Explanation:

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.

Jun 2, 2016

The boiling point of methanol at 25 mmHg is -8 °C.

Explanation:

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"