Question

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"`?

Answer 1

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.

Answer 2

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"`