# 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"?

Jun 2, 2016

The $\text{normal boiling point of } C {H}_{3} O H$ is $64.7$ ""^@C. You might have to restate this question.

#### Explanation:

What do I mean by $\text{normal boiling point}$? 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 $m m \cdot H g$ and $\text{inches} \cdot H g$ are units of length, but $760 \cdot m m \cdot H g$ and $25 \text{ inches Hg}$, 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)|)))" "

${P}_{1} = \text{25 mmHg";color(white)(ll) T_1 = "?}$
${P}_{2} = \text{760 mmHg"; T_2 = "(64.70 + 273.15) K" = "337.85 K}$

Δ_"vap"H = "35.21 kJ·mol"^"-1"; $R = \text{8.314 J·K"^"-1""mol"^"-1}$

 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")

$3.414 = \frac{\text{4235 K}}{T} _ 1 - 12.535$

$15.949 = \frac{\text{4235 K}}{T} _ 1$

${T}_{1} = \text{4235 K"/15.949 = "265.5 K" = "-7.6 °C}$