# What is the new boiling temperature for bromine, if the pressure changes from atmospheric pressure to "76000 Pa"?

## To what temperature do we need to heat up bromine in order for it to boil on 76 000 Pa. At atmospheric pressure bromine boils on 58 degrees. Enthalpy of vaporization for bromine is $\text{194.26 J/g}$.

Jan 1, 2017

I got $\textcolor{b l u e}{{T}_{b 2} \approx {50.5}^{\circ} \text{C}}$, when I used ${T}_{b 1} = {58.8}^{\circ} \text{C}$ and your $\Delta {H}_{\text{vap}}$ of $\text{194.26 J/g}$, or $\Delta {\overline{H}}_{\text{vap" = "31.04 kJ/mol}}$. But that's not the $\Delta {\overline{H}}_{\text{vap}}$ I see here.

• If you use ${T}_{b 1} = {58}^{\circ} \text{C}$ and $\Delta {\overline{H}}_{\text{vap" = "31.04 kJ/mol}}$, you should get ${T}_{b 2} \approx {49.8}^{\circ} \text{C}$.
• If you use ${T}_{b 1} = {58}^{\circ} \text{C}$ and $\Delta {\overline{H}}_{\text{vap" = "29.96 kJ/mol}}$, you would get ${T}_{b 2} \approx {49.5}^{\circ} \text{C}$.
• If you use ${T}_{b 1} = {58.8}^{\circ} \text{C}$ and $\Delta {\overline{H}}_{\text{vap" = "29.96 kJ/mol}}$, you would get ${T}_{b 2} \approx {50.2}^{\circ} \text{C}$.
This demonstrates how good the assumption is that $\Delta {H}_{\text{vap}}$ doesn't vary in this temperature range.

So basically, you should get somewhere between ${49.5}^{\circ} \text{C}$ and ${50.5}^{\circ} \text{C}$.

Since we need to find a new temperature given two pressures, one temperature, and an enthalpy of vaporization $\Delta {H}_{\text{vap}}$, we should turn to the Clausius-Clapeyron equation in its natural logarithm form.

$\boldsymbol{\ln \left({P}_{2} / {P}_{1}\right) = \ln \left(\frac{P \left({T}_{b 2}\right)}{P \left({T}_{b 1}\right)}\right) = - \frac{\Delta {\overline{H}}_{\text{vap}}}{R} \left[\frac{1}{T} _ \left(b 2\right) - \frac{1}{T} _ \left(b 1\right)\right]}$,

where $P \left({T}_{b i}\right)$ is the vapor pressure of the substance at some specified boiling point ${T}_{b i}$, $\Delta {\overline{H}}_{\text{vap}}$ is the molar enthalpy of vaporization, in $\text{kJ"/"mol}$, and $R = \text{0.008314472 kJ/mol"cdot"K}$ is the universal gas constant.

To use it, we're solving for ${T}_{b 2}$. It's generally easier to plug in numbers early, actually.

Recall that atmospheric pressure is $\cancel{\text{1 atm" xx "101325 Pa"/cancel"atm" = "101325 Pa}}$. So, ${P}_{1} = \text{101325 Pa}$, and:

ln("76000 Pa"/("101325 Pa")) = -(194.26 xx 10^(-3) "kJ"/cancel"g" xx (2xx79.904 cancel"g")/("mol Br"_2))/("0.008314472 kJ/mol"cdot"K")[1/T_(b2) - 1/(58.8 + "273.15 K")]

$\implies \ln \left(0.7501\right) = - \left(31.044 \cancel{\text{kJ/mol")/(0.008314472 cancel"kJ/mol"cdot"K")[1/T_(b2) - 1/"331.95 K}}\right]$

= -"3733.77 K"[1/T_(b2) - 0.0030125 "K"^(-1)]

$\implies - \ln \frac{0.7501}{\text{3733.77 K" = 1/T_(b2) - 0.0030125 "K}} ^ \left(- 1\right)$

$\implies - \ln \frac{0.7501}{\text{3733.77 K" + 0.0030125 "K}} ^ \left(- 1\right) = \frac{1}{T} _ \left(b 2\right)$

$\implies {T}_{b 2} \approx \text{323.68 K}$,

or about $\textcolor{b l u e}{{50.5}^{\circ} \text{C}}$.

This should make physical sense because we say that boiling is easier at higher altitudes, at which the pressure is lower.

Lower pressure makes it easier to achieve the required vapor pressure, and thus, it is easier to boil at higher altitudes.

I derive the Clausius-Clapeyron equation below, if you want to take a look.

DISCLAIMER: DERIVATION BELOW!

I never remember the Clausius-Clapeyron equation, but I do remember the Clapeyron equation, so we can derive it by comparing with the Clapeyron equation:

$\boldsymbol{\frac{\mathrm{dP}}{\mathrm{dT}} = \frac{\Delta {H}_{\text{vap}}}{T \Delta {V}^{\left(l\right) \to \left(g\right)}}}$

where $\Delta {V}^{\left(l\right) \to \left(g\right)}$ is the change in volume due to vaporization, and $T$ is temperature in $\text{K}$. This equation tells you the slope of the liquid-vapor coexistence curve on a phase diagram.

Since a gas is much more compressible than a liquid, we can say that $\Delta {V}^{\left(l\right) \to \left(g\right)} \approx {V}_{g a s}$. Therefore:

$\frac{\mathrm{dP}}{\mathrm{dT}} \approx \frac{\Delta {H}_{\text{vap}}}{T {V}_{g a s}}$

If the gas is assumed ideal, then we can further substitute the ideal gas law to get that ${V}_{g a s} = \frac{n R {T}_{b}}{P}$, and:

$\frac{\mathrm{dP}}{\mathrm{dT}} \approx \frac{\Delta {H}_{\text{vap}} P}{n R {T}_{b}^{2}}$

$= \frac{\Delta {\overline{H}}_{\text{vap}} P}{R {T}_{b}^{2}}$

Recall that the Clausius-Clapeyron equation has $\frac{\mathrm{dl} n P}{\mathrm{dT}}$ on the left side. From the chain rule, $\frac{\mathrm{dl} n P}{\mathrm{dT}} = \frac{\mathrm{dl} n P}{\mathrm{dP}} \frac{\mathrm{dP}}{\mathrm{dT}} = \frac{1}{P} \frac{\mathrm{dP}}{\mathrm{dT}}$.

So, we obtain the Clausius-Clapeyron equation:

$\frac{1}{P} \frac{\mathrm{dP}}{\mathrm{dT}} = \textcolor{g r e e n}{\frac{\mathrm{dl} n P}{\mathrm{dT}} = \frac{\Delta {\overline{H}}_{\text{vap}}}{R {T}_{b}^{2}}}$

where DeltabarH_"vap" = (DeltaH_"vap")/n is the molar enthalpy of vaporization of ${\text{Br}}_{2} \left(l\right)$.

Now, when we integrate both sides (the left side varies from ${P}_{1}$ to ${P}_{2}$, and the right side varies from ${T}_{b 1}$ to ${T}_{b 2}$), and assume that $\Delta \overline{H}$ varies little in this temperature range, we get:

${\int}_{{P}_{1}}^{{P}_{2}} \mathrm{dl} n P$

$= {\int}_{{T}_{b 1}}^{{T}_{b 2}} \frac{\Delta {\overline{H}}_{\text{vap")/(RT_b^2)dT_b = (DeltabarH_"vap}}}{R} {\int}_{{T}_{b 1}}^{{T}_{b 2}} \frac{1}{{T}_{b}^{2}} {\mathrm{dT}}_{b}$

$\implies \boldsymbol{\ln \left({P}_{2} / {P}_{1}\right) = - \frac{\Delta {\overline{H}}_{\text{vap}}}{R} \left[\frac{1}{T} _ \left(b 2\right) - \frac{1}{T} _ \left(b 1\right)\right]}$

This is the equation we'd use to solve this problem.