# Question e3d6d

Apr 11, 2015

I would say that the difference in boiling points is due to an increase in external pressure.

A liquid boils when its vapor pressure is equal to the external pressure - more often than not, this is the atmospheric pressure.

So, at normal atmospheric pressure, or 1 atm, ethanol boils at ${78.3}^{\circ} \text{C}$. In this case, the heat you provide is enough to make the vapor pressure equal the atmospheric pressure $\to$ boiling takes place.

If the atmospheric pressure is higher, the vapor pressure will have to be higher as well, since boiling occurs only when vapor pressure equals atmospheric pressure.

As a result, you'll need to supply more heat to the sample, which means it will boil at a higher temperature.

In your case, the first sample boils at ${79}^{\circ} \text{C}$, which means that the atmospheric pressure is very, very close to 1 atm, while the second sample boils at ${81}^{\circ} \text{C}$, which implies that the atmosperic pressure is a little higher than 1 atm.

SIDE NOTE

You can determine this mathematically by using the Clausius-Clapeyron equation

$\ln \left({P}_{2} / {P}_{1}\right) = \frac{\Delta {H}_{\text{vap}}}{R} \cdot \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)$, where

${P}_{1}$, ${P}_{2}$ - the vapor pressure at boiling point ${79}^{\circ} \text{C}$ and ${81}^{\circ} \text{C}$, respectively;
$\Delta {H}_{\text{vap}}$ - the enthalpy of vaporization for ethanol - $\text{38.56 kJ/mol}$;
${T}_{1}$, ${T}_{2}$ - the boiling points of the first and of the second sample - expressed in Kelvin.

For the sake of argument, let's assume that, at 1 atm, ethanol boils at ${79}^{\circ} \text{C}$. Since ${T}_{2}$ is bigger than ${T}_{1}$, you'd expect ${P}_{2}$ to be bigger than ${P}_{1}$. Indeed it is

ln(P_2/1) = (38560cancel("J")/cancel("mol"))/(8.314cancel("J")/(cancel("K")cancel("mol"))) * (1/(352.15) - 1/(354.15))cancel("K")#

$\ln \left({P}_{2}\right) = 0.06917 \implies {P}_{2} = \text{1.072 atm}$

Higher atmospheric pressure, higher boling point.