Question #e3d6d

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.

![http://www.800mainstreet.com/08/0008-0013-vapor_pres.html]()

So, at normal atmospheric pressure, or 1 atm, ethanol boils at `78.3^@"C"`. In this case, the heat you provide is enough to make the vapor pressure equal the atmospheric pressure `->` 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^@"C"`, which means that the atmospheric pressure is very, very close to 1 atm, while the second sample boils at `81^@"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(P_2/P_1) = (DeltaH_"vap")/R * (1/T_1 - 1/T_2)`, where

`P_1`, `P_2` - the vapor pressure at boiling point `79^@"C"` and `81^@"C"`, respectively;
`DeltaH_"vap"` - the enthalpy of vaporization for ethanol - `"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^@"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(P_2) = 0.06917 => P_2 = "1.072 atm"`

Higher atmospheric pressure, higher boling point.