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# Boiling point dependence on the altitude above sea level

Timur2015-07-30 09:13:10

After the creation of a of pressure calculator Pressure units converter and an atmospheric pressure calculator Barometric leveling I wanted to know how to calculate the boiling point according to the altitude. I've discovered that at a higher altitude water boils at a lower temperature. But what's that temperature?

This task consists of two stages - establish the atmospheric pressure dependence on the altitude and dependence of boiling point to pressure.

Boiling is a phase transition of the first order( water changes it's physical state from liquid to gas).

Phase transition of the first order is described by Clapeyron equation:
$\frac{dP}{dT}=\frac{q_{12}}{T(v_2-v_1)}$,
where
$q_{12}$ -the specific heat of the phase transition, which is numerically equal to the amount of heat received by a unit of mass for the phase transition.

$T$ - phase transition temperature
$v_2 - v_1$ - change of the specific volume in the transition

Clasius simplified the Clapeyron equation for the case of evaporation and sublimation, assuming that

1. The vapor obeys the law of ideal gas
2. The specific volume of fluid is much smaller than the specific volume of steam
It comes from the paragraph 1 that the state of vapor can be described by the Mendeleev-Clapeyron equation
$PV=\frac{M}{\mu} RT$,
and from the 2nd paragraph - the specific volume of fluid $v_1$can be neglected.
Thus, Clapeyron equation takes the following form
$\frac{dP}{dT}=\frac{q_{12}}{Tv}$,
where the specific volume can be expressed through
$v=\frac{V}{M}=\frac{RT}{P\mu}$,
and finally
$\frac{dP}{dT}=\frac{q_{12}\mu P}{RT^2}$
separating the variables, we obtain
$\frac{1}{P}dP=\frac{q_{12}\mu }{RT^2}dT$

By integrating the left part $P_1$ to $P_2$ and the right part from $T_1$ to $T_2$ i.e. from one point $(P_1,T_1)$ to another $(P_2,T_2)$, lying on the line liquid-vapor equilibrium, we obtain the following equation
$lnP_2-lnP_1=\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})$
called the Clausius-Clapeyron equation.

Actually, that is the desired dependence of the boiling temperature of the pressure

Here are some more transformations
$ln\frac{P_2}{P_1}=\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})$
$\frac{P_2}{P_1}=e^{\frac{q_{12}\mu }{R}(\frac{1}{T_1}-\frac{1}{T_2})}$,
where
$\mu$ - molar mass of the water, 18 gram/mol
$R$ -universal gas constant 8.31 J/(mol K)
$q_{12}$ - specific heat of water vaporisation $2.3$
10^6 J / kg

Now we have to to establish the dependence of the altitude to the atmospheric pressure. Here we will use the barometric formula (we don't have any other anyways):
$P=P_0e^{\frac{-\mu gh}{RT}}$
or
$\frac{P}{P_0}=e^{\frac{-\mu gh}{RT}}$,
where
$\mu$ - molar mass of the air , 29 gram/mol
$R$ - universal gas constant, 8.31 J/(mol K)
$g$ - acceleration of gravity, 9.81 m/(s
s)
$T$ - air temperature

We will mark the value relating to the air with index v and relating to the water with index h.
By equating and getting rid of the exponent, we will get
$-\frac{\mu_v gh}{RT_v}=\frac{q_{12}\mu_h }{R}(\frac{1}{T_0}-\frac{1}{T_h})$

And the final formula is
$T_h=\frac{T_0T_vq_{12}\mu_h}{q_{12}\mu_hT_v+\mu_vghT_0}$

Now, a small fly in the ointment - the actual air pressure don't follow the barometric formula as that with high altitude difference air temperature can not be considered as permanent. Moreover, the gravitational acceleration depends on the geographical latitude and atmospheric pressure - and also on the concentration of water vapor. So we can have only the estimate results with this formula.
Anyway, it won't be very different from the tables that you can find on the Internet

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