The latent heat of vaporization of water is about $2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$ and the vapor density is $0.598 \mathrm{~kg} / \mathrm{m}^{3}$ at $100^{\circ} \mathrm{C}$. Find the rate of change of the boiling temperature with altitude near sea level in ${ }^{\circ} \mathrm{C}$ per $\mathrm{km}$. Assume the temperature of the air is $300 \mathrm{~K}$.

(Density of air at $0^{\circ} \mathrm{C}$ and 1 atm is $1.29 \mathrm{~kg} / \mathrm{m}^{3}$ ).
The Boltzmann distribution gives the pressure change with height:

$$
p(z)=p(0) \exp -\frac{m g z}{k T},
$$

where $p(0)$ is the pressure at sea level $z=0, m$ is the molecular weight of air, and $T_{0}=300 \mathrm{~K}$ is the temperature of the atmosphere. The ClausiusClapeyron equation can be written as

$$
\frac{d p}{d T}=\frac{L}{T\left(V_{2}-V_{1}\right)}=\frac{L}{T M\left(\frac{1}{\rho_{2}}-\frac{1}{\rho_{1}}\right)}=\frac{\alpha}{T} .
$$

with $\rho_{1}=1000 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{2}=0.598 \mathrm{~kg} / \mathrm{m}^{3}$ and $L / M=2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$, we have

$$
\alpha=\frac{L \rho_{1} \rho_{2}}{M\left(\rho_{1}-\rho_{2}\right)}=1.40 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} .
$$

So the rate of change of the boiling point with height is

$$
\frac{d T}{d z}=\frac{d T}{d p} \cdot \frac{d p}{d z}=\frac{T}{\alpha} \cdot\left(\frac{-m g}{k T_{0}}\right) p(z) .
$$

Using the equation of state for ideal gas $p=\rho k T_{0} / m$, we have near the sea level

$$
\frac{d T}{d z}=-\rho g T(0) / \alpha,
$$

where $\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}$ is the density of air, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ and $T(0)=100^{\circ} \mathrm{C}$.

Thus $\frac{d T}{d z}=-0.87^{\circ} \mathrm{C} / \mathrm{km}$.