Subfield: Thermodynamics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Mountains have two sides: windward and leeward. The windward side faces the wind and typically receives warm, moist air, often from an ocean. As wind hits a mountain, it is forced upward and begins to move towards the leeward side. During social distancing, Rishab decides to cross a mountain from the windward side to the leeward side of the mountain. What he finds is that the air around him has warmed when he is on the leeward side of the mountain.

Let us investigate this effect. Consider the warm, moist air mass colliding with the mountain and moving upwards on the mountain. Disregard heat exchange with the air mass and the mountain. Let the humidity of the air on the windward side correspond to a partial vapor pressure $0.5 \mathrm{kPa}$ at $100.2 \mathrm{kPa}$ and have a molar mass of $\mu_{a}=28 \mathrm{~g} / \mathrm{mole}$. The air predominantly consists of diatomic molecules of oxygen and nitrogen. Assume the mountain to be very high which means that at the very top of the mountain, all of the moisture in the air condenses and falls as precipitation. Let the precipitation have a heat of vaporization $L=2.4 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}$ and molar mass $\mu_{p}=18.01 \mathrm{~g} / \mathrm{mole}$. Calculate the total change in temperature from the windward side to the leeward side in degrees Celsius.

Solution:
We use the first law of thermodynamics to solve this problem. For diatomic molecules, the internal energy per mole is given by $\frac{5}{2} R T$. If the molar mass of the air is $\mu_{a}$, then we have that the change in internal energy of the air is given by
$$
\Delta U=\frac{5}{2} \frac{M}{\mu_{a}} R \Delta T
$$

We also note that the total work performed by the gas is

$$
W=P_{2} V_{2}-P_{1} V_{1}
$$

since the process is adiabatic, we can use the ideal gas equation $P V=\nu R T=(M / \mu) R T$ to express the total work as

$$
W=\frac{M}{\mu_{a}} R \Delta T
$$

The heat that is taken away during condensation at the top of the mountain is given by $Q=L \Delta m$ where $\Delta m$ is the total mass of the precipitation. According to the ideal gas law, we have that

$$
P V_{1}=\frac{\Delta m}{\mu_{p}} R T_{1}, \quad P_{1} V_{1}=\frac{M}{\mu_{a}} R T_{1}
$$

recombining these equations and equating them gives us

$$
\begin{aligned}
\frac{\Delta m}{\mu_{p} P} R T_{1} & =\frac{M}{\mu_{a} P_{1}} R T_{1} \\
\Delta m & =M \frac{\mu_{p} P}{\mu_{a} P_{1}}
\end{aligned}
$$



Therefore,

$$
Q=L M \frac{\mu_{a} P}{\mu_{p} P_{1}}
$$

We finally can now use the first law of thermodynamics

$$
Q=\Delta U+W \Longrightarrow L M \frac{\mu_{p} P}{\mu_{a} P_{1}}=\frac{5}{2} \frac{M}{\mu_{a}} R \Delta T+\frac{M}{\mu_{a}} R \Delta T
$$

We then simplify this equation to get

$$
\begin{aligned}
L M \frac{\mu_{p} P}{\mu_{a} P_{1}} & =\frac{7}{2} \frac{M}{\mu_{a}} R \Delta T \\
\Delta T & =\frac{2}{7} \frac{L \mu_{p} P}{R P_{1}}=7.41 \mathrm{~K}
\end{aligned}
$$

A simpler approach could be to assume that the number of moles of water vapour in the atmosphere is equal to number of moles of water condensed. Then, the mass of precipitated water is $\mu_{p} n \frac{P}{P_{1}}$, where $n$ is the number of moles of air. Thus,

$$
\mu_{p} n \frac{P}{P_{1}} L=\frac{7}{2} n R \Delta T
$$

Final answer: 7.41 K