Physics

Physics · thermodynamics

Problem

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 0.5kPa0.5 \mathrm{kPa} at 100.2kPa100.2 \mathrm{kPa} and have a molar mass of μa=28 g/mole\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⋅106 J/kgL=2.4 \cdot 10^{6} \mathrm{~J} / \mathrm{kg} and molar mass μp=18.01 g/mole\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. Given constants: - Proton mass, mp=1.67⋅10−27 kgm_{p}=1.67 \cdot 10^{-27} \mathrm{~kg} - Neutron mass, mn=1.67⋅10−27 kgm_{n}=1.67 \cdot 10^{-27} \mathrm{~kg} - Electron mass, me=9.11⋅10−31 kgm_{e}=9.11 \cdot 10^{-31} \mathrm{~kg} - Avogadro's constant, N0=6.02⋅1023 mol−1N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1} - Universal gas constant, R=8.31 J/(mol⋅K)R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) - Boltzmann's constant, kB=1.38⋅10−23 J/Kk_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K} - Electron charge magnitude, e=1.60⋅10−19Ce=1.60 \cdot 10^{-19} \mathrm{C} - 1 electron volt, 1eV=1.60⋅10−19 J1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J} - Speed of light, c=3.00⋅108 m/sc=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s} - Universal Gravitational constant, G=6.67⋅10−11( N⋅m2)/kg2G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} - Solar Mass M⊙=1.988⋅1030 kgM_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} - Acceleration due to gravity, g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^{2} - 1 unified atomic mass unit, 1u=1.66⋅10−27 kg=931MeV/c21 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} - Planck's constant, h=6.63⋅10−34 J⋅s=4.41⋅10−15eV⋅sh=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} - Permittivity of free space, ϵ0=8.85⋅10−12C2/(N⋅m2)\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) - Coulomb's law constant, k=14πϵ0=8.99⋅109( N⋅m2)/C2k=\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, μ0=4π⋅10−7 T⋅m/A\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} - Magnetic constant, μ04π=1⋅10−7( T⋅m)/A\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} - 1 atmospheric pressure, 1 atm=1.01⋅105 N/m2=1.01⋅105 Pa1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} - Wien's displacement constant, b=2.9b=2.9. 10−3 m⋅K10^{-3} \mathrm{~m} \cdot \mathrm{K} - Stefan-Boltzmann constant, σ=5.67⋅10−8 W/m2/K4\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}

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