Atmospheric physics 6: Mount Everest

Classical Mechanics Level 3

What is the air pressure at the summit of Mount Everest at an altitude z = 8848 m z = 8848\,\text{m} above sea level? Use the barometric formula for the isothermal atmosphere from the last section. Assume that at sea level there is an air pressure of P 0 = 1013 hPa P_0 = 1013 \,\text{hPa} . The temperature of the atmosphere is T = 288 K T = 288\,\text{ K} . The molar mass of air is M = 29 g / mol M = 29\,\text{g}/\text{mol}

P ≈ 128 hPa P \approx 128 \,\text{hPa} P ≈ 354 hPa P \approx 354 \,\text{hPa} P ≈ 512 hPa P \approx 512 \,\text{hPa} P ≈ 256 hPa P \approx 256 \,\text{hPa} P ≈ 432 hPa P \approx 432 \,\text{hPa}

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

With the given values for temperature and molar mass, the scale height results to H = R T M g = 8.314 J / mol K ⋅ 288 K 0.029 kg / mol ⋅ 9.81 m / s 2 ≈ 8417 m H = \frac{R T}{M g} = \frac{8.314 \,\text{J}/\text{mol}\,\text{K} \cdot 288 \,\text{K}}{0.029 \,\text{kg}/\text{mol} \cdot 9.81 \,\text{m}/\text{s}^2} \approx 8417 \,\text{m} which corresponds to the height of the highest mountains. For the summit of the Mount Everest we get an air pressure P = P 0 exp ⁡ ( − z / H ) = 1013 hPa ⋅ exp ⁡ ( − 8848 / 8417 ) ≈ 354 hPa P = P_0 \exp ( - z/H) = 1013\,\text{hPa} \cdot \exp(- 8848/8417) \approx 354 \,\text{hPa}

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