Assume that the earth’s atmosphere has a constant temperature of 298K (not an accurate assumption). Estimate the pressure at 8900m above sea level (roughly the altitude of Mount Everest). Assume that air is a single ideal gas with a molar mass of 0.029 kg mol^−1.

Question

Assume that the earth’s atmosphere has a constant temperature of 298K (not an accurate assumption). Estimate the pressure at 8900m above sea level (roughly the altitude of Mount Everest). Assume that air is a single ideal gas with a molar mass of [latex]0.029 kg mol^{−1}[/latex].

Accepted Answer

Representing the high-altitude state as 2 and the sea-level state as 1, the probability ratio is

[latex]\frac{p_{2}}{p_{1}} =\frac{N_{2}}{N_{1}} =\frac{P_{2}}{P_{1}} =\frac{e^{-(k_{2}+V_{2})/k_{B}T}}{e^{-(k_{1}+V_{1})/k_{B}T}} [/latex]

The average kinetic energy of a molecule depends only on the temperature, which we have assumed to be independent of altitude. The kinetic energy terms will cancel after averaging:

[latex]\frac{P_{2}}{P_{1}} =exp\Bigl(\frac{V_{2}-V_{1}}{k_{B}T} \Bigr)[/latex]

Assuming that the ideal gas law is valid, the pressure is proportional to the number of molecules per unit volume at constant temperature and is proportional to the population given by the Boltzmann distribution.

[latex]V_{2}-V_{1}=\frac{\bigl(0.029 kg mol^{−1}\bigr)\bigl(9.80ms^{2}\bigr)(8900m)}{6.022×10^{23} mol^{−1}} =4.2×10^{−21} J[/latex]

If P=1.00 atm at sea level, and if we assume that air is an ideal gas:

[latex]\frac{P}{1.00 atm} =exp\Biggl(\frac{−4.2 × 10^{−21} J}{\bigl(1.3807 × 10^{−23} J K^{−1}\bigr)(298K)} \Biggr)=0.36[/latex]

Question 9.14: Assume that the earth’s atmosphere has a constant temperatur...

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