#### Problem Of the week: 12th February 2024

#### Michael Wu

Hundreds of years ago, scientists believed that air was a pure substance which, alongside earth, wind and fire, made up all matter on Earth. Nowadays, we know that air is a mixture of gases.

1 mole of air contains:

0.78 moles of nitrogen

0.21 moles of oxygen

0.01 moles of argon

The first layer of the Earth’s atmosphere, called the troposphere, ends approximately 10 km above the ground. At standard room temperature and pressure, 1 mole of gas occupies a volume of 24 dm^{3}. Assuming that the earth is a perfect sphere with a radius of 6371 km and that the gas in the troposphere is at standard room temperature and pressure, estimate the mass of the air in the troposphere.

First, the mass of one mole of air can be calculated. Using a familiar equation studied at GCSE level, mass = molar mass x no. of moles; therefore

mass of N_{2} = 0.78 x 28.01 = 21.85g

mass of O_{2} = 0.21 x 32 = 6.72g

mass of Ar = 0.01 x 39.95 = 0.4g

Adding these masses:

21.85 + 6.72 + 0.4 = 28.97g

The volume of the troposphere can also be calculated, given the fact both the Earth and the troposphere are perfect spheres. Using the equation:

V = 4/3 x r^{3} x 𝛑

Volume of Earth: 4/3 x (6.371 x 10^{7})^{3} x 𝛑 = 1.0832 x 10^{24} dm^{3}

Volume of Troposphere + Earth: 4/3 x (6.381 x 10^{7})^{3} x 𝛑 = 1.0883 x 10^{24} dm^{3}

(Note the conversion from km ^{3} to dm^{3} for subsequent calculations)

Therefore, the volume of the troposphere = (1.0883 x 10^{24}) – (1.0832 x 10^{24}) = 5.11 x 10^{21} dm^{3}

As we are given that the gas is at standard room temperature and pressure, we can divide the volume of air in the troposphere by 24 to calculate the number of moles of air in the troposphere:

(5.11 x 10^{21})/24 = 2.13 x 10^{20} mol

Therefore, the mass of air in the troposphere is:

2.13 x 10^{20} x 28.97 = 6.17 x 10^{21}g