Problem Of the Week: 12th February 2024
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
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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 dm3. 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.
Hint #1 As the molar distribution of air is given, try and work the mass of 1 mole of air. Then try and see where to go next with the information given in the problem. Hint #2 Try and work out the volume of the troposphere, remember that both the Earth and the troposphere are perfect spheres. Solution
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 N2 = 0.78 x 28.01 = 21.85g
mass of O2 = 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 r3 x 𝛑
Volume of Earth: 4/3 x (6.371 x 107)3 x 𝛑 = 1.0832 x 1024 dm3
Volume of Troposphere + Earth: 4/3 x (6.381 x 107)3 x 𝛑 = 1.0883 x 1024 dm3
(Note the conversion from km 3 to dm3 for subsequent calculations)
Therefore, the volume of the troposphere = (1.0883 x 1024) – (1.0832 x 1024) = 5.11 x 1021 dm3
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 1021)/24 = 2.13 x 1020 mol
Therefore, the mass of air in the troposphere is:
2.13 x 1020 x 28.97 = 6.17 x 1021g
