The genie is out of the bottle 2
What is the change of the total entropy in the process of releasing 1 mole "smoke" into the atmosphere? Assume that "smoke" has 10% mole fraction of CO and the rest is other gases; the atmosphere has mole fraction (400ppm) of carbon dioxide. The smoke is at room temperature and atmospheric pressure when it is released. Give your result in units of J/K.
Notes: The gas constant R is 8.314 J / mol K . The "other gases" in the smoke is mostly made of N , but for simplicity we will assume that this part of the smoke has the same composition as the outside air.
The answer is 3.8.
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1 solution
First consider the process of mixing two containers of gases. Each container has two different gases: container #1 has n 1 A mole of gas A and n 1 B mole of gas B. Similarly container #2 has n 2 A mole of gas A and n 2 B mole of gas B. The mole fraction of gas A in container #1 is x 1 A = n 1 A / n 1 , where n 1 = n 1 A + n 1 B is the total amount of gas in that container. Similarly we can define x 1 B = n 1 B / n 1 , x 2 A = n 2 A / n 2 and x 2 B = n 2 B / n 2 . Note that x 1 B = 1 − x 1 A and x 2 B = 1 − x 2 A .
The total amount of gas is n = n 1 + n 2 , made up of gas A of amount n A = n 1 A + n 2 A and gas B of amount n B = n 1 B + n 2 B . The corresponding mole fractions are x A = n A / n and x B = n B / n . The pressure p and the temperature T is the same for both containers.
In the process of mixing the components expand into the volume available to them in the new, joint space, and according to textbooks the entropy production per unit amount of gas in each process is
Δ S = n R ln p i / p f
where p i and p f are the initial and final partial pressure of the gas. The partial pressures scale with the mole fractions. For example, if we consider n 1 A it has an initial partial pressure of p i = p x 1 A and final partial pressure of p f = p x A , so the entropy production is n 1 A [ ln x 1 A / x A ] . Adding up all terms:
Δ S t o t / R = n 1 A ln x 1 A / x A + n 2 A ln x 2 A / x A + n 1 B ln x 1 B / x B + n 2 B ln x 2 B / x B
This is a general formula for mixing two containers of two component gases. Let us take the limit of n 2 A > > n 1 A , n 2 B > > n 1 B . In this case the composition of the gas in the larger container (#2) does not change and n 2 A = n 2 , n 2 B = n 2 , and the corresponding logarthmic terms are equal to zero. We get
Δ S t o t / R = n 1 A ln x 1 A / x A + n 1 B ln x 1 B / x B
Using x 1 A = 0 . 1 mole (for CO 2 ), x 1 B = 0 . 9 mole (for air) in the smoke and x 2 A = 4 × 1 0 − 4 mole (for CO 2 ), x 2 B = 1 mole for the outside atmosphere we get
Δ S = 3 . 8 1 J/K