Calcium carbonate decomposes at 12000 C to form carbon dioxide and calcium oxide. If 25 liters of carbon dioxide are collected at 12750 C, what will the volume of this gas be after it cools to 250 C?

1L

Explanation:

Data obtained from the question include:

Initial volume (V1) = 25L

Initial temperature (T1) = 12750°C

Final temperature (T2) = 250°C

Final volume (V2) =..?

Next we shall convert from celsius to Kelvin temperature. This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T1) = 12750°C

Initial temperature (T1) = 12750°C + 273 = 13023K

Final temperature (T2) = 250°C

Final temperature (T2) = 250°C + 273 = 523K

Finally, we can obtain the new volume of the gas by using Charles' law equation as shown below:

V1/T1 = V2/T2

25/13023 = V2/523

Cross multiply to express in linear form

13023 x V2 = 25 x 523

Divide both side by 13023

V2 = 25 x 523 / 13023

V2 = 1L

Therefore, the new volume of the gas is 1L.

The volume of this gas after it cools to 250°C is equal to 0.490 Liter.

Given the following data:

To determine the volume of this gas after it cools to 250°C, we would apply Charles's law:

Mathematically, Charles law is given by the formula;

Where;

Substituting the given parameters into the formula, we have;

Final volume = 0.490 Liter.

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