Question

8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

Answer 1

The gas' new volume will be `V_2 = 7.28` L.

This problem is a simple application of Charles' law

`V_1/T_1 = V_2/T_2`,

which states that the volume of a gas is directly proportional to its temperature(measured in `K` - very important).

So, we have an initial volume equal to `V_1 = 8.00L`, an initial temperature of `T_1 = (273.15+60) = 333.15K`, and a final temperature of `T_2 = (273.15 + 30) = 303.15K`, which gets us

`V_2 = T_2/T_1 * V_1 = 303.15/333.15 * 8.00 = 7.28L` -> a decrease in temperature is correlated with a decrease in volume, just as expected.

Note, however, what would have happened with degrees Celsius instead of K:

`V_2 = 30/60 * 8.00 = 4.00 L`, a result significantly different...