Question

If `7/5 L` of a gas at room temperature exerts a pressure of `6 kPa` on its container, what pressure will the gas exert if the container's volume changes to `2/3 L`?

Answer 1

The gas will exert a pressure of `63/5` kPa.

Explanation:

Let's begin by identifying our known and unknown variables.

The first volume we have is `7/5` L, the first pressure is `6kPa` and the second volume is `2/3L`. Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law:

The letters i and f represent the initial and final conditions. All we have to do is rearrange the equation to solve for the final pressure.

We do this by dividing both sides by `V_f` in order to get `P_f` by itself like so:
`P_f=(P_ixxV_i)/V_f`

Now all we do is plug in the values and we're done!

`P_f=(6\kPa xx 7/5\ cancel"L")/(2/3\cancel"L")` = `63/5kPa`