Question

If `17/3 L` of a gas at room temperature exerts a pressure of `24 kPa` on its container, what pressure will the gas exert if the container's volume changes to `19/4 L`?

Answer 1

The gas will exert a pressure of `544/19` kPa.

Explanation:

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

The first volume we have is `17/3`L, the first pressure is `24kPa` and the second volume is `19/4L`. 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=(24\kPa xx 17/3\ cancel"L")/(19/4\cancel"L")` = `544/19kPa`