Question

If `3 L` of a gas at room temperature exerts a pressure of `2 kPa` on its container, what pressure will the gas exert if it the container's volume changes to `1 L`?

Answer 1

The gas will exert a pressure of `6kpa`

Explanation:

Let's identify the known and unknown variables:

`color(yellow)("Knowns:")`
- Initial Volume
- Final Volume
- Initial Pressure

`color(orange)("Unknowns:")`
- Final Pressure

We can obtain the answer using Boyle's Law:

The letters i and f represent the initial and final conditions, respectively.

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 this:
`P_f=(P_ixxV_i)/V_f`

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

`P_f= (2kPa xx 3cancel"L")/(1\cancel"L")` = `6kPa`