# 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?

Sep 9, 2016

The gas will exert a pressure of $6 k p a$

#### Explanation:

Let's identify the known and unknown variables:

$\textcolor{y e l l o w}{\text{Knowns:}}$
- Initial Volume
- Final Volume
- Initial Pressure

$\textcolor{\mathmr{and} a n \ge}{\text{Unknowns:}}$
- 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:
${P}_{f} = \frac{{P}_{i} \times {V}_{i}}{V} _ f$

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

${P}_{f} = \left(2 k P a \times 3 \cancel{\text{L")/(1\cancel"L}}\right)$ = $6 k P a$