# If 8 L of a gas at room temperature exerts a pressure of 28 kPa on its container, what pressure will the gas exert if the container's volume changes to 7 L?

Sep 27, 2016

$32 k P a$

#### Explanation:

Let's identify the known and unknown variables:

$\textcolor{v i o \le t}{\text{Knowns:}}$
- 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 numbers 1 and 2 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 ${V}_{2}$ in order to get ${P}_{2}$ by itself like this:
${P}_{2} = \frac{{P}_{1} \times {V}_{1}}{V} _ 2$

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

${P}_{2} = \left(28 k P a \times 8 \cancel{\text{L")/(7\cancel"L}}\right)$ = $32 k P a$