# If 13/4 L of a gas at room temperature exerts a pressure of 16 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/12 L?

Aug 20, 2016

The gas will exert a pressure of $\frac{624}{5} k P a$

#### Explanation:

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

The first volume we have is $\frac{13}{4}$L, the first pressure is $16 k P a$ and the second volume is $\frac{5}{12} L$. 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, 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 so:
${P}_{f} = \frac{{P}_{i} \times {V}_{i}}{V} _ f$

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

${P}_{f} = \left(16 \setminus k P a \times \frac{13}{4} \setminus \cancel{\text{L")/(5/12\cancel"L}}\right)$ = $\frac{624}{5} k P a$