# If 2 L of a gas at room temperature exerts a pressure of 72 kPa on its container, what pressure will the gas exert if the container's volume changes to 7 L?

Aug 3, 2016

The new pressure is $21 k P a$

#### Explanation:

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

The first volume we have is $2$ L, the first pressure is $72 k P a$ and the second volume is $7$. Our only unknown is the second pressure.

The answer can be ascertained by using Boyle's Law:

All we have to do is rearrange the equation to solve for the final pressure.

We do this by dividing both sides by ${V}_{2}$ to get ${P}_{2}$ by itself like this:
${P}_{2} = \frac{{P}_{1} \times {V}_{1}}{V} _ 2$

${P}_{2} = \left(72 \setminus k P a \times 2 \setminus \cancel{\text{L")/(7\cancel"L}}\right)$ = $21 k P a$