# Question #34022

Feb 23, 2017

Here's how you can solve this problem.

#### Explanation:

Notice that the problem doesn't mention temperature and number of moles of gas, which means that you can assume that they are constant.

This implies that you can use the equation for Boyle's Law to find the volume of the gas at a pressure of $\text{2.00 atm}$

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}}}$

Here

• ${P}_{1}$ and ${V}_{1}$ are the pressure and volume of the gas at an initial state
• ${P}_{2}$ and ${V}_{2}$ are the pressure and volume of the gas at a final state

What you have to do here is to rearrange the above equation to solve for ${V}_{2}$

${P}_{1} \cdot {V}_{1} = {P}_{2} \cdot {V}_{2} \implies {V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

At this point, you would plug in your values and find the value of ${V}_{2}$. The new volume of the gas is expressed in milliliters and must have three sig figs.

So, to sum this up, Boyle's Law tells you that when temperature and number of moles of gas are kept constant, the pressure and the volume of the gas have an inverse relationship.

This means that when pressure increases, the volume of the gas must decrease by the factor. Similarly, when pressure increases, the volume of the gas must increase by the same factor.

In your case, the pressure decreases from $\text{8.00 atm}$ to $\text{2.00 atm}$, so you should expect the volume to increase.

${P}_{2} < {P}_{1} \implies {V}_{2} > {V}_{1}$