# Question d794a

Aug 3, 2016

$\text{3.9 L}$

#### Explanation:

The problem wants to test if you're familiar with the fact that the pressure and volume of a gas have an inverse relationship when temperature and number of moles are kept constant, as described by Boyle's Law.

Simply put, when temperature and number of moles of gas, which essentially means that quantity of gas, are kept constant, increasing the pressure will cause a decrease in volume.

Similarly, decreasing the pressure will cause an increase in volume.

Mathematically, you can express this relationship as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} {V}_{1} = {P}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

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

In your case, the pressure of the gas is increasing from $\text{108 kPa}$ to $\text{224 kPa}$, which means that you should expect the final volume of the gas to be smaller than $\text{8 L}$.

You will thus have

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

Plug in your values to find

V_2 = (108 color(red)(cancel(color(black)("kPa"))))/(224color(red)(cancel(color(black)("kPa")))) * "8 L" = "3.857 L"#

I'll leave this rounded to two sig figs, but keep in mind that you only have one sig fig for the initial volume of the gas

${V}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{3.9 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$