# If 9/4 L of a gas at room temperature exerts a pressure of 15 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/3 L?

May 23, 2018

$20.25$ kilopascals

#### Explanation:

For constant temperature and number of moles of gas, we use Boyle's law, which states that:

$P \propto \frac{1}{V}$ or ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

where ${P}_{1} , {P}_{2}$ are the initial and final pressure of the gas, and ${V}_{1} , {V}_{2}$ are the initial and final volume of the gas.

So, we get:

$\frac{9}{4} \setminus \text{L"*15 \ "kPa"=P_2*5/3 \ "L}$

$\therefore {P}_{2} = \left(\frac{9}{4} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L"*15 \ "kPa")/(5/3color(red)cancelcolor(black)"L}}}}\right)$

$= \frac{81}{4} \setminus \text{kPa}$

$= 20.25 \setminus \text{kPa}$