# If 8 L of a gas at room temperature exerts a pressure of 42 kPa on its container, what pressure will the gas exert if the container's volume changes to 9 L?

For a given quantity of gas at constant temperature, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$. And thus the pressure should DECREASE is the volume INCREASES:
${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \left(42 \cdot \text{kPa"xx8*cancel"L")/(9*cancel"L}\right)$
=??"kPa"?