# If 9/4 L of a gas at room temperature exerts a pressure of 16 kPa on its container, what pressure will the gas exert if the container's volume changes to 9/8 L?

Sep 27, 2016

The final pressure will be 32 kPa.

#### Explanation:

This is a problem involving Boyle's law, which states that the volume of a gas at constant temperature and amount, is inversely proportional to the pressure. This means that when the volume increases, the pressure will decrease.

The equation for solving this kind of problem is:

Known
${P}_{1} = \text{16 kPa}$
${V}_{1} = \text{9/4 L}$
${V}_{2} = \text{9/8 L}$

Unknown
${P}_{2}$

Rearrange the equation to isolate ${P}_{2}$. Substitute the known values into the equation and solve.

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

P_2=(16"kPa"*9/4cancel"L")/(9/8cancel"L")="32 kPa"

Note: Notice that the volume decreased by half, and the volume increased 2x.