# If 2/5 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 7/3 L?

May 10, 2017

Before starting the calculation, it's worth thinking about what we expect the answer to be. We have the same amount of gas, in a container that is getting a lot bigger: from $\frac{2}{5}$ (40%) of a liter to $\frac{7}{3}$ (233%) of a liter.

With the same amount of gas in a larger container, we expect the pressure to be lower than the initial $15$ $k P a$.

There are a number of possible approaches to doing the calculation. My own favourite way is to use the Combined Gas Law and then remove the one that we're not using:

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

In this case, we are not changing temperatures, so we can remove that:

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

Rearranging to make ${P}_{2}$ the subject:

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

Now we can substitute in the values from the question:

${P}_{2} = \frac{15 \times \left(\frac{2}{5}\right)}{\frac{7}{3}} = 2.6$ $k P a$