# If 3 L of a gas at room temperature exerts a pressure of 7 kPa on its container, what pressure will the gas exert if it the container's volume changes to 2/3 L?

May 15, 2016

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{7 k P a \cdot 3 L}{\frac{2}{3} L} = 31.5 k P a$

#### Explanation:

This question is asking us to use things we know about gasses to find the final condition after a change of one of the variables. The ideal gas law can help us:

$P V = n R T$

Where $P$ is pressure the gas exerts on its container, $V$ is the volume of the container, $n$ is the number of moles of gas particles, $R$ is the universal gas constant, and $T$ is the temperature. In our question, the only things that are changing are the pressure and the volume which means that everything else is constant, i.e.

$P V = \text{constant}$

Since we have two conditions, call them 1 and 2, we can use this expression to solve for them, i.e. in our initial conditions we have

${P}_{1} {V}_{1} = \text{constant}$

and in our final state we have

${P}_{2} {V}_{2} = \text{constant}$

the trick is to understand that the constant is the same in both cases (none of $n$, $R$, or $T$ are changing), which gives us:

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

We can solve this for ${P}_{2}$ since that is what the question is asking us for:

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{7 k P a \cdot 3 L}{\frac{2}{3} L} = 31.5 k P a$