# If 3/7 L of a gas at room temperature exerts a pressure of 24 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/4 L?

Aug 18, 2016

We use the Ideal Gas Law to create an equality for the two conditions and solve the the unknown quantity:

$P = \frac{288}{35} k P a$

#### Explanation:

To answer this type of question, we can start with the Ideal Gas Law and decide from there which things are constant and which are changing. Rearranging the equation to put all of the constants on one side, we can then create an equality for the two conditions (before and after). Sounds simple - let's begin! We start with the gas law which states:

$P V = n R T$

In this case, we know from the question that the volume, $V$, and the pressure, $P$, are changing. We can safely assume that the remaining quantities (number of moles of gas particles, $n$, and the temperature, $T$) are constants, as is the constant, $R$. So we can write:

$P V = \text{constant}$

Which means that if we change one of these quantities, the other will change in such a way that the product of the two remains the same. In equation form:

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

Since our question is asking for the new pressure when we change the volume, let's solve for that and then we can plug in the values:

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{\frac{3}{7} L \cdot 24 k P a}{\frac{5}{4} L} = \frac{288}{35} k P a$