Question

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`?

Answer 1

`P_2= (P_1V_1)/V_2 = (7kPa * 3L)/(2/3 L)=31.5kPa`

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:

`PV=nRT`

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.

`PV = "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_1V_1 = "constant"`

and in our final state we have

`P_2V_2 = "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_1V_1 = P_2V_2`

We can solve this for `P_2` since that is what the question is asking us for:

`P_2= (P_1V_1)/V_2 = (7kPa * 3L)/(2/3 L)=31.5kPa`