If 9/7 L of a gas at room temperature exerts a pressure of 8 kPa on its container, what pressure will the gas exert if the container's volume changes to 3/4 L?

Apr 2, 2016

The final pressure is 13.7 kPa

Explanation:

This question can be solved using Boyle's Law, which states that for a fixed mass of gas its pressure is inversely proportional to its volume at constant temperature.

Therefore we can write the following:
$p V =$constant

Since the product of pressure and volume is equal to a constant that means the product of initial p and V will be equal to the product of final p and V.
⇒ p_1V_1 = p_2V_2

The question requires us to calculate ${p}_{2}$, so rearrange the equation above for ${p}_{2}$:
⇒ p_2 = p_1V_1/V_2*

Now substitute in the values:
p_2 = 8 × (9//7) / (3//4) = 8 × 9/7 × 4/3 = 13.7 kPa

As I did not convert the prefix of the initial pressure my answer calculates into the same prefix, i.e. the initial pressure was given in kilo pascals so my answer is in kilo pascals.

*Notice the structure of the rearranged equation, the final pressure is equal to the initial pressure multiplied by a ratio of volumes. So the units of the left hand side of the equation are pressure units and so are the units of the right hand side. That's because in a ratio of volumes the units cancel out. With questions on the gas laws you will notice that if you correctly rearrange the equations you will have a ratio multiplied by another quantity. This is a helpful quick check that you did the rearranging correctly.