Question

The volume of a gas is 0.250 L at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa, assuming the temperature remains constant?

Answer 1

`P_1V_1=P_2V_2` at constant temperature.

Explanation:

Thus `V_2` `=` `(P_1V_1)/P_2`.

The good thing about Boyle's law is that here, given the proportionality, we don't have to bother about converting units; we could use pints, and bushels, and pounds and shillings and pence if we liked.

Thus `V_2` `=` `(340*cancel(kPa)xx0.250*L)/(50*cancel(kPa))` `=` `??L`.

Clearly volume will INCREASE almost sevenfold, as we would expect if we reduce the pressure.

Answer 2

The final volume of this gas will be 1.7L.

Explanation:

Assuming this is an ideal gas, and that the number of molecules n of this gas remains constant, and given that the temperature is also constant, we have then a situation where the Boyle-Mariotte Law applies.

This law basically states that pressure P and volume V of a gas are inversely proportional when the number of molecules and temperature are constant:
`P prop 1/V`
That is, in these conditions, pressure times volume is constant :
`PV = k`
So if we compare two separate situations, we conclude that
`P_1V_1 = P_2V_2`

Now we can solve this problem. We know that the initial pressure is `340.0kPa`, the initial volume is `0.250L`, and the final pressure is `50.0kPa`:

`340.0kPa*0.250L = 50.0kPa*V_2`

If we cancel out the pressure units, we end up with a volume unit, which is precisely what we want:
`340.0color(red)cancel(kPa)*0.250L = 50.0color(red)cancel(kPa)*V_2`

So the final volume of this gas, at 50kPa, will be:
`85L = 50.0V_2`
`V_2 = (85/50)L`

`V_2 = 1.7L`

This is in accordance with Boyle-Mariotte, which implies that an increase in pressure results in a decrease in volume, and vice-versa.
Hope this helped!