# 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?

Apr 30, 2016

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ at constant temperature.

#### Explanation:

Thus ${V}_{2}$ $=$ $\frac{{P}_{1} {V}_{1}}{P} _ 2$.

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 ${V}_{2}$ $=$ $\frac{340 \cdot \cancel{k P a} \times 0.250 \cdot L}{50 \cdot \cancel{k P a}}$ $=$ ??L.

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

Apr 30, 2016

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:
$P \propto \frac{1}{V}$
That is, in these conditions, pressure times volume is constant :
$P V = k$
So if we compare two separate situations, we conclude that
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

Now we can solve this problem. We know that the initial pressure is $340.0 k P a$, the initial volume is $0.250 L$, and the final pressure is $50.0 k P a$:

$340.0 k P a \cdot 0.250 L = 50.0 k P a \cdot {V}_{2}$

If we cancel out the pressure units, we end up with a volume unit, which is precisely what we want:
$340.0 \textcolor{red}{\cancel{k P a}} \cdot 0.250 L = 50.0 \textcolor{red}{\cancel{k P a}} \cdot {V}_{2}$

So the final volume of this gas, at 50kPa, will be:
$85 L = 50.0 {V}_{2}$
${V}_{2} = \left(\frac{85}{50}\right) L$

${V}_{2} = 1.7 L$

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!