# A 12.0 L sample of argon gas has a pressure of 28.0 atm. What volume would this gas occupy at 9.70 atm?

Jun 6, 2016

The volume that this gas occupies is $34.6 L$

#### Explanation:

Let's start off with identifying our known and unknown variables.
The first volume we have is 12.0 L, the first pressure is 28.0 atm, and the second pressure is 9.70 atm. Our only unknown is the second volume.

We can obtain the answer using Boyle's Law which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.

The equation we use is ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the volume.

We do this by dividing both sides by ${P}_{2}$ in order to get ${V}_{2}$ by itself like so:
${P}_{1} {V}_{1}$ $\div$${P}_{2}$ = ${V}_{2}$

Now all we do is plug and chug!
(28.0 cancel (atm)) (12.0 L) $\div$ (9.70 cancel (atm)) = 34.6 L

Jun 6, 2016

${V}_{2} \equiv 34.6 L$

#### Explanation:

Use the combined gas low equation.

$\frac{{P}_{1} \times {V}_{1}}{T} _ 1 = \frac{{P}_{2} \times {V}_{2}}{T} _ 2$

${T}_{1} = {T}_{2}$ since the temperature remains constant.

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

${V}_{2} = \frac{{P}_{1} \times {V}_{1}}{P} _ 2$

${V}_{2} = \frac{28.0 \setminus a t m \times 12.0 \setminus L}{9.70 \setminus a t m}$

${V}_{2} \equiv 34.6 L$