# If 3/5 L of a gas at room temperature exerts a pressure of 15 kPa on its container, what pressure will the gas exert if the container's volume changes to 7/2 L?

Jun 24, 2016

The new pressure is $\frac{18}{7} k P a$

#### Explanation:

Let's begin by identifying our known and unknown variables.
The first volume we have is $\frac{3}{5}$ L, the first pressure is $15 k P a$ and the second volume is $\frac{7}{2} L$. Our only unknown is the second pressure.

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 pressure.

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

Now all we do is plug in the values and BAM we're done!
${P}_{2} = \left(15 \setminus k P a \times \frac{3}{5} \setminus \cancel{\text{L")/(7/2\cancel"L}}\right)$ = $\frac{18}{7} k P a$