# If 17/3 L of a gas at room temperature exerts a pressure of 24 kPa on its container, what pressure will the gas exert if the container's volume changes to 19/4 L?

Aug 19, 2016

The gas will exert a pressure of $\frac{544}{19}$ kPa.

#### Explanation:

Let's begin by identifying our known and unknown variables.

The first volume we have is $\frac{17}{3}$L, the first pressure is $24 k P a$ and the second volume is $\frac{19}{4} L$. Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law:

The letters i and f represent the initial and final conditions. All we have to do is rearrange the equation to solve for the final pressure.

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

Now all we do is plug in the values and we're done!

${P}_{f} = \left(24 \setminus k P a \times \frac{17}{3} \setminus \cancel{\text{L")/(19/4\cancel"L}}\right)$ = $\frac{544}{19} k P a$