# If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

Aug 21, 2016

From Boyle's Law, $\text{pressure"xx"volume"="constant}$

${P}_{2} \cong \frac{2}{7} \times 0.12 \cdot a t m$

#### Explanation:

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

Thus ${V}_{2}$ $=$ $\frac{{P}_{1} {V}_{1}}{P} _ 2$ $=$ $\frac{2.31 \cdot L \times 0.12 \cdot a t m}{7.14 \cdot L}$.

The advantage of using Boyle's Law is that I can use any units I like, $\text{pints}$, $\text{bars}$, $\text{mm Hg}$, $\text{foot pounds}$, $\text{gallons}$ so long as I am consistent.