# The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant?

Using Boyle's law, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, ${P}_{2} \cong 0.40 \cdot a t m$.
From Boyle's law, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.
${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2$ $=$ $\frac{1.20 \cdot a t m \times 2.25 \cdot \cancel{L}}{6.50 \cdot \cancel{L}}$ $=$ ??atm