A gas occupies 11.2 liters at .860 atm. What is the pressure if the volume becomes 15.0 L?

${P}_{1} {V}_{1}$ $=$ ${P}_{2} {V}_{2} \text{ at constant } T$. Boyle's Law
${P}_{2}$ $=$ $\frac{{P}_{1} {V}_{1}}{V} _ 2$ $=$ $\frac{0.860 \cdot a t m \times 11.2 \cdot \cancel{L}}{15.0 \cdot \cancel{L}}$ $\cong 0.6 \cdot a t m$.