A piston of 11*L volume under 42*kPa pressure is compressed to a volume of 8*L; what is the new pressure?

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$. ${P}_{2} \cong 58 \cdot k P a$
${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2$ $=$ $\frac{42 \cdot k P a \times 11 \cdot L}{8 \cdot L}$ $=$ ??