Answer:
[tex]P_2=1.1x10^6Pa[/tex]
Explanation:
Hello.
In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:
[tex]P_2=\frac{P_1V_1}{V_2}[/tex]
Consider that the given initial pressure is also equal to Pa:
[tex]P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa[/tex]
Which stands for a pressure increase when volume decreases.
Regards.
