Answer:
![P_2=404 kPa](https://img.qammunity.org/2021/formulas/chemistry/college/g4ldjn5lndnkegloa3wk8pf22uh3543mkl.png)
Step-by-step explanation:
Hello,
In this case, the Boyle's is mathematically defined via:
![P_1V_1=P_2V_2](https://img.qammunity.org/2021/formulas/physics/high-school/z2nkrx5zmdtkkms997yjs4edj5t3o0sadf.png)
Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:
![V_2=(1)/(4) V_1](https://img.qammunity.org/2021/formulas/chemistry/college/jdhupn4899yo23wcd10j0vpbopm04csxus.png)
We can compute the new pressure:
![P_2=(P_1V_1)/(V_2) =(P_1V_1)/((1)/(4) V_1) =(101kPa*V_1)/((1)/(4) V_1) \\\\P_2=4*101kPa\\\\\\P_2=404 kPa](https://img.qammunity.org/2021/formulas/chemistry/college/wt4egimo4gwclcoldih72gnn36i07jaeuo.png)
Which means the pressure is increased by a factor of four.
Regards.