Answer:
Final pressure will be equal to 394.94 kPa
Step-by-step explanation:
It is given volume
![V_1=55L](https://img.qammunity.org/2021/formulas/chemistry/high-school/djyrf7n0pmn1ibrz7evktuij4784pjh71w.png)
Temperature of the gas
![T_1=235K](https://img.qammunity.org/2021/formulas/chemistry/high-school/fnfla6lntnppz44yeek2nzboamzx33863o.png)
Initial pressure exerted
![P_1=225kPa](https://img.qammunity.org/2021/formulas/chemistry/high-school/zfcg1mcrcmns1y6llu50puoe7km3pzr4ha.png)
Now volume is decreased to
![V_2=42L](https://img.qammunity.org/2021/formulas/chemistry/high-school/qz2xr123ug910oqzytqqw32b4yxbc5lka0.png)
And temperature is increases to
![T_2=315K](https://img.qammunity.org/2021/formulas/chemistry/high-school/xq5laeysrtt7474ux33xbkrwx0p6fqgq6u.png)
By using gas equation
![(P_1V_1)/(T_1)=(P_2V_2)/(T_2)](https://img.qammunity.org/2021/formulas/physics/high-school/xkw5w11w5j6h3n0i0p5108jnbt1i78v4rv.png)
![(225* 55)/(235)=(P_2* 42)/(315)](https://img.qammunity.org/2021/formulas/chemistry/high-school/yln2ss228agiuj55bfz4mdjvn8gxlzeeui.png)
![P_2=394.94kPa](https://img.qammunity.org/2021/formulas/chemistry/high-school/c2fh1w9daqfppwv1d0lruo946md8l88h2v.png)
So final pressure will be equal to 394.94 kPa