Answer:
Final pressure will be equal to 394.94 kPa
Explanation:
It is given volume [tex]V_1=55L[/tex]
Temperature of the gas [tex]T_1=235K[/tex]
Initial pressure exerted [tex]P_1=225kPa[/tex]
Now volume is decreased to [tex]V_2=42L[/tex]
And temperature is increases to [tex]T_2=315K[/tex]
By using gas equation
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
[tex]\frac{225\times 55}{235}=\frac{P_2\times 42}{315}[/tex]
[tex]P_2=394.94kPa[/tex]
So final pressure will be equal to 394.94 kPa
