Answer:
[tex]P_2=1.48atm[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer this problem is about the application of the Gay-Lussac's gas law to relate the initial and final pressure and temperature as shown below:
[tex]\frac{P_2}{T_2} =\frac{P_1}{T_1\\ }[/tex]
Thus, solving for the final pressure, P2, and using the temperatures in Kelvins, we obtain:
[tex]P_2 =\frac{P_1T_2}{T_1 }\\\\P_2 =\frac{1atm*403.15K}{273.15K}\\\\P_2=1.48atm[/tex]
Regards!
