Answer:
[tex]p_2=1.17atm[/tex]
Explanation:
Hello!
In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:
[tex]\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\[/tex]
Thus, for the initial pressure and temperature in kelvins the final temperature in kelvins, we compute the final pressure as:
[tex]p_2=\frac{p_1T_2}{T_1} \\\\p_2=\frac{1.12atm*310.15K}{298.15K}\\\\p_2=1.17atm[/tex]
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