Answer:
[tex]T_2=-125.58\°C[/tex]
Explanation:
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In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:
[tex]\frac{T_1}{P_1} =\frac{T_2}{P_2}[/tex]
Whereas solving for the final temperature T2, we get:
[tex]T_2=\frac{T_1P_2}{P_1}[/tex]
Thus, we plug in the given data (temperature in Kelvins) to obtain:
[tex]T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C[/tex]
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