Answer:
[tex]T_2=-53.15\°C[/tex]
Explanation:
Hello there!
In this case, in agreement to the Charles' law which states that the volume and temperature relate to each other via a directly proportional relationship at constant pressure, we can write:
[tex]\frac{T_2}{V_2} =\frac{T_1}{V_1}[/tex]
Thus, given V1, V2 and T1 (310 in Kelvin), we can solve for the final temperature as shown below:
[tex]T_2=\frac{T_1V_2}{V_1} \\\\T_2=\frac{310K*32.5mL}{45.8mL}\\\\T_2=220 K-273.15K\\\\T_2=-53.15\°C[/tex]
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