Answer: [tex]472.22 \°C[/tex]
Explanation:
According to the Ideal Gas Law for an isobaric process (at a constant pressure):
[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex] (1)
Where:
[tex]V_{1}=5 L[/tex] is the initial volume of the sample
[tex]T_{1}=25\°C + 273.15=298.15 K[/tex] is the initial temperature of the sample in Kelvin
[tex]V_{2}=12.5 L[/tex] is the final volume of the sample
[tex]T_{2}[/tex] is the final temperature of the sample
So, we have to find [tex]T_{2}[/tex] from (1):
[tex]T_{2}=V_{2}\frac{T_{1}}{V_{1}}[/tex] (2)
[tex]T_{2}=12.5 L\frac{298.15 K}{5 L}[/tex] (3)
[tex]T_{2}=745.37 K[/tex] (4)
Transforming this result to Celsius:
[tex]T_{2}=745.37 K-273.15=472.22 \°C[/tex] This is the final temperature in Celsius.
