Answer:
392.20K
Explanation:
-We apply Charles' Law which states that for an ideal gas at constant pressure, its volume is directly proportional to it's temperature:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
-Given that:
[tex]V_1=4.87L, V_2=6.08\\\\T_1=41.0\textdegree C=314.15\ K\\T_2[/tex]
#We substitute in the ratio formula above and calculate for [tex]T_2[/tex]:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\T_2=\frac{V_2T_1}{V_1}\\\\=\frac{6.08\ l\times 314.15\ K}{4.87\ L}\\\\\\=392.20\ K[/tex]
Hence, the temperature of the balloon at a volume of 6.08L is 392.20K
