Answer:
[tex]1.22\cdot 10^3 L[/tex]
Explanation:
We can solve the problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas is directly proportional to its temperature:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where here we have
[tex]V_1 = 1.09\cdot 10^3 L[/tex] is the initial volume
[tex]T_1 = 23^{\circ}+ 273 = 296 K[/tex] is the initial temperature
[tex]V_2[/tex] is the final volume
[tex]T_2 = 59^{\circ}+ 273 =332 K[/tex] is the final temperature
Solving for V2, we find
[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(1.09 \cdot 10^3 L)(332 K)}{296 K}=1.22\cdot 10^3 L[/tex]
