Answer: [tex]2.69\ L[/tex]
Explanation:
Given
The initial volume [tex]V_1=2.5\ L[/tex]
Temperature [tex]T_1=298\ K[/tex]
Final temperature [tex]T_2=321\ K[/tex]
Suppose pressure is constant for this process
[tex]\therefore \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]
Insert the values
[tex]\Rightarrow \dfrac{2.5}{298}=\dfrac{V_2}{321}\\\\\Rightarrow V_2=\dfrac{321}{298}\times 2.5\\\\\Rightarrow V_2=2.69\ L[/tex]
