Answer:
The temperature at 581.2 mL volume was 242.8 K.
Explanation:
Using Charle's law
[tex]\frac {V_1}{T_1}=\frac {V_2}{T_2}[/tex]
Given ,
V₁ = 581.2 mL
V₂ = 830.5 mL
T₁ = ?
T₂ = 347 K
Using above equation as:
[tex]\frac {581.2\ mL}{T_1}=\frac {830.5\ mL}{347\ K}[/tex]
[tex]T_1=\frac{581.2\times 347}{830.5}\ K=242.8\ K[/tex]
The temperature at 581.2 mL volume was 242.8 K.
