The volume of 45.7L will be at temperature 16.72°C
Explanation:
Given:
Volume, V₁ = 45.7L
V₂ = 33.9L
Temperature, T₂ = 12.4°C
T₁ = ?
According to the Charle's law:
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Substituting the value we get:
[tex]\frac{45.7}{T_1} = \frac{33.9}{12.4}\\\\T_1 = \frac{45.7 X 12.4}{33.9} \\\\T_1 = 16.72^oC[/tex]
Therefore, the volume of 45.7L will be at temperature 16.72°C
