Answer
1) Charle's law
2) The new volume at -4.0 °C = 1.8 L
Explanation
1) Charle's law
2)Given:
Initial volume of the soda bottle, V₁ = 2.0 L
Initial temperature, T₁ = 25.0 °C
Final temperature, T₂ = -4.0 °C.
The Charle's law equation is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]What to find:
The new volume, V₂ at -4.0 °C
Before plugging the values of the given parameters into the equation, the temperature must be converted from °C to K.
T₁ = 25.0 °C = 25 + 273.15 K = 298.15 K
T₂ = -4.0 °C = -4.0 + 273.15 K = 269.15 K
Therefore, to get the new volume, V₂, put V₁ = 2.0 L, T₁ = 298.15 K, and T₂ = 269.15 K:
[tex]\begin{gathered} \frac{2.0L}{298.15K}=\frac{V_2}{269.15K} \\ \\ Cross\text{ }multiply \\ \\ V_2\times298.15K=2.0L\times269.15K \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }298.15K \\ \\ \frac{V_2\times298.15K}{298.15K}=\frac{2.0L\times269.15K}{298.15} \\ \\ V_2=1.8\text{ }L \end{gathered}[/tex]Hence, the new volume, V₂ at -4.0 °C = 1.8 L