Answer :
Answer
-50.6 °C
Explanation
Given:
Initial volume, V₁ = 18.0 L
Initial teperature, T₁ = 35.0°C = (35. + 273.15 K) = 308.15 K
Final volume, V₂ = 13.0 L
What to find:
The finaltemperatrure, T₂ in °C
Step-by-step solution:
The final temperature, T₂ can be calculated using Charle's lawfequation.
[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \\ \Rightarrow T_2=\frac{T_1V_2}{V_1} \end{gathered}[/tex]Plugging the values of the given parameters into the formula, we have;
[tex]T_2=\frac{308.15\text{ }K\times13.0\text{ }L}{18.0\text{ }L}=222.55\text{ }K[/tex]The final temperature, T₂ in °C = (222.55 - 273.15 ) = -50.6 °C